Electrochemisty Lecture Video Presentation #3 Electrolytic Cells

Summary

Highlights

• Electrolytic cells require energy input, unlike spontaneous voltaic cells 🔋.
• Using Faraday's law to determine the stoichiometry of redox reactions in electrolytic cells 📐.
• Electrolysis of water produces hydrogen and oxygen gases, showcasing stoichiometric principles 💧.
• Difference between voltaic and electrolytic cells in terms of cell structures and reactions 🧪.
• Sacrificial anodes used to prevent oxidation and corrosion in metals 🛡️.

Key Takeaways

• Electrolytic cells need an external power source to drive non-spontaneous reactions ⚡.
• Voltaic cells vs. electrolytic cells: spontaneous versus forced processes 🔄.
• Faraday's law helps calculate the charge flow in an electrochemical reaction 📊.
• The role of electrolytic processes in metal plating and ore separation 🏗️.
• Understanding corrosion and how sacrificial anodes can prevent it ⚙️.

Overview

Chapters

• 00:00 - 00:30: Introduction to Electrolytic Cells The chapter introduces electrolytic cells, which are a key concept in electrochemistry. It signals the end of a series of lectures on electrochemistry, marking a milestone for the year. The focus on electrolytic cells highlights their nature as non-thermodynamically favored processes, requiring external intervention to proceed.
• 00:30 - 01:00: Understanding Electrochemical Cells The chapter titled 'Understanding Electrochemical Cells' covers the process of creating and analyzing electrochemical cells using a power source. It teaches how to calculate the amount of charge flowing through the cell based on the changes in reactants and products. The chapter explains the use of Faraday's law to determine the stoichiometry of a redox reaction, focusing on variables such as the number of electrons transferred, the mass of material deposited or removed, current, time elapsed, and the charge of ionic species.
• 01:00 - 01:30: Calculating Charge and Using Faraday's Law The chapter focuses on understanding the calculation of electrical charge using the equation i = q/t, where 'i' is current, 'q' is charge, and 't' is time. It also explores Faraday's Law and the prediction of electrolysis products in electrolytic cells. These cells require an external current to drive a chemical reaction since the cell potential is negative, leading to a positive Gibbs free energy (ΔG), indicating that the reaction is non-spontaneous.
• 01:30 - 02:00: Electrolysis and Cell Diagrams The chapter discusses the concept of electrolysis, a process that requires the supply of electrical energy to induce a chemical reaction, as these reactions do not occur spontaneously. This process is utilized for separating ores and plating metals, such as applying chrome coatings to car bumpers. Important distinctions in this process are noted.
• 02:00 - 02:30: Differences between Galvanic and Electrolytic Cells In this chapter, the differences between galvanic and electrolytic cells are discussed. The main distinction highlighted is that voltaic (or galvanic) cells operate spontaneously or thermodynamically favorable, whereas electrolytic cells require an external force such as an electron pump, battery, or direct current source to function. While a voltaic cell is divided into two half cells to generate electricity, an electrolytic cell can occur either in a single or multiple containers.
• 02:30 - 03:00: Predicting Electrolysis Products and Standard Cell Potentials This chapter discusses the distinction between voltaic (or galvanic) cells and electrolytic cells, highlighting that a voltaic cell acts as a battery, while an electrolytic cell requires a battery to operate. The main principles of electrochemical cells are explained, including the occurrence of oxidation at the anode and reduction at the cathode, as well as the differences in electrode polarities.
• 03:00 - 03:30: Using the Nernst Equation for Small Ion Concentrations This chapter focuses on the application of the Nernst equation specifically for small ion concentrations. It explores the concepts of galvanic and electrolytic cells. The content highlights the process of electron flow from the anode to the cathode, explaining their roles in different settings. It stresses on remembering that in electrolytic cells, the positive anode (EPA) is reversed in comparison to voltaic cells. Solutions often use inert electrodes for these processes. The chapter concludes with a brief discussion on the delta G value for voltaic cells.
