Electro Chemistry - One Shot Lecture | CHAMPIONS - JEE/NEET CRASH COURSE 2022

Summary

Highlights

• Electrochemistry finds applications in everyday life, from battery usage to rust prevention. 🔌
• Difference between galvanic and electrolytic cells explained with smartphone analogy. 📱
• Understanding Gibbs free energy helps in identifying reaction spontaneity. 🔄
• Electrochemical cells explained through detailed reactions at electrodes. ⚗️
• Tips for mastering numerical problems related to electrolytic conductance. 💪
• Using the Nernst equation to calculate cell emfs at non-standard conditions. 🔍
• Simplified approach to Faraday’s laws and electrolysis calculations. 🚀

Key Takeaways

• Electrochemistry is all around us, from batteries to corrosion! 🔋
• Understand the difference between galvanic and electrolytic cells. 💡
• Gibbs free energy is key to understanding spontaneous reactions. 🔑
• Reduction and oxidation happen at cathodes and anodes, respectively. 🔄
• Master the concept of electrolytic conductance for numerical problems. 📊
• Learn to apply Nernst equation for emf calculations. ⚙️
• Faraday's laws make electrolysis problems simple. 🚀
• Coulombs, faradays, and moles: Know their relationships! 🧮
• Practice and more practice makes perfect in electrochemistry! 🏆

Overview

Chapters

• 00:00 - 00:30: Introduction and Welcome The chapter begins with a greeting to the students as part of the champion series in chemistry. The instructor announces the focus for the session, which is the highly requested chapter on electrochemistry. The importance of the topic and the key areas to be covered will be explored in the session.
• 00:30 - 01:00: Importance of Electrochemistry This chapter introduces the importance of electrochemistry. It begins by setting the expectation of addressing crucial questions, primarily focusing on why it is essential to study electrochemistry. The chapter then aims to identify the applications of electrochemistry in everyday life, emphasizing the significance of understanding its role before delving deeper into the topic.
• 01:00 - 04:00: Applications of Electrochemistry The chapter discusses the pervasive use of batteries in everyday life. Students interact with batteries through various common activities such as playing video games, watching television, and using remotes. The chapter emphasizes the importance and omnipresence of batteries across multiple applications, illustrating their fundamental role in powering devices such as microphones among others.
• 04:00 - 06:00: Basics of Electrochemistry The chapter discusses the basics of electrochemistry, highlighting its integral role in various everyday applications, such as batteries and vehicles. The conversation acknowledges different types of batteries, those that are rechargeable and those meant for single use disposal, and hints at a broader context where electrochemistry plays a crucial role in daily commuting and transportation.
• 06:00 - 10:30: Electrochemical Cells The chapter titled 'Electrochemical Cells' begins with an engaging question about how students are accessing the lecture, implying a modern learning environment using phones and laptops. It seems to focus on a lecture format, indicating a conversational teaching style. The instructor appears enthusiastic about the subject of electrochemistry, as they repeatedly say 'next' to perhaps encourage progression through the material. Although the transcript suggests a momentary diversion from the main topic, the intention is clear to guide the students back to electrochemistry consistently. Overall, the chapter introduces the setup of learning electrochemical concepts, with an emphasis on maintaining focus on the topic despite potential digressions.
• 10:30 - 17:00: Galvanic Cells Explained The chapter 'Galvanic Cells Explained' delves into the fundamental aspects of electrochemistry, beginning with a relatable comparison to drones. It highlights how the principles of electrochemistry underpin various technologies, including those used in national security, photography, and everyday lighting. Through this analogy, the chapter aims to elucidate the practical applications and the pervasive influence of electrochemistry in modern technology.
• 17:00 - 24:00: Electrolytic Cells The chapter introduces the concept of electrochemistry and its practical applications, such as corrosion. It emphasizes the significance of understanding electrochemistry to explain phenomena like the rusting of old cars and methods to prevent corrosion. The session promises to cover these aspects in detail, highlighting the importance and excitement of learning about electrochemistry.
• 24:00 - 30:00: Electrolytic Cells Example This chapter, titled 'Electrolytic Cells Example,' covers the concept of electroplating and its relationship with electrochemistry. The discussion includes the example of golden spoons, which, despite their name, only have a layer of gold plating. This plating process is an application of electrochemical principles. The chapter also touches on the role of electrochemistry in various batteries, such as those used in electronic devices, further illustrating the pervasive application of electrochemical concepts and reactions.
• 30:00 - 33:40: The Concept of Conductance The chapter explores the concept of conductance, emphasizing its prevalence in everyday life through electrochemistry. It aims to heighten awareness of how numerous daily activities and objects function using electricity and electrochemical principles. The chapter encourages students to develop a new perspective on the role of electrochemistry in the world around them, altering their thinking process to recognize and understand these concepts in various contexts.
• 33:40 - 39:00: Conductivity and Units The chapter titled 'Conductivity and Units' begins with an introduction to electrochemistry, focusing on two major parts. The first key aspect discussed is electrolytic conductance, which determines whether a compound or electrolyte can conduct electricity, and to what extent it can conduct electricity. This section appears to set the foundational understanding for exploring electrical conductivity in different substances.
• 39:00 - 41:00: Molar and Equivalent Conductivity This chapter focuses on the concept of molar and equivalent conductivity, particularly emphasizing its importance from a numerical and examination perspective. The key takeaway for students is the need to memorize relevant formulas and understand their application to solve numerical problems effectively. Additionally, the chapter hints at a transition to an upcoming topic: the electrochemical cell, which is described as a primary and conceptual aspect of the subject.
• 41:00 - 45:00: Strong vs Weak Electrolytes This chapter discusses the concept of electrochemical cells, emphasizing both the conceptual and numerical aspects of electrochemistry. It highlights the focus on understanding the fundamental questions posed during examinations, particularly from this concept of electrochemistry. The chapter aims to provide a comprehensive discussion about electrochemical cells, strengthening the students' grasp of the subject.
• 45:00 - 48:00: Kohlrausch’s Law The chapter titled 'Kohlrausch’s Law' begins with a transition from discussing thermodynamics to electrolytic conductance. The instructor engages with students by asking them about their understanding of chemical energy, indicating that it will be a key concept in the chapter.
• 48:00 - 52:00: Electrochemistry Review The chapter "Electrochemistry Review" discusses the concept of chemical energy within the context of chemical reactions. It emphasizes the understanding that chemical energy is the energy that changes during a reaction, often represented by delta H (enthalpy change). This explanation links electrochemistry concepts with thermodynamics, suggesting a foundational review of how energy transformations occur in reactions.
• 52:00 - 60:00: Question Practice The chapter titled 'Question Practice' focuses on clearing misconceptions related to chemical energy. The instructor acknowledges a common incorrect answer given by students and clarifies the concept by explaining that delta H refers to enthalpy, which is a change in enthalpy, rather than chemical energy. The inclusion of 'my dear students' multiple times suggests a supportive and encouraging approach to teaching.

Electro Chemistry - One Shot Lecture | CHAMPIONS - JEE/NEET CRASH COURSE 2022 Transcription

• 00:00 - 00:30 hello students welcome to the another session of chemistry today my dear students in champion series we are going to discuss the most awaited chapter the chapter which all of you wanted electrochemistry yes i read all your comments you all wanted electrochemistry to be the next chapter so i bought it for you so my dear students today we are going to study everything about electrochemistry why is it important what are the important topics from which
• 00:30 - 01:00 we expect most of the questions and the most important part what is the need to study electrochemistry so my dear students get ready for the amazing session right let's move forward what are we going to discuss first of all let's see wherever where all in our day-to-day life we use electrochemistry before starting a chapter we should know right where all we can see the applications of electrochemistry so my
• 01:00 - 01:30 dear students we are going to see that first right yes so uh you all use batteries right yes my dear students play the you all of you do play video games yes all of you do watch television yes so you use remotes right yes what is the most important thing that you use in them it is your batteries right yes so yes my dear students batteries so use you use batteries everywhere all right i am using this mic i am using battery we
• 01:30 - 02:00 do charge them we don't charge them there are some batteries which are charged which there are some batteries which are used only once and you just throw them back right yes so all of this everything is surrounded by electrochemistry right yes this is based on electrochemistry next my dear students we commute from one place to other yes you use cars yes you do use that is also based on electrochemistry my dear students so we are going to discuss about that as well perfect let's
• 02:00 - 02:30 move forward what other things are there are you watching my lecture of course ma'am we are so how are you watching that on your phones on your laptops yes my dear students you might be watching there only what is this this is again based on electrochemistry next next next next next i hope you're studying right yes i hope you're starting this is nothing related to electrochemistry but yes there is something which is again related to
• 02:30 - 03:00 electrochemistry that is your drone you uh have heard about drone nowadays it is like uh trending nowadays everywhere be it your uh national security purposes be it anyone's marriage be it any photo shoots you use drones for various other reasons right throne is based on the electrochemistry concepts right next if we talk about the lights my dear students the room you're sitting in the lights which you are using all these lights are based on the concepts of
• 03:00 - 03:30 electro chemistry right yes what other have you seen these types of cars the old cars with you know red red rust on it yes what is that what is that that is also based on the electrochemistry concepts my dear students how corrosion of iron takes place and how can we protect it everything is based on electrochemistry so we are going to discuss everything today and it is going to be an amazing session you've heard
• 03:30 - 04:00 about uh you've heard about uh golden spoons right yes are they made of gold no they have only a plating of gold on them and that is known as your electroplating concept and that electro plating concept is also based on electro chemistry next if we talk about the batteries the in water batteries the batteries which are used in various other electronic items all the batteries are based on the electrochemistry concept
• 04:00 - 04:30 right yes ma'am in our day-to-day life there are hundreds of things which involve electrochemistry and we were not aware of that but now my dear students wherever you will go you will definitely see electrochemistry concepts all around you right yes this would be the most fascinating part of your life you'll you'll try to understand that how is this particular thing using electricity how is this particular thing using electrochemistry concepts now your thinking process is going to change all
• 04:30 - 05:00 together so let's start this session my dear students first of all we are going to talk about electrochemistry in two major parts what are those two major parts first and the most important is my dear students electrolytic conductance it will tell you that whether a particular compound whether a particular electrolyte will conduct electricity or not if it conducts then how much can it conduct the electricity right yes so these this part this particular part is actually
• 05:00 - 05:30 this particular part is actually based on the numerical point of view again i am telling you let's be logical we are going to talk about the exam point of view so yes this is your numerical point um part okay so most of the questions which you are asked from this part is numerical okay yes you just need to remember the formulas you should know how to apply them and you are good to go right perfect next which is the main and the conceptual part that is your electrochemical cell
• 05:30 - 06:00 electrochemical cell is your conceptual part so yes numerical questions are also asked yes it's not that numerical questions are not as numerical part is also there but the conceptual questions which are majorly asked from electrochemistry are from this particular concept okay yes ma'am perfect now let's move forward we'll discuss everything about it first my dear students we are going to talk about the electrochemical cell and
• 06:00 - 06:30 then we'll move forward with the electrolytic conductance okay perfect now let's move forward the first and the most important question that i would like to know is you all have studied thermodynamics yes ma'am we have studied right my dear students so my question to you is we generally talk about the chemical energy right in this particular chapter you are going to use a word which is chemical energy if i ask you what do you understand by chemical energy what will you say
• 06:30 - 07:00 then chemical energy would be some energy which would be changing during the reaction right yes now go to the thermodynamics part and just think about it what was what was your chemical energy what could be your chemical energy man there's a reaction a which is going to be and ma'am you are saying that chemical energy would be any energy which is changing during the chemical reaction and we know that the changing energy is delta h so ma'am this is your chemical energy this might be answer
• 07:00 - 07:30 this is your answer yes and if you have answered this my dear students i am really proud of you because this answer is absolutely wrong yes so my dear students this is not what is known as your chemical energy actually delta h is your enthalpy it is your change in enthalpy my dear students it is not considered to be a chemical energy when you say chemical energy you
• 07:30 - 08:00 actually mean gibbs free energy so chemical energy so chemical energy of a reaction of a reaction is known as its gibbs free energy gibbs free energy so whenever let's say it as delta g so my dear students
• 08:00 - 08:30 whenever you are asked about gibbs free energy or whenever you are asked about chemical energy you should understand that it is actually denoting your gibbs energy and not your enthalpy okay the most important thing which you should remember right let's move forward then now now we are going to talk about the electrochemical cell now i just want to tell you that when you talk about the electrochemical cells you have two major type of cells
• 08:30 - 09:00 what are those two major types of cells the first would be your galvanic cell and the other would be your electrolytic cell okay perfect i hope you got it now my dear students i want to make you understand these two particular cells in
• 09:00 - 09:30 a very interesting way in a very very super interesting way but for that you have to you have to imagine things with me okay yes so i am going to tell you a story and with that story you will understand both the concepts so now let's start imagine not a story in fact it is your day-to-day work which you do and i'm sure you use that do you all use smartphones of course you use smartphones right yes so we are going to understand these two types of cells with that now my dear students first and the
• 09:30 - 10:00 most important thing which i want to tell you is that galvanic cells actually have spontaneous reactions they have spontaneous reactions what does what does spontaneous mean my dear students spontaneous means anything that is happening on its own okay and vamp these electrolytic cells are non-spontaneous non spontaneous okay what are non-spontaneous my dear students non-spontaneous means any reactions
• 10:00 - 10:30 which are not done on their own but you apply some force and the reactions could be done for example you want to take your water from the bottom level to the upper level how will you do that you will have to use motor yes you use motor to pull the uh to take away take your water up right yes so there you are using an external force that is a non-spontaneous process which you are using external force and doing it right that is actually a non-spontaneous process now when your
• 10:30 - 11:00 water is at some height and you want that water to go down it will go on its own yes you don't need any external forces that is your spontaneous process so i hope you understand the difference between spontaneous and non-spontaneous in non-spontaneous you actually need external force but for spontaneous you don't need any external forces okay now if you remember the thermodynamics part then my dear students can you tell me that gibbs free energy is a sign for spontaneous processes and non-spontaneous processes
• 11:00 - 11:30 i hope you can tell me remember when you talk about the thermodynamics when you think about the thermodynamics you studied you will say ma'am for a non-spontaneous process which is this the gibbs free energy change that is delta g is always a positive quantity whenever delta g's value was positive it was a non-spontaneous process and whenever it was a spontaneous process delta g will always be a negative value so for galvanic cell gibbs free energy would be
• 11:30 - 12:00 negative and for electron electrolytic cell delta g would be positive right perfect till here no issues now let's imagine the smartphone okay now my dear students all of you all of you use phones right yes ma'am when when your phone is discharged when your phone is discharged what do you do what do you do man we just get it charged yes when you are getting it charged you are using some external force yes ma'am you
• 12:00 - 12:30 are using electricity right yes that means this when you are charging your phone after discharging that particular part is your non-spontaneous process and you are actually what are you doing you are actually using external force for getting it charged yes ma'am so charging is the non-spontaneous process so charging is a non-spontaneous process and mom in charging what are we doing we are actually using that electrical
