Chemical Science Oxidation Reduction - Lecture 25

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Principles of Chemical Science/nVideo Lectures - Lecture 25/nTopics covered: /nOxidation/Reduction/nInstructor: /nProf. Catherine Drennan/nTranscript - Lecture 25/nAll right./nToday we are continuing in oxidation-reduction. And so we started this a little bit last week. And we talked briefly about balancing equations which, as you recall, is really about counting. It is not an intellectually challenging topic and just requires making sure that you counted everything appropriately./nYou have to be a little bit careful doing it, but it is really not too hard. Being able to balance oxidation-reduction equations then allows you to go on to talk about the things that we will be talking about today./nToday we are going to cover electrochemical cells, Faraday's law and the relationship between delta G and cell potential. If everyone could quiet down a little bit, I can hear a lot of noise in the background./nI realize you are still settling down and pulling out your notes. As I have mentioned all the way through the lectures I have given so far, you have to remember all the things you have learned because they are going to come back again./nToday we will be back to thermodynamics and talking about spontaneous reactions in delta G. This all ties in together. These units are really more applications of things that you have already learned in the first half of the course./nAnd currently we are in Chapter 12. Electrochemical cell. What is an electrochemical cell? Well, it is any device in which you have an electric current. An electric current is just the flow of electrons through a circuit./nAnd the current is either produced by a spontaneous reaction. See where delta G is going to come into this? It is either produced by spontaneous reaction or used to bring about a non-spontaneous reaction./nYou can use electric chemical cell to force a non-spontaneous reaction to go. A battery, which most of you are familiar with and use in many of the gadgets that you have, is really just a collection of these electric chemical cells./nAnd, as we will see perhaps at the end of today's lecture or next time, all these principles that you are learning about oxidation-reduction apply to oxidation-reduction reactions that happen in your body./nThe same principles that apply for a battery apply for chemical reactions that are in your body that are important for human health. We will talk about that as well. Here is an example of an electrochemical cell./nThis is a pretty simple electrochemical cell. We have two beakers here. And we have electrodes, which are connected by wires, and they are connected to something that can read the current that is coming through the system./nThere is also a salt bridge, which we will talk about in a few minutes. This would be a fairly simple one that one could construct. And I like to joke about what people would say in the old days when I go and do fundraising stuff for MIT and meet alumni that graduated from this place a really long time ago./nThey probably say, well, before we could even do our problem sets, we had to make our own batteries to be able to get some electricity to see by. Your life is a bit simpler now. You should hear some of the old stories of what MIT students had to go through./nAll right. This is a schematic view of what you just saw. And we will walk through and talk about all the different parts of the electrochemical cell. This I drew actually myself. I think actually I stole maybe parts of it, or did I draw the whole thing? I cannot quite remember./nBut, again, this is my depiction of the beaker. As I have mentioned in the past, art was really not my strength. This is a beaker over here filled with liquid. In this case, we have ions dissolved in the solution./nWe have zinc plus two and sulfate dissolved. Another beaker over here. In this beaker we have a zinc electrode, and this is at the anode. And so this is a piece of zinc solid stuck into this beaker where you have this solution./nThere is a wire connecting two electrodes. On the other side we have the cathode, and this is made of copper solid. And it is sitting in a beaker that has copper plus two and also sulfate in it. And so let's look at what would happen if this reaction where to run./nOver here at the anode we have zinc solid being oxidized to zinc plus two, so the reaction is shown down here. This electrode actually gets consumed during the reaction. You actually have solid being converted to zinc plus two in solution, so you have a change in phase./nAnd as that happens, two electrons are generated from that reaction. And the electrons then will go through the wire and the current will be red. They will go through the wire over to the copper solid./nAnd over here they will pass into this beaker and the electrons will find the copper plus two, as shown in this equation down here, and this will form copper solid. You are actually going to have extra copper plating onto the electrode./nThis electrode is going to get bigger. And this one will be consumed in the course of the reaction. And so this is a little picture over here of electrons coming into the copper solid, finding the copper plus two, and the copper plus two then is reduced at the cathode to copper solid./nAnd the blowup on the other side shows the copper solid, once electrons leave, say the zinc solid, electrons leave, you form zinc plus two. These are sort of blowups of the reactions that are happening at the cathode and anode./nAnd so as this happens, you will see there will be a change in the charge. Here we have a zinc solid or