• 03:30 - 04:00: Electrolytic vs Voltaic Cells - Tin and Copper Reaction In this chapter, the discussion centers around electrolytic and voltaic cells. Specifically, it covers the reaction between tin and copper. Key points include the fact that these cells are thermodynamically favored processes with an E cell value greater than zero. Oxidation occurs at the anode, which has a negative polarity, while the cathode, being the positive electrode, is where reduction takes place. Important terms such as 'anox' (anode oxidation) and 'red cat' (reduction cathode) are highlighted in the context of electrolytic cells.
• 04:00 - 04:30: Industrial Electrolysis of Molten Salts The chapter focuses on the process of industrial electrolysis involving molten salts. It highlights key thermodynamic principles including the significance of delta G and electrochemical cell potential (E cell) values. A positive delta G and a negative E cell value indicate an unfavored thermodynamic process. Oxidation occurs at the anode, maintaining its role in electrolysis, but with noteworthy changes in electrode sign.
• 04:30 - 05:00: Electrolysis of Mixtures and Periodic Trends The chapter discusses the topic of electrolysis of mixtures and explores periodic trends. It highlights the change in sign for the cathode in electrolytic cells, noting that while reduction still occurs at the cathode, it is now considered negative. The text suggests that an explanation for the change in sign will be provided and mentions the relationship to the equilibrium constant (k value), indicating that for electrolytic cells, the k value would be greater than one.
• 05:00 - 05:30: Faraday's Law and Calculations The chapter on 'Faraday's Law and Calculations' discusses the electrolysis of water, focusing on the formation of hydrogen (H2) and oxygen (O2) gases. It highlights the stoichiometric difference in the amounts of gas produced, where twice as much hydrogen gas is formed compared to oxygen gas. This difference is related to the oxidation half-reaction occurring during the process.
• 05:30 - 06:00: Example on Faraday's Law - Aluminum Electrolysis The chapter titled 'Example on Faraday's Law - Aluminum Electrolysis' explains the electrochemical reactions occurring within an electrolytic cell during aluminum electrolysis. It describes the formation of water molecules into hydrogen ions, oxygen, and the release of electrons as part of the process. A key observation made is that in the oxidation half-reaction, the pH level decreases, marked by an increase in hydrogen ions, indicating where oxidation is taking place within the cell.
• 06:00 - 06:30: Corrosion of Metals and Iron The chapter discusses the corrosion of metals, focusing on the chemical reactions involved, particularly the reduction half-reaction where water and electrons form hydrogen gas and hydroxide ions. It notes an increase in the concentration of H+ and a potential rise in pH during the reduction process, emphasizing that these do not align with standard cell potentials.
• 06:30 - 07:00: Sacrificial Anodes for Corrosion Prevention This chapter discusses the role of sacrificial anodes in preventing corrosion. It emphasizes the importance of standard cell potentials for corrosion prevention, explaining that the molarity of the solution must be 1.0 molar under standard conditions. The chapter briefly revisits concepts from previous lessons, particularly touching on the behavior of pure water, including the initial concentrations of H3O+ and OH- as learned in the acid-base unit.
• 07:00 - 07:30: Conclusion and Upcoming Topics The chapter focuses on wrapping up the current discussion and introduces topics for the upcoming discussions. It revisits the use of the Nernst equation to determine cell voltage in reactions involving pure water and small ion concentrations. Additionally, it touches upon the necessity of introducing a salt or electrolyte in such cases to facilitate the reaction.