• 12:30 - 13:00 energy and that electrical energy is actually getting converted into your smartphone's chemical energy i hope you understand that yes ma'am electrical energy which is into the in the plug that is getting converted into your chemical energy so what can we say for electrolytic cell or for non-spontaneous things we can say that ma'am hair electrical energy electrical energy is actually getting converted into
• 13:00 - 13:30 chemical energy it is getting converted into chemical energy right perfect now let's talk about this process now let's talk about galvanic cell ma'am galvanic cell is a spontaneous process yes my dear students once your phone is charged now you are using your phone when you're sitting on your bed and you're using your or your phone are you using some external force no ma'am it's charged and you are using
• 13:30 - 14:00 it without any external force what are you doing now my dear students at the time of discharging at the time of discharging it is acting as a galvanic cell it is acting as a galvanic cell because it is a spontaneous and what's happening ma'am electrical energy the uh sorry the chemical energy which is stored in my phone which i just converted electrical energy to chemical energy of phone this chemical energy is actually getting converted into electrical energy again your phone you uh whenever you
• 14:00 - 14:30 switch your phone on you see the lights yes what's happening there there you can see that ma'am chemical energy is actually getting converted into electrical energy and that is a spontaneous process right so i hope you got this idea so now we can write that for galvanic cell ma'am chemical energy of my phone is getting converted into electrical energy
• 14:30 - 15:00 so these are the three major points which you actually need to know about these two particular types of cells if you know then you are good to go okay my dear students yes so i hope you got this yes you have to just remember the story and you can write all these three uh points about these cells right yes i hope you got the point now let's move forward perfect my dear students now you have electrochemical cell you can have galvanic cell you can have electrolytic cell and now we are just
• 15:00 - 15:30 going to summarize what all we have learnt first is my electrolytic cell mam i know that for electrolytic cell non-spontaneous process first second is mam delta g will be positive obviously and ma'am what is happening here which form of energy is getting converted if i talk about electrolytic cell it is non-spontaneous non-spontaneous means charging charging means electrical to chemicals so electrical energy is getting converted into chemical
• 15:30 - 16:00 energy now let's talk about galvanic cell if you talk about galvanic cell and then my dear students this would be a spontaneous process if spontaneous then delta g would be negative and what can i say it is a discharging process the chemical energy of the phone is getting converted into electrical energy so chemical energy is getting converted into electrical energy right yes perfect so this is the major thing which we have just understood now let's
• 16:00 - 16:30 read the points now you can very easily understand every point in electrolytic cell this means your non-spontaneous process in electrolytic cell the delta g for non-spontaneous yes delta g for non-spontaneous chemical reaction is provided by the electro electrical energy perfect yes now example you have been given next my dear students the decrease in gibbs energy decrease in gibbs energy means delta g is negative galvanic cell in galvanic cell or mydeo students
• 16:30 - 17:00 another name of galvanic cell is voltaic cell so voltaic cell is converted to electrical energy right yes i hope you are getting all these points the three points which i have made you understand here are actually written in these two lines if you read them thoroughly you'll understand all the all the lines right yes perfect now we have gone through every part till now now let's move forward and study even more
• 17:00 - 17:30 things perfect now my dear students we are going to study about the cell's complete apparatus now okay so the first cell which we are going to discuss about is your electrolytic cell so the first is electro lytic cell so what do i know about electrolytic cell let's just write the points which we know about electrolytic cell
• 17:30 - 18:00 electrolytic cell means ma'am non-spontaneous process though so the first thing is non spontaneous first point second point was ma'am if non spontaneous then delta g would be positive then ma'am delta g would be positive the other thing which i know is that if it is a non-spontaneous process means it is actually the story of charging and when it is charging ma'am i can say that electrical energy is getting converted
• 18:00 - 18:30 into chemical energy perfect and this is actually the setup of your electrolytic cell since electrolytic cell have non-spontaneous processes so they are connected by a battery so they need a battery an external force that is the reason they have a battery now my dear students you know a battery has a positive terminal and a negative terminal the electrode which would be connected to the positive terminal would be positive and the electrode which which would be connected to the negative
• 18:30 - 19:00 terminal would be negative so logical anna yes ma'am so my dear students i just gave you the sign of this positive and this negative just for a few minutes what you have to understand is i'm just telling you uh just a second you don't know which is your cathode which is your anode i'll be making you understand which is your cathode and which is your anode okay so now what are you supposed to understand the positive terminals electrode would be positive
• 19:00 - 19:30 the negative terminals electrode would be negative and the electrolyte used here is actually your nacl na plus and cl minus okay i know in electrolytic cell the reactions are non-spontaneous and you know the charges of your electrodes as well right positive or negative whatever it is but you know the their charges so now how are you going to understand this concept what you have to understand is since n a is positive where will it go
• 19:30 - 20:00 opposites attract i hope you know that positive will try to go to the negative part and negative would try to go to the positive part so if mam n a is plus then my dear students it will try to go to the negative electrode which is the negative electrode man this is the negative electrode right if you talk about the negative ion negative ion would try to go towards the positive electrode so mam c l minus is negative it will try to go to the positive electrode so n a plus will go this side
• 20:00 - 20:30 and c l minus would go this side yet i don't know that which is my anode and which is my cathode we know that there are two electrodes one is a node and one is cathode which is the node which is cathode i don't know we are going to understand that as well okay so till now i know n a plus will go to the negative and c l minus would go to the positive what are the reactions that will take place my dear students i hope you know that redox reactions are actually taking place in electrochemical electrochemistry what are redox
• 20:30 - 21:00 reactions you've learned in your tenth ma'am it is the oxidation and reduction reactions happening together right yes so if n a plus goes to the negative terminal c if n a plus goes to the negative terminal obviously it is short of an electron and since this electrode is negatively charged this electrode can give it ah give na plus an electron so n a plus will actually accept an electron from the negative electrode
• 21:00 - 21:30 so here electron you will get n a solid easy yes next my dear student if you talk about cl minus what will cl minus 2 it has an extra electron and it is going to the positive terminal which is actually short of electrons so cl minus will actually give the electrons to the positive electrode so when cl minus will give electrons then you will get cl2 plus electron right yes these are the
• 21:30 - 22:00 two reactions which i can just understand which will actually take place okay but i still don't know which is anode and which is cathode from here you need to understand one very very important thing that what is a anode my dear students in electrodes always and always anode is that electrode where oxidation takes place always and always oxidation will take place on anode okay
• 22:00 - 22:30 and cathode is the electrode where always and always reduction takes place okay so i can say anode where oxidation takes place where oxidation takes place okay always and if i talk about if i talk about cathode what can i say where reduction takes place
• 22:30 - 23:00 okay ma'am perfect no issues now we have to just figure out where oxidation is taking place and where reduction is taking place what is the oxidation my dear students what is oxidation ma'am oxidation is actually oxidation is actually my dear students loss of electron always and always loss of electron is considered to be oxidation and what is reduction and what
• 23:00 - 23:30 is reduction it is always and always gain of electron it is gain of electron okay yes so mam oxidation will be loss of electron and reduction would be gain of electrons and i know that oxidation will always take place on a node and reduction will always take place on cathode okay now you have to come to these reactions and just see where oxidation is taking
• 23:30 - 24:00 place wherever it is wherever there is oxidation it is your anode and wherever it is reduction it will be your cathode so now let's figure it out n a plus plus electron n a is actually taking an electron n is actually taking an electron and getting it converted to n a solid so ma'am any plus is actually gaining electron so my dear students here you can say that it is gain if it is gain i can say it is reduction
• 24:00 - 24:30 and since it is reduction reduction always takes place at cathode so mam this is cathode similarly if we talk about this reaction cl minus is actually giving an electron and getting converted to half cl2 so mamat is actually losing electrons so you can say it is loss so ma'am this would be oxidation and where does oxidation takes place only and only at a node so this will be
• 24:30 - 25:00 your anode this will be your anode okay yes perfect now now just relate just relate this cathode and anode to your setup n a plus is actually gaining an electron this na plus is going here and it is actually gaining an electron if it is gaining an electron here then i can say that this particular part is your cathode and if you see it is your cathode right yes ma'am and cathode is negatively charged so i can say that
• 25:00 - 25:30 cathode is so i can say that cathode is negatively charged in an electrochemical cell similarly if we talk about anode cl minus is actually losing its electron on this electrode i hope you can see cl minus is going to this electrode and mammoth is losing electrons so this is actually your anode yes and my dear students you can see that anode is your positive electrode so the charge of anode would
• 25:30 - 26:00 be positive so you've learned hundreds of things just by understanding the concept right yes so we just find out found out that in a electrochemical cell a cathode is always a negative negatively charged electrode and a anode is always a positively charged electrode always remember that oxidation will always take place at a node and reduction will always take place at cathode right yes since it is the electrochemical cell a non-spontaneous cell so you need a battery okay yes so
• 26:00 - 26:30 this is the whole setup which i hope you just understood perfect my dear students so this is all about electrochemical cell which you actually need to know perfect let's move forward let's move forward and understand what we have just discussed okay the electrode where oxidation happens is known as anode the electrode where reduction happens is known as cathode electrodes are the metallic plates always the exchange of electrons
• 26:30 - 27:00 whenever you lose electron or you gain electron everything just takes place on the surface of the electrodes it is never possible that electrons can be exchanged in the liquid solution no it is not possible always and always will take place at the surface of the electrode okay let's move forward [Music] we always assume equal solution we always assume equal solution if i have considered nacl solution then that is an equal solution pre electrons cannot remain in the solution if there are free electrons they will not remain on the solution
• 27:00 - 27:30 they are present on the surface of electrode the redox changes can take place at the surface of electrode only the redox reaction will take place on the surface of the electrode perfect let's move forward next is your galvanic cell now we are going to talk about the galvanic cell okay yes so uh let me just have water and then we'll move forward to the galvanic cell okay till then what you have to do is write everything you know about galvanic cell till now okay
• 27:30 - 28:00 yes just write it let's talk about galvanic cell ma'am the things which i know about galvanic cell are it has spontaneous processes the first thing that i know is it has spontaneous processes second if it is spontaneous then i can say that delta g will be negative no issues yes and also i know that when i talk about
• 28:00 - 28:30 the spontaneous uh galvanic cell that means that ma'am uh i am talking about the discharging process and when discharging actually happens i can tell you that the chemical energy is actually getting converted into electrical energy okay these are the three things which i know i am going to tell you a few more things my dear students galvanic cell is also known by two other names what are those other names one is
• 28:30 - 29:00 your voltic cell and the other is your denial cell both of them actually means your galvanic cell only okay perfect now my dear students since i am telling you that and you also know that uh galvanic cell is a spontaneous process so do you need a battery for it no you don't need a battery for it the setup for galvanic cell is always like this the setup for galvanic cell is always like this where
• 29:00 - 29:30 you have two different where you have my dear students two different containers right and what all you have in it i'll just tell you here you have cu so4 sorry zn so4 solution and since this is blue color so this would be your cu so4 solution obviously both are equals i told you
• 29:30 - 30:00 about that right this will have 2 plus this would be 2 minus this would be 2 plus and this would be 2 minus okay now let's understand everything about it we don't need a battery for it yes but again it it will have to electrode the node and a cathode but now we have to understand which would be anode and which would be cathode we have to find out the uh signs of the anode and cathode as well okay so how are we going to study about it now the most important thing which i want to tell you here is that in
• 30:00 - 30:30 galvanic cell always then always spontaneous processes take place right the reactions the processes which are happening on its own you don't need an external force and when you talk about the spontaneous processes you have to trust ma'am and you have to understand that for the time being i'll give you the reason as well why are they spontaneous and why not but for the time being just trust me and trust me that zinc always
• 30:30 - 31:00 zinc always wants to go to the zn2 plus and it always wants to give two electrons away okay my dear students this is a spontaneous process for the time being please understand it please accept it i'll give you the reason why is it spontaneous in next 10 to 15 minutes i i assure you about that okay till then you can trust me now i hope you try you can trust me right let's move forward link to zinc two plus plus two electrons my dear students this is a spontaneous process and copper
• 31:00 - 31:30 always wants to accept two electron and get converted into copper solid and this is also a spontaneous process these two are your spontaneous processes i'll give you the reason again don't worry i am here i will give you every reason but for the time being just trust me now ma'am we got it till here now you have to find out your cathode and anode okay yes imagine my dear students link is getting
• 31:30 - 32:00 converted into zn2 plus and it is losing electron so what is happening here loss of electron and loss of electron is actually oxidation right so one this process is oxidation and oxidation always and always takes place at anode right yes so ma'am this would be my anode i got my anode you can get your cathode as well what is happening here mammoth is actually gaining electrons so my dear students gaining electron means you can
• 32:00 - 32:30 write it here gain of electron and it is reduction and it is reduction so mam reduction always and always takes place at cathode right yes so this would be your cathode now let's go to the setup now if we see the setup what can i say zinc is getting converted into zn to plus this means that my dear students this electrode is actually of zinc and
• 32:30 - 33:00 this electrode is actually of copper in your galvanic cells the aqueous whichever the aqueous solution is your electrode would be of that particular metal only okay so this will always be zinc and this will always be copper now i know that i know that what is the reaction happening ma'am zinc is getting converted into zn2 plus right yes so this means that my dear students actually the zinc from your electrode is getting converted into your aqueous solution initially your zinc was this
• 33:00 - 33:30 thick and slowly and slowly since thick was zinc was getting converted into zn2 plus it just got to be shrinking like this right yes so i can say that slowly and slowly zinc electrode would be vanishing would be going uh would be worn out right yes now let's talk about copper copper is actually gaining two electrons and it is getting converted into copper solid this means that this co2 plus is actually
• 33:30 - 34:00 going and getting deposited on the copper electrode so initially copper electrode was like this and slowly and slowly the copper electrode would be gaining weight right yes so it will be getting thicker and thicker so i got my two very major points that here zinc get will be uh deposited into the solution and eventually my zinc electrode would be turning into uh thin and thin and thin now if i talk about the copper electrode
• 34:00 - 34:30 it would be uh addition of electro copper more and more and so it would result in the thickening of the electrode right yes and also i can say that the zinc electrode imagine your zinc electrode was actually your anode so i can say that i can say that this is my anode and i can say that copper one is my cathode perfect we just found out our cathode and node but yet we don't know the signs we don't know the positive and