zinc zero going to zinc plus two. To balance that charge there is a salt bridge, and you will have some kind of ion come in to neutralize the charge./nAnd here we have chloride minus. On the other side, we are going from copper plus two to copper zero. And so you will have a positively charged ion coming in to help neutralize that change in charge./nOver here something is being oxidized at the anode. Anode is where you have oxidation. And something is being reduced at the cathode. Cathode is where you have reduction. We could write that cell in a little bit more simple fashion./nInstead of having that picture, we can convey to others what that picture showed us by this equation here. We have the two reactions going on. We have the zinc reaction, which is shown on one side./nAnd that reaction was the zinc solid going to the zinc plus two. And this single line here represents a change in phase, a phase boundary. Solid going to an ion dissolved in aqueous solution. The two lines right here represent the salt bridge./nThis equation then is telling you one reaction going on in one beaker separated by the salt bridge. This would be the reaction going on in the other beaker. And so over here we have copper plus two aqueous with a single line for the phase boundary going to copper solid./nIf you see this kind of notation, you can envision that electric chemical cell with the two beakers, the wire and the salt bridge. And this is a reaction going on on one side and the reaction going on on the other side./nAs I mentioned, in that particular electric chemical cell, zinc is being consumed. Zinc solid is being consumed. It is being converted to zinc plus two. Whereas, copper solid is being deposited onto the electrode./nIn that case, you have copper plus two aqueous going to copper solid. That is being deposited. And the amount of zinc that is being consumed and the amount of copper that is deposited is going to be proportional to the charge that passes through the system or to the electrons that pass through the system./nAnd that makes sense, because if you have a zinc solid going to zinc plus two, you have two electrons formed which would then go through to the other side. The amount of charge that is going to go through has to do with the amount of zinc that is consumed, which would also then have to do with the amount of copper that is deposited when the electrons get to the other side./nAnd so this is a law. We don't have very many laws in chemistry. A lot of times we have theories that help to explain observed information. Chemistry really is an experimental science, so we observe a lot of things and then we try to rationalize why they are true./nThere are not a whole lot of laws, but this is one of them. All right. Let's just kind of look at little movies of this procedure of what is happening. And here we are going to see oxidation at the anode./nWe can pretend this is our solid electrode here. And we have, say, the zinc. And here is the aqueous part, so these are water molecules. And you are going to see electrons leave, go up. And then you are going to have the solid be oxidized to, in this case, zinc plus two, and it will go into solution./nLet me just play this for you. You can actually see the electrons leave in this movie, which is interesting. There they go. There are the electrons leaving. You've freed up some of the solid and now you have freed up again./nNow we have zinc plus two. We had zinc solid, but now we have consumed some of our electrode and it is in solution. Now you know what you are looking for, I will just play that one more time. The electrons are heading off into the wires and you are consuming your electrode./nAll right. Let's go the other direction. The electrons that left and passed through the wire are now coming down to the cathode. And so those electrons can come in and reduce, say, copper plus two forming copper solid./nWe have copper plus two. Here come the electrons, here come the electrons, and so now you plated those two copper plus two ions. They have become copper solid. And now your cathode has expanded. You have added more metal onto it./nIt has been deposited or plated onto this electrode. And, again, the amounts are going to be equal. Faraday's law would tell you that it is all going to be proportional, the amount that is consumed and the amount that is produced is all going to be equal to the number of electrons that pass through the system./nLet me just show you one more example of why this could be useful in terms of the plating. In this little experiment, we have a steel spoon. The steel spoon will act like an electrode and get plated with copper./nThere is the copper wire. That is pretty simple. You can see how it can get deposited on top of this, which was acting as an electrode in this case. All right. As you probably have figured out, yeah? You often will have to figure that kind of thing out./nIn the example that I showed you, it was set up to work like that. But, as we will talk about, it depends if it is going to be a spontaneous reaction or a non-spontaneous reaction. If you have a cell where it is a spontaneous reaction, you can figure out which reaction will occur at the cathode and which will occur at the anode./nThere is only way you can put those if you have particular ingredients for it to be spontaneous. And so we will actually talk about that at the end of the lecture. In that particular case it was just stated as this is the reaction happening here and here, but which will go which place will depend on whether you want it to be spontaneous or non-spontaneous./nThat is a good question and we will definitely come to that. In terms of