Electrochemisty Lecture Video Presentation #3 Electrolytic Cells Transcription

• 00:00 - 00:30 good morning from a stormy mr anderson's house uh we're going to conclude the last of our electrochemistry lectures today uh which will be the last for the year congratulations you made it through almost uh but we're going to focus on electrolytic cells in this presentation so electrolytic cells again are the non-uh thermodynamically favored processes they are processes that will require an external
• 00:30 - 01:00 power source in order to create that electrochemical cell in this video you'll learn how to calculate the amount of charge that flows based on the changes in the amounts of reactants and products in an electrochemical cell will use faraday's law to determine the stoichiometry of a redox reaction occurring in an electrochemical cell with respect to the number of electrons transferred the mass of the material deposited on or removed from an electrode the current the time elapsed and the charge of the ionic species
• 01:00 - 01:30 with an equation i equals q over t we'll also learn how to predict the products of the electrolysis in this process as well for the electrolytic cells we are forcing a current through a cell to produce a chemical change for which the cell potential is negative so if you remember we would have a positive delta g in the situations where the cell potential is a negative value and so that would mean that it would not
• 01:30 - 02:00 happen on its own so energy must be supplied in this case electrical energy must be supplied to the reaction in order to make it take place we use this process as a way to separate ores from each other and played out metals so if you've ever heard about like chrome bumpers on old old cars that's the plating process that they would use to make the chrome metal on the bumpers there is a couple of very important differences
• 02:00 - 02:30 between the voltaic or the galvanic cells in an electrolytic cell number one the voltaic cells again are spontaneous or thermodynamically favorable and the electrolytic cells are forced to occur by using an electron pump or a battery or any direct current source dc source the voltaic cell is separated into two half cells to generate electricity an electrolytic cell can occur in a single container it can occur in multiple containers but
• 02:30 - 03:00 more often than not it happens in a single container so from a visual standpoint with the diagrams for the cells that's a very easy way to differentiate between the two a voltaic or galvanic cell is a battery and an electrolytic cell needs a battery you still have the oxidation occurring at the anode and the reduction occurring at the cathode but the polarities of the electrodes get
• 03:00 - 03:30 reversed the cathode will become a negative and the anode becomes a positive so you can remember epa electrolytic positive anode the electrons will still flow from the anode to the cathode and usually because we're dealing with solutions they will occur using inert electrodes let's look at a quick comparison of voltaic and electrolytic cells first off your voltaic cells will have a delta g value
• 03:30 - 04:00 less than zero and will have an e cell value of greater than zero meaning again that they are thermodynamically favored processes the electrodes you have the anode where oxidation occurs and the polarity of the electrode is a negative value the cathode is your positive electrode and it will be where reduction takes place so you still have the anox and the red cat coming into play here for your electrolytic cells
• 04:00 - 04:30 you would have your delta g value being greater than zero your electrochemical cell value being less than zero so you'd have a negative value for e cell positive value for delta g which means a thermodynamically unfavored process your anode is where oxidation takes place so it still has the oxidation at the anode but you should hopefully notice again that the sign for the electrode has now switched to a
• 04:30 - 05:00 positive the cathode is still where reduction takes place but it is also switched signs to now being a negative i'll explain why that is the case in just a few moments but also just kind of remember the last part of that is our relation to k so i'm going to pop this in here and our k value in this situation for your electrolytic cells your k value would be greater than one and your electrolytic
• 05:00 - 05:30 cells your k value would be less than one all right so now we're going to look at the electrolysis of water during this process you can see that we have bubbles of h2 gas and o2 gas that are forming but please notice the difference in the amounts stoichiometrically speaking we have approximately twice as much of the hydrogen gas that is formed compared to the oxygen gas when we look at the processes that are taking place the oxidation half reaction
• 05:30 - 06:00 within the cell is two h2o forming four h plus plus o2 and four electrons being released notice here again i have four h pluses so one of the little signature pieces that you could see for where the oxidation is taking place in that electrolytic cell is to look for a lowering of the ph so the ph will go down in the oxidation half reaction as the
• 06:00 - 06:30 concentration of h plus increases when we look at the reduction half reaction i have two h2o plus four electrons forming two h2 gas plus two o h minus so again at this point where we have the reduction taking place we could see an increase in the ph of the cell as well now just understand these will not correspond to regular standard cell potentials