• 34:30 - 35:00 the negative signs but you will just give me the signs as well imagine now you have to imagine again my dear students this zinc is getting converted into zinc two plus okay and i told you that zinc cat can be get converted into zinc two plus by giving two electrons okay and these two electrons cannot reside in the solution electrons never resides in the solution they are always present on the electrodes so this zinc will give its electron here will give its electron here will give its electron here and
• 35:00 - 35:30 eventually go into the solution so everything is giving its electron on the electrode and getting going into the solution so slowly and slowly i can say that this electrode would be negative in sign there will be a lot of negative charge a lot of electrons so ma'am this electrode this anode is actually your negatively charged okay now let's understand about cathode what is cathode doing my dear students the copper two plus in the solution is actually accepting the
• 35:30 - 36:00 electrons and getting converted into copper solid it is accepting the electrons and getting converted into copper solid so what can we say ma'am we can say that this copper two plus is getting into copper to copper solid by accepting electrons from where is it taking the electrons it is taking the electrons from the electrode because in the solution there are no electrons present so if anything is accepting the electrons then what is the sign that this electrode will get this electrode
• 36:00 - 36:30 is actually giving out electrons when it is giving out electrons it is getting converted into positively charged so this is your positively charged so what can i say ma'am you can say that your cathode is positively charged perfect yes and one now what will happen my dear students at one particular point of time at one particular point of time there will be excess of negative charge there will be excess of positive charge
• 36:30 - 37:00 now what are we supposed to do now what will we do obviously the reaction will stop but we don't want it to be stopped what can we do we can connect these two we can connect these two if you connect these two this is actually short of electrons and this has a lot of electrons what will happen electrons will start moving from here to this side and eventually this will get electrons and the work would be done and this will also be happy because it has
• 37:00 - 37:30 excess of electrons which he could help right yes so the electron flow actually takes place from anode to cathode so the next point which we just came to know is that electron flow in galvanic cell from anode to cathode from anode to cathode right easy perfect my dear students now we understood this in physics you just uh in physics you've learned that uh
• 37:30 - 38:00 electricity always flows in the opposite direction to the flow of electrons so what can you say we can say that flow of current would be from cathode to a node from cathode to anode right yes why it would be opposite to the sign of flow of electrons perfect we just understood this as well
• 38:00 - 38:30 now now my dear students what i want to tell you one more important thing is that when you talk about this solution can you say that zn2 plus concentration is increasing and so4 2 minus concentration is comparatively decreasing since zinc is getting converted into z and two plus then you can say no because that in two plus concentration is so increasing but so4 two minus is intact so eventually this will be more and this would be less similarly if we talk about
• 38:30 - 39:00 here copper two plus is getting deposited on the electrode so my copper two plus concentration is decreasing and so4 two minus is intact this means that here so4 two minus concentration is actually a lot as compared to copper two plus right here my ds ruler students solution is a negatively charged and hair solution is positively charged here your solution is negatively charged here your solution is positively charged how can we neglect
• 39:00 - 39:30 how can we sort this out this you can sort out by using a salt bridge a salt bridge if i connect these two solutions somehow and if i somehow connect these two solutions and nullify their positive and negative charges to a great extent the solution would be for uh neutral kind of neutral right yes so this is the role of salt bridge what does salt bridge do it neutralizes the solution right yes this is very
• 39:30 - 40:00 important that it neutralizes the solution right perfect it is made of kc l agar agar solution i'll write it here agar solution so what does salt bridge do brings electro neutrality
• 40:00 - 40:30 it brings electro neutrality to the solutions okay yes so this is everything this is actually everything about your cells about your galvanic cell now only one thing is left that anode is negatively charged so we can write here that anode is negatively charged and cathode is positively charged so we can
• 40:30 - 41:00 write here that cathode is positively charged so just understand the story and you can write everything about yourself and only these questions are asked about conceptual questions are asked about these cells nothing else can be asked so i hope you understood this and it really helped you right yes now we will move forward perfect my dear students you can write all this and then we'll move forward okay let's move forward my dear students
• 41:00 - 41:30 next is the everything that we've learned to summarize it we are here now what you have to understand here is that my dear students that you just remember the sign of any one electrode any one electrode and you can find out signs of other three electrodes like this okay how can you do that let's understand if we talk about the galvanic cell and electrolytic cell i have learned the sign of anode of an electro galvanic
• 41:30 - 42:00 cell anode of an uh galvanic cell is always and always negatively charged right we just learned so if i know only this then i can say that this will be negatively charged if my dear students a node is negatively charged then my obviously galvanic self-cathode will be positively charged so my dear students this will be positively charged always and always remember that since this is a spontaneous process and this is a non-spontaneous process then the science of the electrodes of your electrolytic cell
• 42:00 - 42:30 will always be opposite of galvanic cell so if this is negative then man this will be positive and if this is positive then this has to be negative so if you know any any one electrodes sign of any one cell you can find out the cell's other three electrode sign in a bit right yes so this is how you are going to study all about your cells let's move forward my dear students now we are supposed to do some question
• 42:30 - 43:00 practice okay so now you have to pause the video and answer these questions yourself it is all based on what we have learned till now the first question is a cell reaction would be spontaneous if the delta g value if the delta g value if the delta g value of the reaction is see it is said that the cell would be spontaneous when will it be spontaneous when it is spontaneous delta g is always negative value so mam negative value
• 43:00 - 43:30 option c will be a correct answer let's move forward next question my dear students next question is which of the following statements is correct you have to tell me which statement is correct so pause the video and do yourself what does this question say cathode is a negative terminal in both galvanic and electrolytic cells i only know this i only know this that in galvanic cell anode
• 43:30 - 44:00 is negative okay i only know this how can i use this concept here cathode is negative terminal in both galvanic and electrolytic cell cathode cannot be of same sign in both the cells so this is not true right second a node is positive terminal in both galvanic and electrolytic cell again any electrode cannot have the same signs in both the cells cathode and anode are negative terminal see cathode is negative terminal
• 44:00 - 44:30 cathode is a negative terminal in electrolytic cell and anode is negative terminal in galvanic cell a node is negative in galvanic cell yes so option c will be correct the other all options would be wrong this is how you are going to solve your questions right perfect amazing let's move forward to the next question next question molten sodium chloride conducts electricity due to the
• 44:30 - 45:00 presence of my dear students we know that electricity is conducted by only charge carriers right you need charge carriers and these charge carriers are majorly of two types three ions or free electrons or free electrons now what can you say what can you say my dear students molten sodium chloride conducts electricity in the presence of my dear students it is molten sodium chloride this means that in the solution
• 45:00 - 45:30 and mam in the solution we cannot have electrons then what is the possibility which is conducting electricity only and only free ions so free ions that is option b will be your correct answer right perfect let's move forward next question next is which of the following statement is correct about galvanic cell uh galvanic is spontaneous i know that if it is spontaneous this means i'm using phone if i'm using phone that i am converting electrical energy to
• 45:30 - 46:00 chemical energy okay let's see sorry i am converting chemical energy to electrical energy right the phone's chemical energy is getting converted into electrical energy so it converts chemical energy to electrical energy yes so option is correct i just got the answer option a it converts electrical energy to chemical energy no it converts metal no no no no no only one option is correct which is option a perfect this is how you are going to solve the questions let's move forward
• 46:00 - 46:30 the function of a salt bridge is to maintain the electrical neutrality of both the half cells yes this is the absolutely correct answer you can read the other options these options would be definitely wrong okay yes only one option is correct which is your option a perfect so now we have studied about the different types of cells right now my dear students i just told you that there are two particular electrodes the one is where oxidation takes place one is the
• 46:30 - 47:00 which where reduction takes place so now ma'am is going to tell you about how can you represent these cells how can you represent these cells obviously there is oxidation taking place there is reduction taking place and i know that oxidation takes place on a node and reduction takes place on cathode but there is a particular cell representation how are you going to represent a cell okay so i know that this is a zinc electrode and
• 47:00 - 47:30 if i know that this is a zinc electrode then i know that hair oxidation will take place since it is the anode so zinc will be converted into z n two plus plus two electrons loss of two electrons and mam since it is copper and it is a cathode this means gain of electron will take place copper two plus will gain two electrons and i will get copper solid right yes perfect now my dear students now you are supposed to represent the cell first and the most
• 47:30 - 48:00 important thing you have done you have just found out the reaction that would be taking place on a node and the reaction that would be taking place on the cathode okay now you have to represent it on the cell how are you going to represent always then always remember that a cell is represented by a salt bridge so i can say that let's write it here always and always represented by a salt bridge these doll these two lines represent your salt
• 48:00 - 48:30 bridge okay this represents your salt bridge next what is happening i'll tell you this side will always be your cathode reaction cathode place and this would be always for your anode okay now my dear students this is the major idea and how are you going to write it i am just going to tell you very very interesting in a very interesting way now you have to write all the cathode
• 48:30 - 49:00 related ions here and all the anode related ions here now what is happening here if i talk about cathode wall apart i have what co2 plus and copper solid always remember the ions always remember the ions are are scared okay they cannot be at last always and always remember irons are scared and solid part are not scared like you have your parents yes uh always your father would be protecting you if you are going on a road if you are traveling on a road if
• 49:00 - 49:30 you're walking on a road your father would never ever uh ask you to go on the road uh those vehicle side he will always be standing on the vehicle side and he will always put you this side right that is he's protecting you the same way my dear students what these solid solid state does is that these act like fathers and these ions are actually scared these cannot be at last okay always and always remember this so if i talk about cathode i have an iron which
• 49:30 - 50:00 is co2 plus and i have a solid which is copper solid so always i can say that now here copper ion will be first right i'll write it equals here and here you can write its concentration and molarity and then put one line phase changing line and then you will write copper solid okay never ever the ions can be at last it will always be solid that is at the last now if we come to the anode what can we
• 50:00 - 50:30 say see in the anode side i have i'm zn2 plus i have zn solid when zn2 plus is an iron again it will be scared so this will be inside and copper zinc solid would be outside so you will be representing it as zinc two plus which is your iron equal state and here you will write its similarity and then you will have zinc solid this is how you are always and always going to represent yourself you will put
• 50:30 - 51:00 a salt bridge always always write cathode reactions this side anode reactions this side ions will be inside and your solid state would be at the end always and always right yes so i hope you got this point this is how you are going to represent a cell so you have a few points you can just read them the electrodes are to be placed at the extreme corners electrodes are always in solid state that is the reason they are always kept at the end the lhs represent
• 51:00 - 51:30 the anode it will be always anode lhs whereas oxidizer oxidation takes place and the rhs would always represent the cathode uh which where reduction takes place the two half cells are usually separated by a salt bridge the two half cells are usually separated by a salt bridge perfect yes very good let's move forward now you have one electrode what if that was a complete cell what if you are given a half cell where only either oxidation is taking place or reduction
• 51:30 - 52:00 is taking place how will you represent if i see this let me tell you that this is a copper sulphate solution so i can see that when in copper you just told me that copper always wants to gain electron and wants to get converted into copper solid this means that ma'am this is a reduction reaction and ma'am this is a cathode right yes so since it is a cathode and i
• 52:00 - 52:30 want to represent this cell i can say that salt bridge would be here ions would be written first so cu 2 plus equal solution and then mam i can put the copper solid here so this would be the half cell representation where a node part would not be written and only your cathode part would be written right perfect so this is how you are going to represent these cells or if the cells are given then you have to find you have to understand that how are you going to write the cell reactions perfect
• 52:30 - 53:00 the metallic electrode is immersed in a solution containing its own cation now let us say you have another you have another electrode which is a gas ion electrode this means this time this time what do you have you have a gas what will happen my dear students let me tell you you have i'll write it here you have h2 gas which is getting converted into h plus plus electron right yes so what can i say here ma'am what can i say here that
• 53:00 - 53:30 this is oxidation taking place this is oxidation taking place this is your gas this is your iron aqueous iron so how are you going to represent this cell mam i know that since electron is on the right hand side since electron is on the right hand side this means that this is the oxidation loss of electron so ma'am this is a node i can say this is node and if i want to represent a cell in anode form i know
• 53:30 - 54:00 that i'll have to put a salt bridge this side and then mam iron would be first because it is scared so h plus iron equals then you will put my dear students gas first iron if if some gas is present then gas would be there this means that h2 gas would be here here you can write its pressure and at last you have to write the solid state but we don't have a solid state here then if any other solid state is not there then you have to write your
• 54:00 - 54:30 electrode let us say this is your platinum electrode you will be given that which electrode you are using if it is a platinum electrode or any solid electrode then you will write platinum solid okay so always then always remember that at the end a solid has to be present be it anything if it is not written then you have to write some inert electrode now platinum is the inert electrode it does not react with anyone so we call it a inode electrode and it will be written at last okay perfect this is how
• 54:30 - 55:00 you are going to this is how you are going to represent a cell right perfect i hope you understand it very good my dear students let's move forward if the oxidized and the reduced part both are ions it is called a redox half cell to represent the redox half cell we need to use electrodes such as platinum palladium because they are inert they do not react with any other electrodes perfect yes let's move forward now my dear students you have a question
• 55:00 - 55:30 you have to reap you have to answer this question okay so mam is getting aside the question says that represent the cell from the following cell representation so first find out which is anode which is cathode write the reactions and then the cell rotation okay so pause the video and try yourself and then ma'am will tell you okay so now what are we supposed to do we
• 55:30 - 56:00 have to first of all find where oxidation and reduction are taking place so ma'am here if i see then h2 is going into 2h plus right my dear students yes so if i have to balance it i can say that two electrons will be added here if this is happening then what is happening my dear students i can say that from h2 two electrons are getting lost are lost right so here i'm loss of electrons so this is oxidation right
• 56:00 - 56:30 this is oxidation and since it is oxidation then obviously this is your anode right yes next my dear students if i see cl2 i can say that cl2 plus two electrons will give me two cl minus charge balanced yes ma'am so what is happening here ma'am here gain of electron is taking place gain of electron since it is gain of electrons so man this is reduction and since this is a reduction what can we say it is your
• 56:30 - 57:00 cathode right yes ma'am everything is done now you only have to represent the cell salt bridge this side cathode this side anode if i talk about cathode what are the ions from only cl minus my dear students always remember that only the ions will be written here only and only the ions would be written not their stoichiometric coefficients so 2 will not be there next my dear students if i see then i have a gas here