Faraday's law, it is proportional. And that means you can do calculations. One could figure out how much zinc would get consumed or how much copper would get deposited if you add a particular current flowing for a particular amount of time./nAnd these are actually fairly simple calculations to do, and we will just go through one of them. The first step would be to figure out how much charge went through the system. We have Q, magnitude of charge, and Coulombs which unit is C./nIt is going to be equal to the current, and that is amps or A. And here is a useful conversion for you to remember, and this will be given to you on a test. Times the amount of time. How much current times time will tell you the magnitude of the charge./nThat is pretty simple, and that equation would be given to you as well. We are told that we had one amp for one hour, and we have converted it to seconds so you can get the right units. And we can find out the magnitude of the charge./nThen in Coulombs because these units multiplied together with get you Coulombs. Then step two is to figure out how many moles of electrons that charge is equal to. And we have Faraday's constant, which will also be provided to you./nAnd its units, using the conversion, Coulomb per mole. And it also is equal to one Faraday, which is usually kind of a squiggly kind of F symbol here. And so then, if we go through and we use Faraday's constant, we can figure out that that number of Coulombs convert two moles./nAnd so you would get 0.0373 moles of electrons. This is just a conversion using Faraday's constant. And then we have to figure out how many moles of the particular element, zinc in this case or copper, are deposited and convert to grams./nWe know now how many electrons went through our system. And so now we just need to figure out for every one mole of zinc consumed how many moles of electrons would have left. And, in a particular example we had, what would it be? How many electrons? Two, because we went from zinc to zinc plus two./nWe are talking about two electrons for every mole of zinc. And then we can calculate moles and convert to grams using the molecular mass of zinc, which we can look up. We get 1.2 grams that are going to be consumed at that particular time with that kind of current passing through the system./nWe can do the same thing with copper. We have the same number of moles of electrons that are going through out system. Now we just need to know for every one mole of copper deposited, how many electrons are required? And that is two again, because we were taking copper plus two to copper solid./nWe need two electrons for every one mole of copper that gets deposited. And then we can use the molecular mass to find out the number of grams. And since these weigh a very similar amount, it is about the same number of grams./nIf there was a very difference in molecular mass then that wouldn't be true. These are not always near each other. All right. That is just how you do one of these problems. They are really pretty simple./nAnd there are some of those at the end, actually, of the problem set. The book has things in a sort of different order, so you will be able to do the first part of the problem set and the end of the problem set at this point./nAll right. Let me just introduce you to a couple of different kinds of electrochemical cells. Up until this point, we have only talked about electrochemical cells where something was deposited or consumed./nBut you can use electrodes that are inert, i.e., they don't get consumed in a reaction. An example of that is platinum electrode. As it says, they are not always consumed or produced. You can have an inert electrode such as a platinum./nIn this case, we have the platinum electrode to transfer the electrons around, but the reaction that is going on is all in solution. This side would be the same. This is the cathode or the reduction reaction./nWe have, again, the copper plus two. Two electrons will come through the system and plate more copper solid onto the electrode here. But on the other side the anode, where you have the oxidation, the oxidation reaction is happening all in solution so the platinum electrode is not involved in that./nIn this case, we have chromium plus two going to chromium plus three with one electron. That is another type of electric chemical cell. And so we can write this and look at the notation for this kind of cell./nIn this case, you put the platinum solid in the equation on this side. We have a single line meaning there is a phase boundary, so we are going from solid to something in solution. And now here is our reaction that happens at the anode./nAnd it has a comma in between. This indicates that chromium plus two is going to chromium plus three, but there is no phase boundary between those so there is a comma. Then on the other side the two lines indicating the salt bridge./nAnd then on the other side we have the copper plus two aqueous, the line that does indicate there is a phase boundary going to copper. This would be for our anode reaction of chromium going to chromium plus three in one electrode, and our cathode reaction, copper plus two plus two electrons going to copper solid./nYou will see this kind of notation in these problems as well. All right. Another example of an inert electrode that sometime is used, the hydrogen electrode. Often you will see this. Actually, in the back of your book you will see this as well./nThe standard hydrogen electrode abbreviated SHE. Often you will see in papers or in your book that they often need to mention how these redox potentials were measure. And it is often said it is measured against SHE./nAnd