• 06:30 - 07:00 because in order to be standard cell potentials remember the molarity of the solution has to be 1.0 molar for those standard conditions i'll write that down here real quick so your standard conditions would be one molar however in this particular instance since we're talking about pure water remember from our acid base unit the h3o plus and the oh minus concentrations initially
• 07:00 - 07:30 so i'll put the little initial sign there would be equal to one times ten to the negative seventh power because we're talking about pure water so as a result we would have to use the nernst equation to essentially figure out what the cell voltage would be for that particular reaction to take place because of the fact that it has such small ion concentrations we also have to have some type of salt that can be added or some type of electrolyte
• 07:30 - 08:00 that will be added to the solution that will not be able to undergo one of those processes so an example for the electrolyte that could be used would be the na2so4 when you think about those two ions i have the na2plus and i have the so42 minus the na2plus or the sorry the na plus that could theoretically add an electron and go to sodium as a
• 08:00 - 08:30 metal however again we would know that this would not occur because the amount of electricity that would be needed to reduce it will be greater than the amount that is required to reduce the water and so that reaction would not take place hopefully that makes sense as well because if i had solid n a forming we know how reactive that is with water and we would have an explosion with the sodium reacting with the water to form the
• 08:30 - 09:00 sodium hydroxide so we know that that would not be able to take place and then the same type of thing the so4 2 minus is already at its maximum oxidation because if you think about sulfur in the sulfate ion it's already a plus 6 for its oxidation number and so it's run out of valence electrons that it could theoretically give away so you always want to find some type of salt or ionic compound acid that could be placed
• 09:00 - 09:30 into that solution that would allow the electricity to flow through easily but not actually have a chance of reacting in the mixture we will now take a look at the tin copper reaction and compare the process as a voltaic cell compared to an electrolytic cell so you'll notice in this first diagram on the left we have the voltaic cell that is present
• 09:30 - 10:00 and inside of that voltaic cell for our basic sketch of the diagram we have the tin solid for the anode that is undergoing oxidation and it will produce the sn2 plus ions into the solution notice initially we started off with a 1 molar concentration but as the reaction progresses you will see an increase in the concentration of the sn2 plus so again think about the nernst equation
• 10:00 - 10:30 and how as that concentration would increase it would cause the e cell of the compartment to begin to lower and that's because as it increases and the copper will decrease you will see a overall lowering of the e cell value until it reaches equilibrium at which point it would be dead for the reduction we have the copper two plus ion reacting with two electrons to form
• 10:30 - 11:00 copper as a solid so remember the cathode is going to increase its in its mass as the reaction progresses and you can see here that again the concentration of the copper two plus ion in the solution would decrease now the overall reaction that we would see for this thermodynamically favored process where we would have 0.48 volts of energy produced we would have sn as a solid plus cu2 plus aqueous forms sn2
• 11:00 - 11:30 plus aqueous and cu as a solid now let's shift gears and focus on the electrolytic cell when we look at the electrolytic cell again we have the process being reversed notice that it says that we have an external source so that would be my battery or direct current that is being applied electron pump however you want to determine but that is going to be supplied right here and so the electrons still
• 11:30 - 12:00 flow in that fat cat format from the anode to the cathode but it's just a little bit different and i wanted to take a moment to just kind of explain why we see the polarities of the electrodes become the opposite of what they are for the voltaic cell so in the initial galvanic cell or voltaic cell you have a buildup of electrons that are occurring on the anode which makes them have that positive value and you notice the direction of the electron flow still from the anode to
• 12:00 - 12:30 the cathode but you'll notice that since the poles are being reversed the flow of the electrons changes direction so i still have the tin being the receiver of the negative charge because it has the excess of the electrons that are going into that electrode initially so when we look at this here that external power source again supplying the electrons with a voltage greater than 0.48 0.48 would be
• 12:30 - 13:00 that minimum threshold that would be required to make the reaction go in the opposite process at 0.48 volts hopefully you would remember the delta g for that process would be zero and we would have a equilibrium type of situation so once we put it greater than .48 volts now we're forcing the electrons to go back into the opposite direction so as i have this buildup of the electrons
• 13:00 - 13:30 into the wire now it's going to take the sn2 plus ions out of the solution as they come in contact with those electrons and they will now start to plate onto the tin electrode at the same time you're going to have the copper that is going to begin to go into the solution so now the copper ions will decrease their concentration while the tin