• 57:00 - 57:30 so next would be your cl2 gas okay since there is no electrode solid state so we have to write a platinum electrode which is in your solid state now let's come back to the anode side anode side mam i have iron which is my h plus iron and if i see i have a gas which is my h2 gas so here it will be h2 gas and then mam's platinum solid electrode and this will be your representation of cell right yes so simple so easy so first whenever you are
• 57:30 - 58:00 asked just write the reaction find the oxidation and reduction write the cathode and the anode and similarly write the sell representation easy i hope it is very easy very good students let's move forward next is the question which would be your homework right we've done one question one question would definitely be your homework okay so you have to do this question yourself and you have to give me the answer in the comment section i i always check
• 58:00 - 58:30 your comments okay yes let's move forward here is the question for the cell reaction this you have to give me the representation of the cell okay so let's see what are the what would be the representation of the cell if i see copper if i see copper solid is getting converted into copper two plus iron okay and cl2 is getting converted into cl minus so now you have to find the oxidation and
• 58:30 - 59:00 reduction copper to copper two plus so this means that plus two so 1 this is oxidation i hope you understand this if you talk about this then cl2 plus 2 electrons will give you cl minus so this would be your reduction this would be your reduction right yes so this is your oxidation this is your reduction so now you have to represent the cell now let's represent the cell mam since this is reduction this would be my cathode and this would be my anode
• 59:00 - 59:30 salt bridge this side cathode this side anode which would is cathode mam ions would be cl minus aqueous then cl2 gas and mam since i don't have any electrode uh do we have a electrode here no your electrodes are not written so you can write any electrode let's say you have some platinum electrode solid okay yes now beside anode my dear students you have
• 59:30 - 60:00 anode for this side copper copper two plus iron first and then you will have copper solid so here you have copper solid which is your electrode solid state you are given so this is this would be your solid state at last now let's find out which is the correct option i have copper solid copper 2 plus iron i have cl minus i and cl2 gas my dear students they should have written some electrode here as well so let's see is
• 60:00 - 60:30 any other option which is exactly same like this copper solid copper solid copper two plus iron copper two plus iron cl minus equals cl minus excess cl2 gas cl2 they have used a copper electro carbon electrode and carbon electrode is also a inert electrode i i wrote platinum electrode not an issue if your option is given copper uh sorry carbon electrode it is also my dear students in not we just learned three electrodes which were inert one was carbon solid as well so yes we can write copper
• 60:30 - 61:00 carbon solid as well so yes option a will be your correct answer right yes so this is how you are going to solve these questions let's move forward to the next part next part my dear students the question says that for the galvanic cell you are given the galvanic cell cell reaction is you have to give the cell reaction okay this is an arrow this is an arrow this is an arrow now how how are we going to do this my dear students
• 61:00 - 61:30 first you have to see the cell representation and then you have to write the cathode and the node and then you have to write eventually the reactions okay so i know that this is salt bridge so this would be cathode and this would be a node right yes now since i know that cathode will have reduction reactions so reduction is ah gain of electron yes ma'am reduction is gain of electrons so fe2 plus will be gaining two electrons and getting
• 61:30 - 62:00 converted into fe solid so mam gain of electrons means fe two plus will be here then only two electrons and you will get fe solid so this would be your cathode reaction similarly if you see anode you have two ions of anode which are fe2 plus and fe three plus you know that at a node oxidation takes place this means that what will you have you will have loss of electron if you talk about loss of electron you can say that fe
• 62:00 - 62:30 2 plus will give you fe 3 plus plus electron this will be your anode reaction right yes when you have two half cell reactions and you want a net cell reaction what you are supposed to do you just add the two reactions right yes and my dear students always remember that in the net cell reaction never ever electrons are present right so first reaction would be fe2 plus plus two electrons will give you fe
• 62:30 - 63:00 solid the other would be fe two plus plus no this will give you fe3 plus plus electron now the problem is that i cannot cancel out the electrons right why because there are two electrons and here only one so you have to multiply this particular reaction complete reaction by two so i will multiply this complete reaction by two if i multiply the complete reaction by two then i will get my net cell reaction redox reaction net cell reaction is the addition of uh cathode and node
• 63:00 - 63:30 reactions two electrons would be cancelled out so two f e two plus plus fe three plus uh two plus would give you three fe my god what am i saying right yes fe two plus plus two fe two plus will give you three fe two plus so 3 fe 2 plus this will give you fe solid and 2 fe 3 plus this will be your net cell reaction this will be your net cell reaction let's see
• 63:30 - 64:00 which is correct 3 fp 2 plus will give me fe solid plus 2 fe 3 plus option b will be your correct answer ok so this is how you are going to find out the net cell reactions very easy very easy my dear students so you should know how to find out the cell reactions if you are given a cell notation or how can you represent the cell notation if you are given a particular reaction both the ways vice
• 64:00 - 64:30 versa you should know perfect yes let's move forward now we ah before discussing this before discussing this my dear students i want to tell you something just a second yes before telling this i want to discuss a very important part with you which is your agents yes we know what is oxidation we know what is reduction but now we are going to understand the concept of oxidation oxidizing agents and reducing agents okay yes so what are oxidizing
• 64:30 - 65:00 agents my dear students first of all you understand the concept of oxidation potential what is oxidation potential it is represented by o e of o p okay what is oxidation potential any any element which has a tendency of getting oxidized
• 65:00 - 65:30 this means that if i want to oxidize this means i will lose electron the one which loses electron shows oxidation so if i have a tendency very high tendency of oxidation of losing electron then my oxidation potential would be very much right so the elements would have which have higher oxidation potential means they have a high tendency of oxidation of losing electrons okay so oxidation potential means elements with
• 65:30 - 66:00 high tendency of losing electron or we could say showing oxidation okay perfect clear no issues let's talk about the other part which is your reduction potential okay what is reduction potential it would be represented by e of rp every
• 66:00 - 66:30 element has a oxidation potential and a reduction potential what is reduction potential if an element has the tendency if an element has a tendency to gain electrons right it will have a tendency to gain electron it will show reduction so its reduction potential will be very high so what can you say that elements with high tendency to
• 66:30 - 67:00 gain electron or showing reduction okay every element has a particular oxidation or reduction potential but if an element has a high oxidation potential it will definitely have a negative low reduction potential it cannot be that an element has a high reduction potential as well also oxidation potential no always and always remember always and always remember that e of reduction potential will always be equal
• 67:00 - 67:30 to minus e of oxidation potential for a particular element okay yes this is actually the relationship between the reduction potential and oxidation potential of the elements perfect i hope you understood till now till now if you are thorough with this now we are going to study something very important which is your oxygen agent
• 67:30 - 68:00 this is a particular point where every student is confused but i don't want you to be confused okay now understand this understand this if you are talking about an agent an agent always works for someone else he never works for himself if you are hiring any agent then obviously he is working for someone else no he is not working for himself yes ma'am so if you talk about the oxidizing agent then the oxidizing agent actually oxidizes
• 68:00 - 68:30 someone else he will not oxidize himself he will oxidize someone else so oxidizing agent are the ones who oxidize the others and if they are oxidizing the others then they get themselves are getting reduced this is the most important part an agent actually works for someone else if he is oxidizing someone else then he himself is getting reduced so if i talk about an oxidizing agent then it itself gets reduced
• 68:30 - 69:00 and oxidizes others okay yes now my dear students if if an oxidizing agent himself is getting reduced then what will be its uh will its oxidation potential be more or its reduction potential would be more now if he himself is getting reduced then the more he will be reduced the more he can oxidize the other yes and
• 69:00 - 69:30 the more he gets reduced that means he should have a high reduction potential so the other point which you need to remember is for an oxidizing agent that it will have high reduction potential so it will have high reduction potential okay perfect let's move forward to the next part which is your reducing agent now you can
• 69:30 - 70:00 understand you can write all the two points on your own about this so pause the video and try yourself first write on your own and then see my approach what does this say if you're talking about a reducing agent means it will reduce someone else and it itself will get oxidized yes ma'am if it itself will get oxidized this means that its own oxidation potential should be more yes ma'am so my dear students reducing agents
• 70:00 - 70:30 reduces others itself gets oxidized okay it reduces others and it itself gets oxidized so it will have high oxidation potential
• 70:30 - 71:00 here yes right so this is the most important part that you need to understand between the oxidation potential reduction potential oxidizing agent and reducing agent okay we will be doing questions on this again but till now you need to understand this concept and i hope you got this now let's move forward now now we are going to talk about the cells emf what is emf before that i want
• 71:00 - 71:30 to show you that how you are using electricity in your day-to-day life you use water you use light you use radios you use laptops computers you use you make sandwiches as well right yes so everywhere my dear students you are using electricity and that is absolutely based on electro chemistry okay let's move forward now uses of electricity in our daily life this is just for fun don't worry this is nothing which is going to be asked in your question this is only and
• 71:30 - 72:00 only for fun okay so just go through this chart now let's move forward which is your emf off sell whenever you talk about the self emf my dear students a self emf is based on majorly four parts what are those four parts one is the nature of your chemical reaction what is the nature of your chemical reaction actually affects the cells emf next is your concentration of your reactants and products what is the concentration of reactants and products actually affects
• 72:00 - 72:30 the emf of a cell we will discuss about that as well next would be the pressure of gases if your reactants and products are gases then their pressures also affect the emf of the cell and then my dear students temp temperature also affects so yes you are generally asked that what are the factors on which your emf of the cell depends so you need to know all these four points perfect let's move forward now you would have heard about satp conditions satp conditions are standard
• 72:30 - 73:00 ambient temperature and pressure conditions where you consider some standard state and you have calculated emfs and you have given them as the standard values for that particular cells what are the standard conditions that is you take one molar concentration what do you do you take one molar concentration of any element whichever you are taking any ion you are taking take its one molar concentration take its temperature to be to
• 73:00 - 73:30 98 kelvin take its temperature to be 298 kelvin and take its pressure to be one atmosphere actually it's one bar but you can take it to be nearly one atmosphere so at satp conditions you've calculated the emf of all the ions and you have given them their standard value which are known as the standard emf of cells but if you change the concentration if you change the temperature or you change the pressure
• 73:30 - 74:00 the emf of the cell would not be standard it would be some emf because it will definitely change it depends on these factors and then you will calculate them as well okay first we will talk about the standard and then we will talk about the further conditions satp conditions if all the reactants and products are present in their respective standard states standard state means one molar concentration 298 kelvin temperature one bar or one atmosphere pressure the corresponding emf known is is the standard emf but if you change any factor the emf of the cell would
• 74:00 - 74:30 change standard emf has the constant value for a particular given reaction okay yes ma'am perfect let's move forward what is next next is calculation of emf how are we going to calculate the emf of the cell that is the major part we are going to calculate the emf of cell if you are given standard that is absolutely right they are that is given in your books but if you change the concentrations my dear students you should know how to
• 74:30 - 75:00 calculate the emf and that is calculated by the nurse equation for that as well you need to write the redox half cells your oxidation half cell your reduction half cell okay so i'll make you understand how to use the nernst equation as well let's move forward case one my dear students you have calculated the emf of the cells you have calculated the emf of the cell if your cells are in their standard states then your emf of e naught of the cell will be calculated as i'll write it
• 75:00 - 75:30 here e naught of cell would be e naught of cathode minus e naught of anode what does this mean i'll tell you here e naught of cell is the cell emf at standard state if you talk about e naught of cathode this means that it is the standard emf of cathode which will be given to you
• 75:30 - 76:00 this will be standard reduction potential i would say this is the reduction potential of cathode and what is e naught of anode this is standard reduction potential of anode this is the standard reduction potential of n always remember my dear students whenever you calculate the e
• 76:00 - 76:30 naught of cell you write e naught of cathode minus e naught of n because you have a anode and a cathode in a net cell reaction yes you are given the standard values of cathode and anode in the question now just remember that in this particular expression you use the reduction potentials of both if you are given the oxidation potential of anode then you have to convert it into the reduction potential by just putting a negative sign i told you that reduction potential is always equal to minus of
• 76:30 - 77:00 reduction potential of oxidation potential so just like this you calculate it and you will get the reduction potential for cathode as well as reduction potential of anode put them here in this expression and you will get the e naught of cell which is the standard emf of the cell okay yes now this standard emf of cell value positive or negative actually tells you about whether your reaction is spontaneous or non-spontaneous how do we find that it is a galvanic cell or a voltage cell if sorry galvanic cell or
• 77:00 - 77:30 electrolytic cell if it is a positive positive value spontaneous process it would be galvanic right so this is how we are going to find out spontaneous reaction for spontaneous reactions my dear students your e naught of cell should be a positive value so reaction with positive emf must be a positive value or if you talk about non-spontaneous then a naught of cell should be negative a negative value of a
• 77:30 - 78:00 reaction emf of a cell would give you a non-spontaneous reaction okay so this is how you can find out whether a reaction would be spontaneous or non-spontaneous by looking at their e naught values just calculate that e naught of cell if it is positive it is spontaneous if it is negative it is non-spontaneous right yes now i just want to tell you that i am blabbering about the reduction potentials and oxidation potentials a
• 78:00 - 78:30 lot but how did we calculate every elements reduction in oxidation potentials how did we calculate it obviously there was some reference you took some reference and you calculated their oxidation and reduction potentials that is your my dear students she what is she she is woman woman empowerment is shown here my dear students since we say that in our day to day life in this world everything is surrounded by what
• 78:30 - 79:00 a woman everything is possible because of a woman we just see that women's can do everything right yes they do everything they make this world even a better better better place for everyone right they are there as mothers there are they are there as house workers they are there as entrepreneurs they are as they're in offices that they do are there as mothers as well right yes
• 79:00 - 79:30 so what can we say everything is roaming around women everything is about women yes so my dear students similarly everything would be with reference to she and what is she it is the woman but the definition would be standard standard hydrogen electrode standard hydrogen electrode so yeah that is a woman but this is the standard
• 79:30 - 80:00 hydrogen electrode i hope you got it i hope you got it okay my dear students yes ma'am so what is standard hydrogen electrode actually scientists what did they do they just took hydrogen as their reference and they uh themselves gave its reduction potential to be zero okay yes so what did they do they took hydrogen at one electrode
• 80:00 - 80:30 so if i write the reaction of the hydrogen that would be h plus plus electron that will give you half h2 this was the reaction at hydrogen electrode sheath standard hydrogen electrode and they arbitrarily gave its value a naught of h plus 2 h 2 equal to 0. they gave this value 0.