that is what that stands for, the standard hydrogen electrode. You can have the hydrogen electrode on either side of the anode or at the cathode. Over here would be an example when it is at the cathode./nAnd, in this case, H+ is being reduced. And so we have H+. And that is aqueous. The line for the phase boundary to H2 gas. And then you have also a phase boundary to the platinum solid, so this is combining the platinum and the hydrogen./nThen, on the other side, if you had it reacting as an anode, in this case H2 is getting oxidized. You have the platinum solid phase boundary, a hydrogen gas, so that would be oxidation state zero, phase boundary to H+ aqueous./nYou can use this sort of combination of the platinum and the hydrogen electrodes as either cathodes or anodes. And so just one example of what this sort of thing would look like. Here we have our electrochemical cell./nNow on this side we just have the zinc electrode, which we mentioned in the beginning, a zinc solid. And then that would be zinc plus two in solution. We have our wire, our salt bridge. And in the other beak, on the other side, we have some glass tubing where you would have the hydrogen gas going through into the system./nYou also have through that the platinum electrode, platinum solid. And you have HCl in your beaker, so that would be H+ in your beaker. Here is the cathode reaction, so you would be going from H+ to H zero gas./nAnd so as you have the reaction over here, zinc solid to zinc plus two, electrons would come along and would convert H+ to H2. And the electrode here is the inert platinum electrode. Here is my attempt to draw little H2 gas bubbles going on./nThis is another type of electrode that you should know the notation for. It doesn't really change how you do problems particularly. But it is important to know what it stands for. Again, this is on the other side, we have already talked about that kind of anode./nAll right. This is, of course, always things that I really like to do, because what I like to do is go back to something you have already learned and make connections between different units of freshman chemistry./nWe are back to delta G. We are going to now talk about cell potential, and this is abbreviated delta E. Cell potential is often called cell voltage and is also called the electron motive force or EMF./nYou will see all of those different words, and they all stand for the same thing, this delta E term, and the connection with free energy. When the electrons flow that creates a potential difference, delta E between the electrodes in the circuit./nAnd so the overall free energy of the cell is related to this cell potential by the following equation. We have delta G for that particular electrochemical cell. It is going to be equal to minus N, which is the number of electrons that are going through the system, times Faraday's constant, which we talked about a few minutes ago, and delta E, the cell potential for that particular cell./nThere is a nice connection between these two, and that will be really useful in doing these problems. We can also talk about standard states here. And the same equation applies whether you are talking about delta G at any particular time and delta E at any particular time or you are talking about the delta G and the delta E in their standard states./nAgain, this means the same thing that it has before. Delta E knot, the knot means that you are at the standard states. And delta G knot is also delta G in the standard state. And units, of course, are important./nWhen you are talking about any kind of delta E or cell potential, you are talking about things in volts as the units. All right. Now what those equations mean is if you calculate a delta E for the cell, the cell voltage or the cell potential, you can figure out whether the reaction will be spontaneous or not because you can use that equation and look at, with that particular delta E, will delta G be negative? Will it be a spontaneous reaction? The first thing we need to learn how to do is how to calculate the cell potentials or the electromotive force, and then we already know how to convert that to delta G./nLet's look at this equation here and calculate what the delta E, the cell potential for this particular electrochemical cell that we have been talking about is. The reaction at the anode, which we mentioned is the zinc reaction./nAnd this is an oxidation. The reaction at the cathode is the reaction with copper, copper plus two and two electrons going to copper solid. And that was our reduction reaction. And so now we can just use a simple equation to calculate delta E knot if we are using the standard reduction potential for the reaction at the cathode minus the standard reduction potential for the couple at the anode./nAnd the point that I want to make here is that in this fairly simple equation, this term, this just E knot here is the standard reduction potential. You don't want to put in something other than that./nDon't mess around with changing signs of reactions. All you want to do is enter a reduction potential here and a reduction potential here. And then this equation will always work for you. And you will see what I mean, I think, a bit more when we put that in./nBut if it is the reduction potential you would be looking up copper two plus to copper solid. Reduction potential you are looking up zinc plus two to zinc solid. You are not caring about which of these reactions is an oxidation and which is a reduction at this point./nYou have figured it out by which is at the cathode. If you know what is at the cathode and you know what is at the anode then you enter in the reduction