• 13:30 - 14:00 increase and as that process takes place the electrons will then flow through the copper anode and will go back towards the power source so you still have that flow of the electrons from the anode into the cathode but again we're flipping the signs on those electrodes so we look at the different half reactions for that electrolytic cell the oxidation half reaction that's going
• 14:00 - 14:30 to be the copper being oxidized into the cu2 plus and the two electrons the reduction half reaction will have the sn2 plus plus two electrons forming sn as a solid and that will give us an overall reaction of the sn2 plus plus the cu forming sn as a solid and the cu2 plus ions so you notice it's the exact opposite process that we had from the
• 14:30 - 15:00 voltaic cell for that particular reaction we're next going to look at a multi-billion dollar process that occurs in industry all the time and that is the electrolysis of pure molten salts this is a way to separate the components of your salts typically your very active metals which are very hard to get alone by themselves if your salt is pure predicting the products is a very straightforward process the cation is
• 15:00 - 15:30 going to be reduced because we're going to force the electrons back into the cation and the anion is going to be oxidized the electrolyte is the molten salt itself and the ions are attracted by the oppositely charged electrodes that are placed into the liquid if we consider the electrolysis of molten calcium chloride you can see here that the chloride ions as a liquid will go to chlorine gas plus two
• 15:30 - 16:00 electrons so you see that the anion is undergoing oxidation and the calcium liquid the calcium ions as a liquid plus two electrons form calcium as a solid so you have the uh cathode undergoing the reduction and that gives you an overall process of the ca2 plus as a liquid plus the 2cl minus as a liquid forming ca as a solid and cl2 gas for the overall reaction so this is one
• 16:00 - 16:30 of the most important types of processes that we have to prepare your very active metals such as calcium magnesium sodium and a lot of your halogens chlorine and bromine as well so now we're going to look at a mixture of molten salts so in this type of situation we will have the electrolyte as a mixture of molten salts that are being electrolyzed to obtain one of the metals how we can tell which
• 16:30 - 17:00 of these species will react at which electrode it's pretty simple the general rule for the electrolytic cell is the more easily reduced species the stronger the oxidizing agent that will react at the cathode so when i look at my different values for the reduction potentials i'm going to choose the one that is the more positive value because again that is going to allow the substance to be more easily produced when i look at the more easily
• 17:00 - 17:30 oxidized species that is going to be the one that's going to react at the anode now unfortunately we can't use the standard e cell values to tell the relative strengths of the oxidizing and the reducing agents and that's because those are referring to the reactions taking place in aqueous solutions so therefore it's the water that is causing the ions to become
• 17:30 - 18:00 free from the metal compound and so we would have in that situation under the standard state conditions you'd have a certain amount of value that would be present to bring back that aqueous ion into the solid however we can look at periodic trends for the prediction of which ions will gain or lose the electrons more easily so in the situation where we do not have the
• 18:00 - 18:30 aqueous ions taking place when we're looking at the metals we want to look at their ionization energies and whichever electron is going to be held more tightly that is going to have the higher ionization energy and if it holds onto the electrons more tightly then it will be more easily reduced and it will form the solid much more readily if you have a very low ionization energy
• 18:30 - 19:00 remember that means that it does not require much energy at all to remove the electron from the atom and so it would very easily undergo the oxidation process so we want to look for high ionization energies for the metals and then if we have the anion the non-metal we want to have its electrons being held less tightly by the atom so when we're looking at the non-metals we want to focus on the
• 19:00 - 19:30 electronegativity values and remember that's a measure of the attraction force between two electrons that are being shared between two atoms the higher the electronegativity value is for a particular non-metal the more strongly it's going to hold on to those electrons and so it will not be able to oxidize as easily so we want to focus on the lower electronegativity value for the anions because again they will be able to release that
• 19:30 - 20:00 electron more readily from the atom so if we just do a quick summary of what product is going to form at the different electrodes experiments have shown that elements that can be prepared electrolytically from aqueous solutions of their salts the cations of the less active metals including gold silver copper chromium platinum and cadmium are reduced to the metal cations of more active metals including those in group 1a and 2a and aluminum from group 3a are not
• 20:00 - 20:30 reduced the water would get reduced to h2 and oh minus instead and that's because their values the amount of energy that would have to be added into the system in order to get the different ions to go into their solid form would require more energy being added into the system compared to how much it would take to reduce the water so that's why they will not undergo that reduction process the