• 80:30 - 81:00 they used this as cathode they used this as cathode and all the other elements were used as anode all the other elements were used at anode right one would be cathode one would be a node you took hydrogen as the cathode and you took anode uh of all the all the other elements right yes when you calculate the cell emfs you can only and only calculate the complete cell you cannot calculate the emf of a particular half cell or cathode or anode so they took hydrogen as the reference value they
• 81:00 - 81:30 took it zero now when you will perform the reaction you can calculate the e naught of cell you know that it is cathode minus anode now cathode is always hydrogen it is always she if i tell you that let us say a naught of cell this is calculated and it is found to be x then you can say x is equal to 0 minus anode and this is how you got the reduction potentials of anode right and this is how you calculated the emfs of all the half cells with reference to hydrogen
• 81:30 - 82:00 you took hydrogen to be zero and you calculated the emfs of all the other cells that is the reason they said standard hydrogen electrode and the standard hydrogen electrodes the reduction potential is considered to be zero and the reduction potential of all the other elements are calculated with reference to the hydrogen electrode okay yes so i hope you got it perfect so this is very important this part is very important
• 82:00 - 82:30 with reference to this all other electrodes were calculated let's move forward now and after after studying all the reduction potentials this series was given to you which is known as your electrochemical series what does this electrochemical series tells you this tells you that as you move top to bottom here is your hydrogen i'll underline it with pink color here is
• 82:30 - 83:00 your hydrogen and you know that its reduction potential is zero okay with reference to this they calculated others reduction potential and they saw that below hydrogen were all the elements whose reduction potentials were having a positive value they had a positive value and the other elements above the hydrogen had their reduction potentials to be less than zero that is their reduction potentials were negative if
• 83:00 - 83:30 their reduction potentials were negative this means that my dear students their oxidation potentials were positive and these elements actually wanted to show oxidation and not reduction right yes so this is the uh conclusion that scientists drew and they made a electrochemical series what can we say that as we move top to bottom reduction potential reduction potential or tendency to show reduction
• 83:30 - 84:00 tendency to show reduction increases okay this is the first point that we can say what can we say the other other thing oxidation potential obviously your oxidation potential will or tendency or tendency to lose electron
• 84:00 - 84:30 or tendency to show oxidation decreases right so for opposite would be from if you go from top to bottom so bottom to top you would say that this would be opposite oxidation tendency would start on increasing right yes these are the two conclusions we drew right next my dear students next what can we conclude from here is third point which you can
• 84:30 - 85:00 conclude is which would be better oxidizing agents if you move from top to bottom you will get better oxidizing agents or lesser uh better reducing agents okay ma'am if i talk about oxidizing agents then oxidizing agent will oxidize the other and itself will get reduced if itself gets reduced then it should have higher reduction potential and moving top to bottom we get higher reduction potential because tendency of reduction is increasing this means that moving top to
• 85:00 - 85:30 bottom we get better oxidizing agents as reduction potential is increasing right yes and the opposite of this would also be true that we get lesser tendency
• 85:30 - 86:00 of reducing agents as oxidation potential is decreasing yes perfect you got this point as well now what is the other thing which we should know now my dear students imagine imagine you have you have a plus and you have copper right yes no no let's say that you have
• 86:00 - 86:30 copper two plus and you have zinc okay now you have to see that whether zinc will replace copper solution or not okay how can you find out this you have zinc somewhere here and you have copper somewhere here okay now you know that zinc has the tendency zinc wants to get converted into zn2 plus it has a better oxidation potential it shows oxidation it wants to lose electron right so mam
• 86:30 - 87:00 zinc wants to go into zn2 plus and as compared to zinc copper wants to get reduced it does not want to get oxidized so this means that mam copper will be converted into copper solid so what happened zinc actually displaced copper zinc actually displaced copper so so if an element is above the other then the above elements can displace the lower elements from their aqueous
• 87:00 - 87:30 solution why because they have a better oxidation potential so this point is very and very very important my dear students which i am going to write here and you have to remember it forever like forever that elements above have tendency to displace elements
• 87:30 - 88:00 below it from there from their aqueous solution as they are more reactive or
• 88:00 - 88:30 they have more oxidation potential okay so this is again a very important point which you need to remember because questions from this point are asked again and again so this is what is known as electrochemical series and i hope now you can answer why we wrote that in galvanic cell zinc is a
• 88:30 - 89:00 zinc to zinc two plus plus two electrons was spontaneous and copper two plus plus 2 electron into copper was spontaneous why these were spontaneous because ma'am zinc has a positive oxidation potential so it wants to get oxidized and mam copper has a positive reduction potential so it wants to get reduced okay yes that is the
• 89:00 - 89:30 reason these two reactions are spontaneous and i hope now my dear students you can understand all these concepts and you need to remember this electrochemical series okay perfect let's move forward let's move forward now you have a question and you have to answer the question okay given the standard electrode potential you are given the standard electrode potentials you have to arrange these metals in their increasing order of reducing power
• 89:30 - 90:00 reducing power means to reduce someone else what is reducing power reducing power means to reduce uh someone else or reducing itself okay this is the most important part now we are going to understand this you better write it you don't get confused to it you write it reducing potential don't get confused right reducing potential okay now my dear students if it is given that
• 90:00 - 90:30 arrange these metals in their increasing order of reducing potentials then what are you how are you going to do this right yes memory increasing order of the reducing potentials reducing potential means that they themselves want to get reduced so the more the redux reduction potential the more would be their value now if i see that this is a reduction potential positive to k these this means that k plus plus electron will give you k so this is reduction this is reduction this is
• 90:30 - 91:00 reduction and this is reduction this actually represents gives you the actual uh direction of your reaction if it is written mg2 plus this means mg2 plus would be here and mg would be here this means that plus two electrons so this is reduction okay so ma'am all of them are given to be your reduction potentials the more the reduction potential the more its value would be so reduction potentials my dear students which is the most positive value when the most positive value is a g plus one so a g
• 91:00 - 91:30 plus 2 a g has the most reducing potential it will be reduced maximum so i'll write only the elements a g my dear students then next is positive value hg then right mam hg and then my dear students this is less negative yes then it would be chromium then it would be magnesium and then potassium okay so
• 91:30 - 92:00 this is how you're going to do now the question is reducing power what if mammoth comes to be reducing power my dear students reducing power means to reduce someone else the similar way as you use reducing agent reducing power would be the same to reduce someone else not to itself but to the somewhat to someone else that would be my power now that what would be my power to reduce someone else that would be a reducing power so you have to
• 92:00 - 92:30 understand the words and accordingly answer the questions okay yes let's move forward next question is depict the galvanic cell in the following cell reaction which of the electrode is negatively charged the carriers of the current in the cell and individual reaction at each electrode can i give this you as your homework i hope we have studied everything and you can do this question yourself my dear students i will give you hints but you have to do it on your own
• 92:30 - 93:00 you are given a cell reaction write the cathode and the anode reactions then which of the electrode is negatively charged you have to find out which of the following is a negatively charged electrode i think you can understand which is your negatively charged electrode yes next would be the carriers of the current in the cell in the cell my dear students the carriers of current are ions outside the setup would be electrons individual reaction id in each electrode you can write write yes so this is your homework you have to do this question
• 93:00 - 93:30 yourself perfect let's move forward again you have a question you have to pause the video and try this question your self what does this question say okay you understand this question ma'am is having water and then we'll start forward let's start
• 93:30 - 94:00 ma'am the standard reduction potential values of three metallic metallications are given the order of reducing power of the correspond corresponding metal is okay so you are given x y and z and you are given reduction potentials this is 0.52 this is minus 3.03 and minus 1.18 now for reducing power my
• 94:00 - 94:30 dear students you understand this i hope you understand this man that for reducing power means to reduce someone else if you are reducing someone else then you are oxidizing yourself so you need the oxidation potential of all these and you know that reduction potential is equal to minus of oxidation potential the oxidation potential will be minus 0.52 plus 3.03 and plus 1.18 right yes now let's see which has a better oxidation potential will
• 94:30 - 95:00 have a better reducing power so ma'am if i see then y has the maximum oxidation potential so it will have the maximum reducing power and then mam zed because that will have 1.18 value so it has a max second maximum oxidation potential so second maximum reduction uh reducing power and so x so y z x y z x that is option one will be your correct answer option one will be your
• 95:00 - 95:30 correct answer so this is how you are going to solve the questions okay yes perfect let's move forward next question next question is emf of a cell is given e naught of the cell is given you have to find out the value okay now my dear students you have to find out the e naught of cell that is standard emf of the cell and you are given the e naught values as well okay i know that e naught
• 95:30 - 96:00 of cell is always equal to the reduction potential of cathode e naught minus e naught reduction potential of anode perfect yes now my dear students if i see this reaction i have to find which is my cathode which is my anode so let's find out a g plus 2 a g i can say that this is reduction yes ma'am so if it is reduction then this is my cathode and copper to copper two plus plus two electrons this is my anode
• 96:00 - 96:30 perfect so now if i see then i am given e naught of cu 2 plus 2 cu cu 2 plus 2 cu means cu 2 plus plus 2 electrons which gives me copper this is reduction so it will be reduction potential what is the reduction potential given 0.34 okay ma'am 0.34 and my dear students if you see then this is your anode so this is the reduction potential of your anode
• 96:30 - 97:00 is given as how much 0.34 right what is your cathode my dear students cathode is a g plus and that you have to find that you have to find okay so e naught of cell is given yes which is 0.46 so mam 0.46 is given cathode i don't know my cathode is a g plus 2 a g minus reduction potential of anode what is anode copper copper one i have been given which is 0.34
• 97:00 - 97:30 so e naught of a g plus 2 a g will be equal to zero point four six plus zero point three four which is equal to zero point four six zero one four three seven and eight zero point eight zero volt 0.80 volt that is option a will be your correct answer so this is how you have to find your cathode than your anode then you have to find their reduction potentials and then you can
• 97:30 - 98:00 find the e naught of the cell this is when you are given everything in these standard states okay let's move forward next questions so my dear students i have done one question this will again be your homework one question i'll do and the other question will be your homework perfect let's move forward now now we are going to use the nernst equation why are we going to use the nernst equation till now i uh helped you
• 98:00 - 98:30 i asked you to calculate the standard emfs standard emfs of cell at one molar concentration and everything what if the temperature and the concentrations get changed will you get the standard emf no ma'am because emf depends on these factors and if they are changed then you will not get standard emf then you are going to use the nernst equation to calculate the cell emf okay so now we are going to study how are we going to calculate these cell emfs okay
• 98:30 - 99:00 here a cell is given and we'll calculate it uh cell emf using nernst equation okay i'll just tell you what is your next equation okay next equation is given by an expression which is e of cell this time this will not be your standard emf so e of cell will be equal to e naught of cell minus 0.059
• 99:00 - 99:30 upon n log products concentrations upon reactance concentration right yes but if you want to make it a better uh this is the actual equation which you are going to use so this is very important you just need to remember this right let's write the expressions e of cell
• 99:30 - 100:00 is the cell emf at any temperature okay and a naught of cell is the standard emf of cell and n is the number of electrons exchanged number of electrons exchanged okay yes
• 100:00 - 100:30 so this is your next equation you first have to calculate that particular cell's standard emfs you will be given you will calculate e naught of cell by e naught of cathode minus anode formula and then you will find out the n value and then you will have to write the products and reactance concentration and eventually you can find the emf of the cell right yes so now let's move forward the most important thing that you are given here is calculation of the actual emf if you are calculating the
• 100:30 - 101:00 emf uh actual emf at standard states at standard states my dear students what can you say that e of cell will be equal to e naught of cell right at standard states emf of the cell would be equal to e naught of cell no issues right next if we talk about the system at equilibrium if you talk about the system at equilibrium what if your system is at equilibrium my dear students when your system is at
• 101:00 - 101:30 equilibrium then your e of cell is equal to zero but your e naught of cell can never be equal to zero always remember this point this is very important very very important point that at equilibrium mam e of cell will be equal to zero but e naught of cell can never be equal to zero similarly i can say that if i use the nernst equation which is e of cell is equal to e naught of cell minus 0.059 upon n
• 101:30 - 102:00 log can i write it q q was used in equilibrium concentrations i hope you understand that is why i am writing it q what is q q is product concentration upon reactants concentrations raise to the power they are stoichiometric coefficients now if this is 0 then you can say that e naught of cell will be equal to 0.059 upon n log q right yes log y q then this would be at equilibrium