potential. Let's do that. And you can look up the standard reduction potentials in your book./nThere is a table in Chapter 12 and there is also a table at the end of the book in Appendix 2. You can look up the values. Here they are written as reduction potentials, and that is what you will see in the table./nThat is the information you can look up. And then you have your equation. And you will enter in the reduction potential for the reaction at the cathode and enter in the reduction potential for the reaction at the anode./nHere you will get out a positive value, 1.103 volts. Again, this is the copper reaction and this is the zinc reaction. The reason why I am sort of trying to jump up and down about the fact that you should enter in reduction potentials is because if you play around and all of a sudden switch the signs then you are going to come out with the wrong answer./nAnd one of the important points about this answer here is, is delta E positive or negative? Because that is going to tell you if the reaction is spontaneous or not. If you do cleaver things and switch the signs for these around you will always come out with the wrong answer there, which then affects all the problems that you do after that./nAs long as you use this equation and always enter in a reduction potential you will be fine and you won't have any of these problems. All right. Is the flow of electrons spontaneous for this particular cell? People are saying yes, and that is the right answer./nAnd the way that we know that is because we remember this equation comparing delta E and delta G. Here if this one is positive then this term will be negative because this is number of moles, which is always positive./nAnd Faraday's constant is always positive. If we have a positive value here then the net sum is negative, this will be negative and that will make it spontaneous. If delta E is positive, delta G will be negative./nAnd this is true, again, if it is delta E knot, a delta G knot, same equation. If delta G is negative, will it be spontaneous? And the answer is yes, back to previous units. In some of the problems in your book that you will work on, they will tell you that this is a spontaneous electrochemical cell./nAnd then you can figure out which reaction had to be at the anode or which one had to be at the cathode for that through calculating the delta E and figuring out if it is positive or negative. Therefore, delta G must be positive or negative, spontaneous or non-spontaneous./nAll right. What do you call it if there is a spontaneous reaction? Here are some terms that you should know. You have a Galvanic cell. And that is an electrochemical cell in which the reaction is spontaneous./nIt can generate an electric current. And your book uses this and will say this is a Galvanic cell. Already then you should know something about that reaction. The other option is an electrolytic cell./nIn this case, energy has to be provided to the circuit to carry out a non-spontaneous reaction. Sometimes you might just want that non-spontaneous reaction to go. And then you will have to supply some form of energy to get that reaction to work./nAnd in the body there are both of these kinds of reactions. Sometimes you would have an oxidation-reduction reaction that is spontaneous, and that could be used to produce some kind of energy. It would be coupled to an energy generating process in the body./nOther times the body needs a reaction that is unfavorable to go. And it has to also supply energy to get that reaction to go. And we will talk about biological examples of these as well. All right./nJust the brief summary for this part. If the cell operates spontaneously that can be determined by the delta E. A positive value means a negative value, so that will be spontaneous. And you can get your delta Es from standard reduction potentials for the half reactions involved./nAll right. Now we are going to think about what it means if we have certain standard reduction potentials. If you have a standard reduction potential that is a large positive number, or if you have a standard reduction potential that is a large negative number, or if you are somewhere in between./nWhat does that tell you about the properties of those particular elements? A large positive standard reduction potential means that the element is easy to reduce. Let's look at an example for this./nAn example, we have F2 gas plus two electrons going to 2F-. This has a standard reduction potential of plus 2.87 volts, and that is a large positive number. What that is going to mean is that this is favorable in the direction it is written./nAnd favorable means it is easy to add electrons. This wants to be reduced to F-. It is happy that way. If it has a large positive sign for the standard reduction potential then delta G will be negative which will be favorable or spontaneous in that direction./nThis is a favorable reaction. F2 is easy to reduce. It wants to be reduced. Is it a good oxidizing agent? Yes. Again, a good oxidizing agent oxidizes other things and is reduced itself. It runs around and tries to oxidize other stuff, which means that it gets reduced, which is what it wants to be./nIt wants to be reduced. This is a good oxidizing agent. If we look at the couple here, F2 / F-, this reduction couple with a large positive value, what we can say is that the oxidized species, that is this one here, the F2, is very oxidizing./nIt is a good oxidizing agent. And this is a table from your book so it is not in your handout, but you can look it up. Here is the page number. And this is the one we just looked at, F2 to F-. It says right on the top oxidized form, which is F2, is