• 20:30 - 21:00 anions that are oxidized we have a concept that we call over voltage which is going to i'll spend some time in class going over that concept but essentially you can just kind of think about the over voltage occurring when you have gases involved in your reaction and basically it's going to require some energy to convert the aqueous ions
• 21:00 - 21:30 into gases and so that extra energy is related to the extra energy is related to the activation energy and so that means that you have a very high activation energy for that particular process to take place and so it just happens that a lot of
• 21:30 - 22:00 times some of those other ions can undergo the oxidation process and the anions can undergo the process more easily than the oxygen can form from the water and so typically the over voltage you're adding an additional 0.4 to 0.6 volts of energy into the system that's required to electrolyze the oxygen and so something
• 22:00 - 22:30 like chlorine can then just sneak in underneath that threshold value for the oxidation and will allow the substance to undergo the oxidation instead of the oxygen the one exception to that is the fluorine and again that gets into some of our periodic trends atomic size all that kind of good stuff so the anions that are not going to be
• 22:30 - 23:00 oxidized includes f2 and then your common oxo anions such as the so42 minus the co32 minus no3 one minus mpo4 three minus and as i had mentioned previously when we're talking about the water electrolyzing that's due to the fact that the central atom for those different polyatomic ions already has its highest oxidation state so if it were to lose any more electrons it would be coming from the core
• 23:00 - 23:30 electrons in one of those inner shells and so therefore that would require a tremendous amount of energy much more than it would take to have the water undergo that oxidation process instead so we're now going to get into faraday's law just use dimensional analysis and you'll be fine but we're going to use this to look at the plating of metals onto some type of electrode the amount of the substance that's being oxidized or reduced at each electrode during
• 23:30 - 24:00 electrolysis is directly proportional to the amount of electricity that's passing through the cell so we know by definition that one volt is equal to one joule per coulomb and one amp is one coulomb per second we measure current in amps but it's symbolized as an i so that equation that i was talking about the very beginning that's where the i is coming into play we have the faraday's constant that is 96 500 or 485
• 24:00 - 24:30 coulombs per mole of electrons so the basic process that you want to go through to solve these problems is you want to figure out your quantity of the charge that's your coulombs that's also where again you need to have your end value from your balanced half reaction so or your balanced redox reaction so we'll put that
• 24:30 - 25:00 here so we have that from our balanced redox reaction okay so again we use faraday's constant to convert that into moles so again taking your current times time that will get you your quantity of charge in coulombs and then you can use the faraday's constant to get that into moles of the electrons you can use the n from your balanced half reaction to figure out how many moles of the substance were oxidized or
• 25:00 - 25:30 reduced and then from there you can use your molar mass to determine the grams of the substance that it will be oxidized or reduced during that certain amount of time let's now look at an example problem for faraday's law by calculating the number of grams of aluminum producing one hour by the electrolysis of molten aluminum chloride if the electrical current is 10 amps a couple things within this problem notice that it is molten aluminum
• 25:30 - 26:00 chloride so it is not aqueous it is going to be just the melted aluminum chloride so the ions themselves would be the electrolytes i have 10 amps involved in this process and remember that that would be the equivalent of saying 10 coulombs per second and we have hours that are given to us for the time so use dimensional analysis for these problems and it will be very easy to solve the other thing that we need to do just
• 26:00 - 26:30 as a starting point is get the reduction equation for the aluminum so we know the number of electrons so i know that i have aluminum as a three plus charge going to have three electrons added to it and that will produce aluminum as the solid for the electroplating process so i now have my value for n equaling 3 so we're now ready to start so i can use the 10 coulombs per second
• 26:30 - 27:00 as a conversion factor so the first thing i want to do is start with my time so i would have one hour for the time and i know that for every one hour we'll just simplify it and say that there's 3600 seconds instead of doing the 60 minutes and 60 seconds in a minute so we can now look here at our conversion factor so i would have the one second uses 10 coulombs of charge and now at this point if you follow
• 27:00 - 27:30 along the little concept map progression i can now use the faraday's constant 96485 or 96 500 depending on what they give you on the ap exam but that is the number of coulombs for one mole of electrons and now at this point we can use this and we can say that we have three moles of electrons that are consumed for the production of one mole of aluminum and
• 27:30 - 28:00 to solve for grams the molar mass of the aluminum will be 26.98 for our grams and so now at this point we are getting approximately 3.35 grams of aluminum