• 102:00 - 102:30 at equilibrium q will be equal to k equilibrium equilibrium concentrations now you have to again remember your equilibrium part so this would be k equilibrium so this will be your expression for in or emf of the cell at equilibrium okay yes this is how you will calculate every thing perfect let's move forward calculations of cell emf now i will make
• 102:30 - 103:00 you understand how you are going to use the nernst equation if you are given a particular cell reaction i gave you the formula i gave you everything but again my dear students you actually need to know how to use it right yes so here you are given some information so what are you going to do first of all whatever we've learned it is going to check all your knowledge here if you have a cell representation i know that this is salt bridge right i know that this is cathode and i know that this is anode
• 103:00 - 103:30 so first of all i have to find out the net cell reaction so first of all we have to write the half cell reactions that is your cathode what will be your cathode mam i know that at cathode reduction takes place that means cu two plus will gain two electrons and get converted into copper and at a node oxidation will take place this means that zinc will be converted into zn two plus plus 2 electrons i have to add the 2 yes ma'am and what will i get 2 electron cancelled so you will get zinc
• 103:30 - 104:00 plus copper 2 plus this will give you zn2 plus plus copper right yes this will be your net cell reaction and if i ask you what is the value of n mam n value is 2 how many electrons have been exchanged 2 so n value would be 2 now if you write q value here what would be q for this particular reaction since copper is a solid its concentration is unity yes we will consider it to be constant because
• 104:00 - 104:30 copper is a solid pure state zinc 2 plus iron concentration zn2 plus ion concentration divided by here if we see that copper 2 plus ion concentration so here copper 2 plus ion concentration now use the nernst equation now we are going to use the nernst equation which is e of cell is equal to e naught of cell minus 0.059 upon n log zinc 2 plus concentration divided by
• 104:30 - 105:00 copper 2 plus constant tradition you have to calculate e naught of cell as well how are you going to calculate the e naught of cell if i see that this is cu 2 plus 2 c u ma'am this is a reduction potential perfect but if you see this my dear students this is your oxidation potential so you have to calculate its reduction potential so e naught of reduction potential would be negative of this value which is minus 0.76 i told you if you want to convert
• 105:00 - 105:30 oxidation potential to negative uh to reduction potential just put a negative sign so minus of 0.76 will give you your reduction potential now if you see i can say that copper part is your cathode copper part is your cathode so this is the e naught reduction potential of cathode which is equal to 0.34 and this is the e naught reduction potential of anode of anode because zinc is your anode right yes so e naught of cell will be
• 105:30 - 106:00 equal to a naught of cell will be equal to e naught reduction potential of cathode minus e naught reduction potential of anode what is the reduction potential of cathode 0.34 so 0.34 minus what is the reduction potential of anode minus 0.76 so minus 0.76 this will be 0.34 plus 0.76 which is equal to 6 4 10 7 8 9 10 and 1 so this is 1.1 volt 1.1
• 106:00 - 106:30 volt so this is what is your e naught of value since you can see e naught value is positive so i can say that this would be a spontaneous process right when delta when e naught value emf is positive we say it it is a uh spontaneous process right yes now we are going to calculate e of cell value of e naught of cell we just calculated 1.1 minus 0.0 my dear students i am just uh round
• 106:30 - 107:00 i am just rounding off 0.059 to 0.06 for making my calculations is the n value i just calculated to be 2 log concentration of z and 2 plus what is the concentration of z into plus given 0.1 molar so ma'am this would be 0.1 molar divided by concentration of co2 plus iron which is 0.01 0.01 this would be into 10 i can cancel
• 107:00 - 107:30 this out no no no no yes perfect so this would be 1.1 minus 0.03 into log 10 into log 10 which is exactly equal to 1 so 1.1 minus 0.03 into 1 this will give me 1.1 minus 0.03 which will be equal to 0.03 right yes
• 107:30 - 108:00 1.07 yeah 1.07 volt will be your emf of the cell so this is how you are going to calculate the emfs of the cell all these steps are very easy but this will need some practice which will help you do these questions very fast okay so you need a practice my dear students again and again i'm saying but this is how you're going to actually solve the questions let's move forward now we are going to calculate the relationship between the standard gibbs
• 108:00 - 108:30 free energy change and the standard emf of the cell so if i tell you the relationship between standard gibbs free energy change and emf of cell then that is equal to delta g naught is equal to minus n f e naught of cell you just have to remember this expression where n is your number of electrons exchanged f is your faraday value of faraday is nine six five zero
• 108:30 - 109:00 zero coulomb this is actually the charge of one mole of electrons charge of one mole of electron okay and if you talk about the e naught of cell then a naught of cell is standard emf of cell okay so this is how you are going to talk about the relationship between delta g naught and emf of these cells
• 109:00 - 109:30 standard emf of the cell right yes so i hope you got this point very good so you can just go through this and then we will move forward okay so now let's move forward what are we going to discuss next next we are going to discuss about the electrolytic stealth till now we were talking about the galvanic cell right yes everything we've discussed about
• 109:30 - 110:00 galvanic cell how are you going to calculate its uh emf how are you going to calculate it gibbs free energy if you're given e naught of cell right yes now my dear students what are we going to discuss is electrolytic cell in electrolytic cell you need to understand the faraday's law of electrolysis i don't want you to learn the faraday's law of electrolysis it has two different different forms of laws but i am going to make you understand this in a very easy way and you will solve each and every question of it in
• 110:00 - 110:30 just seconds right yes so my dear students what faraday's law actually tells you is that the charge on one electron what is the charge on one electron ma'am it is 1.6 into 10 raised to the power minus 19 coulombs okay what i ask you is what is the charge on one mole of electrons ma'am if charge of one electron is this just multiply it by avogadro number you will get the charge on one mole of electrons this would be 1.6 into 10 raised to the power minus 19 into avogadro number 6.02 into 10 raised
• 110:30 - 111:00 to the power 23 and when you multiply these values you get a value nearly equal to nine six five zero zero coulombs okay my dear students and this particular thing is known as one faraday this is actually known as one faraday what is one faraday faraday denotes you as the charge of one mole of electrons what is faraday charge of one
• 111:00 - 111:30 mole of electron is known as one faraday and its value is nine six five zero zero coulomb this means that mam if you have two moles of electron then you will have two faraday charge if you have three moles of electron you will have three faraday of charge if you have four moles of electron you will have four faraday of charge can i say that that the moles of electron will always be equal to the number of faradays yes if i tell you that i have nine faraday of charge i have nine faraday of charge this means
• 111:30 - 112:00 that i have nine moles of electron that is the reason i have nine faraday of charge right yes so the number of faradays actually gives you the moles of your electrons and this is the most important part which you will be using in your questions to solve your questions fast okay yes let's move forward my dear students i'll give you an example of using this concept here these type of questions which you actually used to do by those faraday's laws that faraday law one and faraday
• 112:00 - 112:30 first faraday's law second faraday's law you don't need them now you only and only need to understand that the number of faradays actually gives you the moles of electron that is the most simplest part and you can solve each and every question let's solve this question now the question says that you have 96 96.5 mph current which was passed through molten nacl for one minute and 40 seconds so here first and the most important thing that we are going to do is
• 112:30 - 113:00 that you have to calculate the charge so go to your physics concept q that is your charge is always equal to i into t where i is your current and t is your time right yes so what is the value of i ma'am it is 96.5 ampere and what is your time it is one minute 40 seconds can you convert your time into seconds only one minute has 60 seconds plus 40 so man this means
• 113:00 - 113:30 100 seconds so basically i have current 96.5 and i have time which is equal to 100 seconds so can i calculate the total charge current 96.5 into time 100 this will give me 9650 coulomb so this is your charge this is your total charge but you don't know how many faradays of charge do you have this is your total charge okay and in one parody you have nine six five zero zero coulomb of
• 113:30 - 114:00 charge so you can write that one you can write that one faraday has nine six five zero zero coulomb charge right yes so how many faradays do 9650 coulombs have mam number of faradays will be equal to given charge
• 114:00 - 114:30 divided by charge in one faraday one faraday which is equal to given charge divided by 96500 coulomb this is how you will always calculate your number of faradays so this came out to be your formula of calculating the number of faradays you can calculate the number of faradays from here yes so let's calculate the number of faradays so number of faradays would be equal to charge is 96500
• 114:30 - 115:00 9650 divided by the one faraday charge i'll get 0.1 right you know that your number of faradays is 0.1 i have just understood this that number of faraday is actually equal to moles of electron so how many moles of electron do i have 0.1 right till now i have calculated the moles of electron what does the question ask me the questions ask me that find
• 115:00 - 115:30 out the amount of sodium deposited at cathode my dear students if your sodium is getting deposited at cathode this means that sodium is showing reduction and how can sodium show reduction at cathode what the reaction will take place gain of electron obviously na plus would be gaining electron and getting converted into sodium solid now you are asked that how much of this solid is getting deposited from here
• 115:30 - 116:00 everything is your mole concept you have moles of electron 0.1 then how many moles of sodium would have been deposited when you had one mole of electron one mole of sodium was deposited when you if you have two moles of electron two moles of sodium will be deposited if you have zero point one moles of electron how many moles of sodium would be deposited zero point one moles so mam moles of sodium
• 116:00 - 116:30 deposited will be equal to 0.1 if you have moles of sodium deposited can you find the mass yes ma'am mass of sodium would be number of moles into molar mass which is 23 so by dear students 2.3 grams of your sodium would have been deposited you just used the concept of faradays you find out the found out the moles of electrons used it in the mole concept way and you just find found out
• 116:30 - 117:00 your answer this was so easy right what does the other part say other part says that and also you have to calculate the volume evolved at anode this means that cl2 is showing you the volume at anode so you have to write the reaction at anode as well what would be the reaction at a node mam loss of electron to cl2 cl minus would give you cl2 plus electron right so here 2 here 2 i hope you understand this concept yes
• 117:00 - 117:30 now still you have moles of electron equal to 0.1 only you have 0.1 moles of electron when you have two moles of electron you one mole of cl2 gases evolved if you have four moles of electron two moles will be evolved it means that i can say that moles of cl2 is always equal to moles of electron into half right it will always be half of moles of electron so this would be equal to zero
• 117:30 - 118:00 point one by two if you have the if you have the moles of a gas how can you find the volume of the gas remember the mole concept multiplied by 22.4 so mam volume of cl2 will be equal to 0.1 moles into 22.4 this will be 11.2 so 1.12 liters of cl2 gas is actually evolved at the anode so my dear students this is how you can
• 118:00 - 118:30 solve each and every question without any problem only and only you have to use your mole concept right yes so this is how you solve the questions related to the faraday's law perfect yes so you have to practice a lot of questions i think you can do that yes very good let's move forward next is my dear students electrolysis now how do we talk about electrolysis you know that you very well know that
• 118:30 - 119:00 your positive ions your positive i'll write positive ions will go to cathode always and forever in electrolytic cell and they will show reduction for electrolytic cells and your negative ions will always go to the anode and they will show oxydation right yes this my dear students we've studied in the starting pattern again and again i'm telling you you have to remember
• 119:00 - 119:30 everything so we studied in the starting part that in electrolytic cell uh positive ions used to go to the cathode and they used to show the reduction part and they used to show the negative part oxidation now my answer my question to you is what happens what will happen if i have two ions in the solution let us say i have h plus iron and let us say i have n a plus ion in a in one solution if i have two ions positive ions in one solution then
• 119:30 - 120:00 obviously at electrode any one ions reduction will take place the other ions reduction will not take place yes the other would stay in the solution and one would go to the cathode the question is out of these two which will go to the cathode which one will go to the cathode that is very important and that is given by this process now we are going to study that part okay so preferential discharge of cations and anions we are going to study that which will be preferred over the other
• 120:00 - 120:30 now you have to understand out of h positive and n a minus if anyone wants to go to the cathode what happens at cathode reduction so the iron which has a greater tendency to get reduced will go to the cathode first obviously so if you talk about h h plus and n a plus what you have to find out you have to find out the one with higher reduction potential the one with the higher reduction potential will go to the cathode and its reduction would take place the one with
• 120:30 - 121:00 the lower reduction potential will not go for this we need to go to the electrochemical series out of h plus and n a plus if we see in the electrochem chemical series ma'am h plus is here and n a plus is somewhere here and i can say that this has a reduction potential more right by because as we move down reduction potential increases so in the positive ions the one which is below will have a more reduction potential than the other and that will get reduced
• 121:00 - 121:30 so out of these two mam h plus will go to the cathode and n a plus will remain in the solution and h will h plus will get reduced right this is how my dear students you find out the electrolysis preferential electrolysis whose which ions will get reduced or oxidized okay so i hope you understand this right the one which is lower will have higher reduction potential and that will get reduced next is for anions right now we talked