strongly oxidizing./nIt has a big positive number. It is at the very top of the table. A big positive number. It likes to be reduced, i.e., it likes to oxidize other things. Now let's go to the bottom of the table and see what we can say about this lithium plus lithium solid couple./nThat had a large negative value for the standard reduction potential, and that means it is hard to reduce. Let's look at that. We have lithium plus one, plus one electron going to lithium solid. And that reduction is listed with a standard reduction potential of minus 3.045 volts./nThat is a large negative number. If it is a large negative number that means it is hard to add electrons. This direction, as written, is not particular favorable. Lithium plus one does not want to go to lithium solid./nIt is hard to add that electron. That direction of the reaction, the forward direction would not be very favorable. Delta G would be positive. It would be a large positive number so it does not want to go in that direction./nIs lithium plus one a good oxidizing agent? No. But it is a good reducing agent. Or, lithium solid is a good reducing agent. Lithium solid likes to reduce other elements. It likes to get oxidized./nThis reaction is favorable in the reverse direction. The lithium solid over here would be very happy to be lithium plus one. Lithium solid is a good reducing agent. If we look again at this couple, the lithium plus one lithium couple with the large negative standard reduction potential, we find that the reduced species here is lithium solid./nThe reduced species is very reducing so it wants to be oxidized itself. It wants to reduce other things. It is a good reducing agent. If we go back to our table down here it says reduced form is strongly reducing./nUp here, on the far corner on the left-hand side, the oxidized form is strongly oxidizing. Down here, on the right-hand side, the reduced form is strongly reducing. You will be asked about various things in between./nIs something a better oxidizing agent than something else? Is something a better reducing agent than something else? And if you remember these parts, oxidized form up here where standard reduction potential E knot is positive, oxidized form is strongly oxidizing./nDown here, where you have a negative number, reduced form is strongly reducing. This makes sense from what you have learned before, actually. And this is another table from your book so I didn't put it in the handout./nWe talked about fluorine over here. We talked about lithium. This has a big positive number over here so it is easy to reduce, these are good oxidizing agents. And on the other side these want to be oxidized./nThey are good reducing agents. Fluorine minus has a noble gas configuration and lithium plus has a noble gas configuration, so it makes sense from some of the trends you talked about earlier in terms of whether something is favorable or not./nWhether it likes to be in the plus on state or prefers to be in the minus one state can all come out of these reduction potential numbers. And you can figure out which direction is favorable, the reduction or the oxidation./nLet's take a look at an example here of this cell potential and think about what are good oxidizing agents and reduction agents. There are these two couples over here we are going to look up. And that is hard to see so I will just go back to this./nWhat is happening in this particular reaction? We have 2Fe3+ and 2e- going to 2Fe2+. Is this a reduction or an oxidation? As written, it is a reduction. That means it would take place at the anode or cathode? At the cathode./nAll right. The other reaction we have is 2I- going to I2 solid plus 2e- because it needs to be balanced, so this would be an oxidation. And so it should take place at the anode. Now we can calculate the cell potential for this particular reaction, the delta E knot for the cell./nAnd that will equal the delta E for the cathode reaction minus the delta E for the anode reaction. The delta E for the cathode reaction, and you were given this in your handouts, we have +770 volts minus +535 volts for the reaction happening at the anode./nAnd so that will equal 0.235 volts. And so will that be spontaneous? Yes, it will be spontaneous. Delta E is positive so delta G will be negative. We will have a spontaneous reaction. Now, let's consider which are the better oxidizing and reducing agents./nFor better oxidizing agent, is it Fe3+ or I2? Which is a better oxidizing agent? We can go back to the chart. Here is the iron and below it is the iodide reaction. Which is better? I think I heard some people say Fe3+./nThat should be a better oxidizing agent. Again, if you look at where they are, the things on this side at the top are the most oxidizing or the best oxidizing agent. Fe3+ is there. It is going to be better./nAnd we can also think about this in terms of the spontaneous reaction. This is spontaneous and the Fe3+ here is being reduced, so it is acting as an oxidizing reagent and that is causing the reaction to be spontaneous./nAnd if we talk about the better reducing agent, the reducing agent is the thing that wants to get oxidized. And if we consider I- or Fe2+ which is better? I- is going to be better. Again, it is down here./nDown here we have the reduced forms that are strongly reducing. This is the reduced form. It is on this end of the table. Also, that is the reaction that is happening here and it is spontaneous. And if you talk about the two reduction potentials you can figure that out./nOK. See you on Wednesday. (less)

Channels: Lectures Chemistry (General)
Tags: Oxidation Reduction

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