• 28:00 - 28:30 all right so now let's talk about the corrosion of metals so we're going to look at the corrosion of iron as an example so first off when we're talking about the corrosion of a metal it is an electrochemical process and basically what it's doing is it is involving oxidation to return the metals to their natural state which is one of their ores so or minerals so when they are involving that process they're going to lose their structural integrity
• 28:30 - 29:00 they're going to lose their nice attractive look for the metals and they're going to look like they are about to fall apart in many cases so we're going to deal with iron as it's one of the most common ones that we see you think about if you have ever lived up north you'll know this in the wintertime again they're always putting salt on the roads to help prevent the freezing up of the water on the
• 29:00 - 29:30 roads and so a lot of times what you'll find is that the cars that are up north are going to have a layer of corrosion along the bottom of the doors where you'll start to see the rusting of the doors on the edges because of all the salt that is present inside of the water that is being splashed up onto the bottom of the car and so when you have the corrosion that is taking place basically the steel surface is not
• 29:30 - 30:00 completely smooth it's got uh not homogeneous because it's an alloy so it's going to have different regions that have some stress points within the metal surface and that's going to cause the iron to be more easily oxidized at some point which we would talk about that being the anodic region and then it has others where the ore will start to build up where it will form the rust
• 30:00 - 30:30 at the cathodic region of the space so when we are looking at these different scenarios here what you will find is that the iron as it gets dissolved in the solution will migrate over towards where the rust is and or is starting to form and it will deposit it as the iron iii oxide on the edge of the metal surface so when we are looking at the region
• 30:30 - 31:00 here you can see this cathodic region where we have the oxygen undergoing the reduction process to form the o2 minus ion and then you have the fe that is in this little pit or the little dent that has been caused uh on the surface of the metal you're going to find that that will again like your anodes you would see that the mass is starting to eat away so you would
• 31:00 - 31:30 have that region of space where the iron is going into the solution reacting with the oxygen that's present and forming the rust on the outside edge of the water and so you end up having that weakened spot where the corrosion is taking place so the last thing that we're going to discuss is the use of sacrificial anodes to prevent iron corrosion
• 31:30 - 32:00 so basically what that means when we have a sacrificial anode is they're going to use a more active metal and that more active metal will undergo the oxidation process instead of the iron so it's easier for the other substance to undergo the oxidation and so think about your [Music] electro chemical potentials it is easier for those substances to undergo that process so they will
• 32:00 - 32:30 undergo that process instead of the metal so it's used as a way to protect the iron inside of the ground in this case for it to not undergo the oxidation process and it will keep its integrity for a longer period of time the magnesium rod that they're using in this case for the pipes the magnesium has a much much larger oxidation potential or much lower more
• 32:30 - 33:00 negative reduction potential than the iron has and so it will be much more readily undergoing oxidation as a result you've seen this before if you've heard the term galvanized for nails they coat the nails in zinc so that way the zinc on the outside will undergo the oxidation process and it will protect the iron that's on the inside of the nail so
• 33:00 - 33:30 make it a much longer lasting nail so that way they can hold up whatever structures it's being used for uh for a much longer period of time and then finally you also uh if you've ever been on a cruise or on a large ship before whether it's for a cruise or if it's like a cargo ship or something like that you would find that underneath the hole you would also have these huge
• 33:30 - 34:00 large rods of magnesium that would be down inside because if you've ever again lived up north and know the effect of salt water on the corrosion of the metals by having the uh magnesium bars tied into the hole of the ship the outside of the ship that will protect the outside of the ship the iron from undergoing the oxidation and it will have the metal bars on the inside that would get eaten away through time
• 34:00 - 34:30 from that electrochemical process and again protecting the integrity of the ship making sure that there is no corrosion on the outside of it and providing weak points where the water could then rush in and sink the ship so that is your basis for your sacrificial anodes is that it allows the metal for the casing of whatever we're looking at it allows it to be protected for a much longer period of time
• 34:30 - 35:00 so that its integrity stands so that wraps up this last video uh covering the electrolytic cells uh we'll spend some more time in class doing some practice problems and i'll take any questions that you guys may have so uh you guys need to just read over the little section on batteries in your textbook for the different types of batteries uh we'll talk about the the down cell and
• 35:00 - 35:30 the halt or the hall rote process for the refining of aluminum from its ores and we'll talk about that in class tomorrow but other than that have a wonderful evening afternoon whenever you're watching it and i will see you guys tomorrow