• 121:30 - 122:00 about the cations we talked about the cations but what will happen for anions what will happen for anions right now if we talk about an ions what will happen my dear students this is the preferential discharge always and always if you have cl minus and oh minus in the preference then always and always cl minus will go to the anode and will get oxidized in if you have cl minus and oh minus in
• 122:00 - 122:30 if you have to choose one out of the two if you have to choose any one out of oh minus and so4 two minus then my dear students obviously o h minus will get oxidized okay so this is your preferential discharge okay yes let's move forward and try some questions now imagine imagine you have equals nacl solution equos means water right so nacl means you have n a plus and you have c l minus equals means you have h plus and o
• 122:30 - 123:00 h minus water i just broke water into h plus and oh minus now my dear students you have to choose out of the two that which will be reduced from n a plus and h plus i know that h plus is below in electric electrochemical series it has a higher reduction potential so man this will be reduced at cathode and you will get a reaction h plus plus electron which will give you h2 gas so mam at cathode what will happen h2 gas will be liberated out of these two cl minus and oh minus
• 123:00 - 123:30 which will be oxidized man we saw that cl minus was before oh minus so obviously it will go to the anode and get oxidized and what will be the reaction cl minus will give you cl2 plus electron so mam cl2 gas will be liberated at the anode right cl2 gas will be liberated at the anode so in this way we can say that in the solution you will get naoh solution which is a basic solution so the solution would be basic and at
• 123:30 - 124:00 cathode you will get h2s and at anode you will get cl2 gas right so this is how you are going to find out all the parts perfect very good let's move forward so this could be your homework you will have to do this as your homework i'll do one more and then we'll move forward next my dear students agno3 solution okay so you have hg plus and n o three minus and you have h plus and you have o h minus so if i talk about a g plus and
• 124:00 - 124:30 h minus a g minus is below in electrochemical series so a g plus will gain electron and give you a g solid and if you talk about these you will say that o h minus will go into the oxidation process and o2 gas will be liberated whenever o2 oh minus is oxidized o2 gas is liberated just remember this now what can you say in the solution you will be left with hno3 and this is a acid so the solution would be acidic in
• 124:30 - 125:00 nature so the solution would be acidic in nature so these are the three things which you need to understand you need to find out how you are going to calculate okay yes so you can just see this and then we will move forward let's move forward now what we are going to discuss next we are going to discuss your electrolytic conductance the second part which i told you that that would be
• 125:00 - 125:30 your numerical point of view part now we are going to discuss about that which is your electrolytic conductance okay so my dear students it is everything about numericals so just understand the concept and you are very good to go just listen to me very carefully if you talk about the resistance which you have learnt in physics resistance is directly proportional to l upon a where l was the length of your electrodes and a is the area between the electrodes okay
• 125:30 - 126:00 so whenever you used to remove the proportionality r would you be equal to rho l upon a you will write l is the distance between electrodes and a is the area between electrodes okay rho is the resistivity and
• 126:00 - 126:30 r is the resistance perfect now what are we going to do what are we going to do my dear students just reciprocate all of them so taking reciprocal taking reciprocal you will get 1 upon r will be equal to 1 upon rho and this will give you a by l okay yes now what i want to tell you is
• 126:30 - 127:00 that 1 upon r is equal to g which is called as conductance this is something new for you so you have to understand this 1 upon r will be called as g and that will be your conductance and one upon row will be known as k which will be your conductivity this will be your conductivity or specific conductance or specific conductance
• 127:00 - 127:30 okay so one upon r is your conductance one upon rho is your conductivity or you may call it as specific conductivity also my dear students your l by a is known as your g star and this is your cell constant these are some expressions which you need to remember which you need to remember so if i just convert all of them into this i can say that this is equal to g this will be equal to k and mam if you
• 127:30 - 128:00 take this a below l you can say that this will be equal to l by a which is equal to g star right so from here you can say k is equal to g into g star and this is a very important expression for you because you get that conductivity of a cell is equal to conductance into the cell constant okay so this is very important for you first write the resistance part take the reciprocal remember these particular symbols conductance conductivity and the
• 128:00 - 128:30 cell constant and you get your expression and you are asked a lot of numericals based on this part okay yes so i hope you understand this very good let's move forward so we found resistance we found resistivity we found conductance we found conductivity of the electrolytic solutions now i told you everything about it now we are going to talk about the units as well okay first we will talk about the units and then we will move forward if
• 128:30 - 129:00 you talk about resistance then the resistance unit would be om or um right yes so ma'am g is equal to 1 upon r so that's unit would be om inverse or you could say mo or you could say um inverse if you talk about resistivity then r is equal to rho a by l so you can say that rho will be equal to r
• 129:00 - 129:30 l by a this will give you the unit of resistivity as ohm meter in ohm centimeter inverse always remember my dear students we always take the distance as centimeter in centimeter okay so the unit of resistive resistivity resistivity would be ohm centimeter inverse then mam what will be the unit for this is important for you okay then what will be the unit for conductivity this is equal to 1 upon resistivity so the unit would be
• 129:30 - 130:00 ohm inverse 20 meter this would be ohm inverse centi meter or you could say that simon centimeter wait there is something which is not right just a second one upon rho rho would be equal to r into a by l rho will be c i did something wrong now
• 130:00 - 130:30 i was feeling that rho will be equal to c we'll write this expression as 1 upon r is equal to 1 upon rho a by l so rho will be equal to rho will be equal to r into a by l r into a by l yes right yes so the unit for resistivity would be ohm centimeter
• 130:30 - 131:00 right its unit would be ohm centimeter so if you find the unit for conductivity that would be equal to 1 upon rho and this will be equal to ohm inverse centimeter inverse right or you can say it as simon centimeter inverse okay so this unit is used a lot which is the simon centimeter inverse so the unit for conductivity is simon centimeter inverse okay we found the units of most of them now g star which is your cell constant which is equal to l by a which
• 131:00 - 131:30 means that centimeter inverse will be the unit for your cell constant which is the g star so this is how you calculate their units my dear students because units are equally important when you talk about the numericals okay and i never ever memorize any of the units i only calculate them on my own so these are important for you so i'll just mark them for you okay i hope you get it perfect let's move forward my dear students if we see then we'll solve a question
• 131:30 - 132:00 and you will understand all the concepts how are we going to use these expressions in the numericals the first question tells you that if k of 0.1 normal kcl is this means that you are given the conductivity as 1.2 into 10 raised to the power minus 2 right simon centimeter inverse perfect and r is 60 ohm and r is given as 60 ohm then find g star then you have to find g star
• 132:00 - 132:30 okay how are we going to calculate this i know very well one thing that k is equal to g into g star we did calculate this yes in the question i have k you have k map you have to calculate g star you have to calculate g star so you need to calculate g what is g g is conductance and how is how can you calculate conductance 1 upon r which is 1 upon 6 t so now just put the values g star will be equal to k upon g k is 1.2
• 132:30 - 133:00 into 10 raised to the power minus 2 and this will be divided by g what is g 1 upon 6 t this will be 1.2 into 10 raised to the power minus 2 into 60 right this will give you your g star value which is your cell constant and cell constants unit would be centimeter inverse right yes this is how you are going to use the expressions in your numericals let's solve one more question and we'll be very clear so now you
• 133:00 - 133:30 multiply this okay i'm not going to multiply it for you because i know my dear students can do this perfect let's move forward next you have a question now solve this question a conductance cell was filled with a 0.02 molar kcl solution which has a conductivity your given conductivity to be 2 into 10 raised to the power minus 3 find the resistance you have to find resistance r whose sigma is 0.1 now i don't know what is sigma but i can find out that
• 133:30 - 134:00 centimeter inverse is given as unit this means that sigma denotes your g star cell constant so g star is given to be 0.1 centimeter inverse resistance you are supposed to calculate how are you going to solve this right mm i know that k is equal to g into g star i am given k and i am given g star so can i calculate g k upon g star which is 2 into 10 raised to the power minus 3 upon 0.1
• 134:00 - 134:30 right yes you will calculate this you know that g is equal to 1 upon r so r will be equal to 1 upon g this means 1 upon of this so 0.1 upon 2 into 10 raised to the power minus 3 will give you your resistance right so this is how you're going to solve the numericals they are so easy they are so easy only you need to remember are the formulas and their inter conversions okay yes more and more practice would really help you let's move forward now there is an
• 134:30 - 135:00 expression which is known as molar conductivity till now we just talked about the conductivity what does molar conductivity say molar conductivity says that the conductivity of one mole of electrolyte if you have only one mole of electrolyte then how are you going to calculate the molar conductivity my dear students for that molar conductivity would be equal to k into thousand upon c where k is the conductivity
• 135:00 - 135:30 and c is the concentration in molarity it will be in molarity mildew student okay yes so k will be your conductivity and this will be your molarity and this would be in simon centimeter inverse then you can calculate your lambda m then you can calculate your molar conductivity okay perfect i hope you understood this similarly if you talk about the equivalent conductivity how are you
• 135:30 - 136:00 going to calculate the equivalent conductivity lambda equivalent will be equal to k into thousand upon n now my dear students here n is your normality okay and this is simon centimeter inverse perfect yes i know that i know very well that uh normality if you have to convert if you have to convert uh if you have to convert norma if you have
• 136:00 - 136:30 to find the relationship between molarity and normality how are you going to find out the relationship use that and then you can calculate the relationship between equivalent conductivity and molar conductivity as well okay you know that molar normality is equal to molarity into n factor we know that yes molarity molar mass divided by n factor n factor will be multiplied so you get molarity into n factor put this value in lambda m
• 136:30 - 137:00 which is k into thousand no put this value here so lambda equivalence will be equal to k into thousand upon molarity into n factor now what can i say ma'am this particular part is actually your lambda m so cross multiply this here so lambda m is equal to lambda equivalent into n factor so if you are given any one value you can find out the other value by just calculating the n factor by just
• 137:00 - 137:30 calculating the n factor right yes perfect very good so this is how you are going to inter convert both of them relationship we've just found ma'am lambda m is equal to lambda equivalent into n factor perfect we just calculated it now we are going to apply that now we are going to apply that okay now imagine for nacl now
• 137:30 - 138:00 a lot of students would have a doubt that how to find the n factor so my dear students always you have to find the n factor as total positive charge or total negative charge either see the positive charge or see the total negative charge if i see here here the total positive charge is plus one here the total positive charge is plus two here total positive charges plus three in this case total positive charge fe has plus three charge on one but you have two fe plus so you will have plus three
• 138:00 - 138:30 into two which is equal to plus six charge okay so n factor here would be one here two here three and here six so total positive charge you have to calculate so the relation would be lambda m would be equal to lambda equivalent into 1 lambda m would be equal to lambda equivalent into 2 lambda m would be equal to lambda equivalent into 3 and lambda m would be equal to lambda equivalent into 6. this is how you are going to calculate
• 138:30 - 139:00 the values perfect no issues very good let's move forward now my dear students i am going to tell you something very very interesting that is that when you talk about the conductance when you talk about the conductance you have strong electrolytes and you have weak electrolytes the strong electrolytes are strong bases strong straws salts and strong acids they have a lot of
• 139:00 - 139:30 conductance and we can calculate their conductance but graphically how can we calculate their conductance by drawing a plot something like this where this gives you your molar conductivity or equivalent conductivity and this gives you root of your concentration this is the root of your concentration right yes when you plot a graph when you plot a graph you get something like this which this is how you get the get your plot
• 139:30 - 140:00 right yes so you say that lambda m is equal to lambda i'll i'll just make you understand this first and then we'll move forward when you talk about the molar conductivity or equivalent conductivity for strong electrolytes this is for strong electrolytes okay so for strong electrolytes what you see is that mam if you if you extrapolate this graph you get this point where it cuts to 0 this means
• 140:00 - 140:30 that there is 0 concentration means 0 i'll write it here this means that 0 concentration maximum dilution right at maximum dilution you get the lambda m infinite or you get the lambda equivalent infinite this is the conductance at infinite dilution where you have the maximum dilution or zero concentration you know that there the
• 140:30 - 141:00 ions will have the maximum conductance so this can be calculated for strong electrolytes because this graph is valid only and only for strong electrolytes but for weak electrolytes this graph is not at all possible for weak electrolytes you get a plot something like this and we see that this whenever we extrapolate it this just goes into the parallel form and we never get the conductance of weak electrolytes at infinite dilution
• 141:00 - 141:30 right so we need to find some way to calculate the uh infinite dilution we need to find out some way to calculate the uh conductance of strong electro of weak electrolytes at infinite dilution okay yes ma'am needs one minute break so that i can have some water and then we'll move forward my dear students you just see this graph till then okay
• 141:30 - 142:00 okay mom the break is over now let's move forward weak electrolyte in case of weak electrolytes the increase in equivalent or molar conductivity is more pronounced with the dilution the degree of ionization of weak electrolyte increases with the increasing dilution which is the austral's dilution law same idea students if you want to find out the conductance then that means you have to
• 142:00 - 142:30 keep the ions away from each other and when you increase the dilution of the solutions you get the conductance okay so for weak electrolytes the more the dilution is done if you see the graph the more the dilution is done see this side dilution is increasing the more the dilution is done you get a better value of molar conductivity but the problem is but the problem is that you cannot find the value at zero uh concentration or infinite dilution then how are we going to calculate that
• 142:30 - 143:00 that is given by the call roche law okay that is given by the coherence law which is the infinite dilution cohorosh law for the independent migration of ions what does the caulrash law says caulrash law says that bring the ions bring the ions and at infinite dilution every ion has its particular conductance for example if you talk about n a plus n a plus will have its particular value be it with nacl be it with naoh na plus at infinite
• 143:00 - 143:30 dilution will have a fixed value just the way you have your pan card your pan card has a fixed value be it you are a friend with one person or with the other person your pan card value won't change the same way the conductance of every iron at infinite dilution never changes be it it is a part of a strong electrolyte or it is a part of a weak electrolyte so called roche law said that if we are able to calculate the independent migrations independent conduct uh conduction values of the
• 143:30 - 144:00 electrolytes of the ions then we can just add them and calculate the values for the strong as well as weak electrolytes at infinite dilution okay so what does this mean how are we going to do this we know that we can calculate infinite dilution values for strong electrolytes so i'll calculate for strong electrolytes and then i'll use maths and then find out the values for weak electrolytes as well how let's talk about it okay see imagine imagine you have strong
• 144:00 - 144:30 electrolytes such as you have strong electrolytes such as naoh you have a strong electrolyte let's say nacl and you have a strong electrolyte let's say hcl okay if you have strong electrolytes all of these three are strong electrolytes then you can calculate their infinite dilutions by that particular graph
• 144:30 - 145:00 extrapolating yes ma'am let's say that its value is 2x this value is y and this value is z now you are supposed to calculate the infinite dilution of h2o which is a weak electrolyte how can we do this how can we do this i am going to teach you this imagine imagine you
• 145:00 - 145:30 diluted them to the maximum level you diluted them to the maximum level the ions are now separate out this means that ma'am lambda m infinite of naoh which is also equal to x is equal to lambda m infinite of n a plus plus lambda m infinite of o h minus right we can say this x is equal to uh infinite dilution of this plus infinite dilution of this yes the other point would be this part lambda m infinite of
• 145:30 - 146:00 nacl will be equal to y and this will be equal to lambda m infinite of n a plus plus lambda m infinite of cl minus right next is hcl that is also a strong electrolyte the lambda m infinite of hcl will be equal to z and this will be equal to lambda m infinite of h plus plus lambda m infinite of c l minus
• 146:00 - 146:30 okay yes these values i am i i know these values but the value which i have to calculate is lambda m infinite of h 2 o which will be equal to lambda m infinite of h plus which will be into 2 no no no no no no no this will be lambda m infinite of h plus plus lambda m infinite of o h minus okay so now what you have to do see these three
• 146:30 - 147:00 equations one two and three how are you going to arrange them so that you get this value you get this value how are you going to do you need h plus you need this one h plus so can i say 3 i will add the equation 3 plus i need o h minus i need o h minus this means that i need this o h minus also so i'll add 1 as well yes now but i don't need this n a plus and i
• 147:00 - 147:30 don't need this nacl so what can i do i can subtract the second equation i'm subtract the second equation i will get what i want which is lambda m infinite of h2o so add three and one and subtract the second equation from that you will get what you want for the weak electrolyte so this is what is known as called roche law so this value would be equal to uh z plus
• 147:30 - 148:00 z plus x minus y so this is how you can calculate the infinite dilution conductance of weak electrolytes using the values of these strong electrolytes and this is what is known as cohors law it is very easy very easy okay yes i hope you understood this perfect now you can study this this is very easy just read the points the equivalent conductance of an electrolyte had its maximum value at infinite dilution it will have maximum value and it is constant for a given electrolyte it is
• 148:00 - 148:30 called the equivalent conductance at infinite dilution any electrolyte strong or weak will be 100 ionized at infinite dilution yes at infinite dilution the inter ionic attraction is zero the ions are so apart from each other that their values are zero that is the reason you get independent conductance of each ion and each ion can migrate independently now each ion can contribute a fixed value towards the equivalence conductance fixed value that is your that pan card value of the electrolyte and this is known as ionic equivalence
• 148:30 - 149:00 conductance okay yes perfect it is constant for a given iron and different for different ions i hope you understand this yes this is the concept we use for call rash law yes so my dear students this was all about the conductance part okay yes there are a few parts in electrochemistry which were related to your cells that is very theoretical part so that is the reason i have not covered that because there are just just two to three reactions which
• 149:00 - 149:30 you have to study and you're good to go so you will read those fuel cells h2o2 cells and those which are written you're in in your ncrt please go through them rest all part ma'am has covered okay so we've discussed everything about the concepts here now let's practice some questions so that we get a little confidence in what we have studied today okay so let's start first is the electrolytic decomposition of dilute sulfuric acid with platinum
• 149:30 - 150:00 electrode cathodic reaction is see it is telling you that you have the electrolytic decomposition of dilute sulphuric acid dilute sulfuric acid means you have h plus and oh minus you have again h2so4 you have h plus and so4 2 minus ion water ions and you have sulfuric acid ions and this is with platinum electrode you have to tell the cathodic reaction what will happen at the cathode okay ma'am cathode reactions will take place
• 150:00 - 150:30 this side and only h plus will go towards this so mam there would be reduction of h plus right there would be reduction of h plus we don't have any other inv in both the cases we have h plus i and so any h plus n can go and its reduction will take place so option a will be your correct answer i hope you got it it is very easy perfect my dear students let's move forward to the next question read the question and try yourself you have to pause the video and do your question yourself the question says that
• 150:30 - 151:00 which one of the following metals cannot be obtained on the electrolysis of equisolution of its salt so if i talk about echo solution this means h plus and oh minus yes and you will have which one of the following metals so basically it is only talking about the positive part so you just remove oh minus you have to talk about this it says that which one of the following metal can not be obtained on electrolysis which part will not be obtained on electrolysis mam which can
• 151:00 - 151:30 never get reduced which will not be reduced the one which are above h plus because the lower the element the more is the reduction potential the more will be the reduction and ma'am the elements which will be above h plus they will never get reduced so out of the following which of the following is above the above h plus if i see that mg plus is the one which is at the top of the electrochemical series so option a will be your correct answer
• 151:30 - 152:00 option a will be your correct answer also my dear students there are two correct answers chromium is also above your hydrogen if you have h plus iron and chromium ion then h plus will get reduced not chromium if you have mg2 plus and h plus then h plus will be reduced not mg2 plus that is the reason they will never be obtained on the reduction but perfect yes so a and d would be a correct answer let's move forward to the next question next question is a solution of potassium
• 152:00 - 152:30 sulphate you have k2 so4 solution in water so this is in water so it is equal solution electrolyzed using inert electrodes the products at the cathode and anode respectively would be this means that you have h plus and oh minus and you have k plus and so4 2 minus so if i remember then i know that out of these out of these man h plus is below in the electrochemical series as compared to k plus so this will go to
• 152:30 - 153:00 the cathode side and you will get h2 gas you will get h2 gas and if you talk about the anode part then i know that o h minus will go to the anode side again remember those that series and you will get o2 gas this means that h2 and o2 gas we will get you have to tell at cathode and then a node at a third gether i'll get h2 and o2 that is option a will be your correct answer right yes let's move forward next question my dear students an echo
• 153:00 - 153:30 solution of na2so4 you have na2so4 solution so you will have n a plus and so forth two minus i n and you will have h plus and o h minus i is electrolyzed using platinum electrode the products at the cathode and anode are respectively so if i see out of these i know that h plus will go to the uh go to the cathode and you will get h2 gas and if you talk about oh minus mam we'll get o2 gas again so h2 and o2 so you have to tell cathode and a node so h2
• 153:30 - 154:00 and o2 again which is option c is your correct answer right so these type of questions are asked a lot of times from you so you have to practice them very well let's move forward next question the electrolysis of a solution resulted in the formation of h2 gas now uh electrolysis took place and you got h2 gas on the cathode okay and o2 gas at the anode so the solution is which of the solution will give you h2 at node and o2 at cathode okay this means that if i
• 154:00 - 154:30 want h2 at cathode i should have h plus here and some ion which will be above in the electrochemical series right and if i talk about oh my o2 gas then i need oh minus ion here and some element which is back of the o h minus ion if i talk about a g plus my dear students a g pluses always and always below the h plus and that is why a g will be reduced not h plus so this will
• 154:30 - 155:00 not this will be nor this will not be possible if you talk about h2 so4 you can write here h plus and so4 minus if i see then h plus will be there then you will get h2 guess and here oh minus you will get s we got our answer option b will be a correct answer still let's see are the other options if you have highly concentrated nacl solution again you will never get this my dear students because when you have a concentrated solution then you get n a at solid state okay and also when you have cl minus and oh minus cl minus will be on the anode
• 155:00 - 155:30 side you will get oxidative oxide on oxidation you will get cl2 gas and not the o2 gas okay next is cu cl2 again cu is below hydrogen so you will get copper on the cathode side not hydrogen on the cathode side right so only one option is correct which is option b perfect let's move forward next question if mercury is used as cathode in the electrolysis of aqueous nacl solution
• 155:30 - 156:00 the iron discharge that cathode is it is the important question my dear students this is an important question which you need to understand that whenever mercury is used as an electrode whenever mercury hg used as an electrode you will always and always get sodium reduced at cathode as it forms amalgam as
• 156:00 - 156:30 it forms amalgam as it forms amalgam that is n a h g it forms amalgam and that is the reason you will always then always get sodium deposited whenever you use mercury as your electrode okay perfect so you will get n a plus discharge at cathode which is n a plus so option b will be your correct answer perfect let's move forward next question which of the following aqueous solution
• 156:30 - 157:00 products metal after electrolysis produces a metal after electrolysis so here we have the iron k plus this will not produce here again you have k plus here you have cu 2 plus and here you have n a plus see all k plus k plus and uh n a plus all these are above hydrogen yes ma'am so they will never be reduced but my dear students copper is below hydrogen so whenever h plus and co2 plus will
• 157:00 - 157:30 come copper will be reduced so correct answer would be option d right yes this is how you are going to solve these questions perfect let's move forward next question is in front of you you have to solve this numerical my dear students what is this numerical the resistance of 0.1 normal solution formic acid is 200 ohms so you are given the resistance as 200 ohm i'll just write the number and cell constant is 2 so you have g
• 157:30 - 158:00 star equal to 2 the equivalent conductivity of 0.1 normal formic acid is how are you going to calculate lambda equivalent is equal to k into 1000 upon normality so basically you have normality but you don't have the conductivity you have to calculate conductivity conductivity is equal to g into g star you have g star but you don't have g you can calculate g as 1 upon r which is equal to 1 upon 200 so 1 upon 200 into 2 which is equal to 1 upon
• 158:00 - 158:30 100 so put this value here this would be equal to 1 upon 100 into 1000 upon normality which is how much 0.1 so just solve this and you will get lambda eq will it let me know what is the correct answer in the comment section okay yes let's move forward to the next question next question my dear students is on your screens now the conductance cell was filled with a 0.02 molar kcl solution which has a specific
• 158:30 - 159:00 conductance of this so my dear students you are given k 2.768 into 10 raised to the power minus 3 if its resistance is 82.4 you are given resistance as 82.4 at 25 degree celsius you have to calculate g star you have to calculate g star it is very easy my dear students k is equal to g into g star and we know that g is equal to 1 upon r into g star and this is k you are given k value you are given r value i hope you can
• 159:00 - 159:30 calculate the g star value yes very easy very easy put the values and solve the question the questions are very easy you don't have to hurry up just think about it and solve the question perfect next question let's see the next question the next question says that electrolytic conductance is due to the movement of now it is asking you about the electrolytic conductance electrolytic conductance you need actually charge carriers for that and i told you that charge carriers are majorly of two types it can be free
• 159:30 - 160:00 electrons or it can be free ions right yes and mam electrons can never be present in the solution it means that only and only ions are present in the solution and electrolytic conductance actually is asking you about the conductance due to electrolyte and their my dear students we will say that only free ions can conduct electricity not the electrons so free ions or ions that is option b will be your correct answer right yes perfect my dear students let's
• 160:00 - 160:30 move forward and with this we have came to the end of our session so i hope you all understood all the concepts that we discussed today my dear students you have to practice a lot of questions on nurse equation on electrolytic conductance to gain confidence in this so do let me know in the comment section how was this session for you was it beneficial or not and we will be meeting in the next session so thank you so much keep studying all the very best