 First of all, let's go over a few administrative details, which are on the website, I believe. I didn't check. Maybe. They might be on the website. They will be on the website, certainly, but are on the transparency or slide or whatever that is over here right now. And that is next week, you're going to get an overdose of electrochemistry. And by the way, as I'm saying this, I'm noticing that I don't seem to see the laser pointer anywhere. But we have double lectures for you next week. So hopefully, you can make it. If not, they should be on the website, but they'll be really good lectures. So you'll want to be here in person for them. We have our normal lecture on Tuesday, and then we'll take like a 15-minute break and we'll dive into a second lecture at 2.45. We'll do a lecture on Wednesday at 2.45, and then the normal lecture on Thursday at 1 o'clock. And then after that, we'll return to our, for the following, return to our regular schedule. However, you will notice up there that I am planning a week from today on giving you a take-home midterm exam. And you'll have until the following Tuesday to, thank you very much, to complete that. And I'll give you a little hint on this right now. My exams are oriented towards, here's some data, do something with it. Tell me what the system is doing, as opposed to the theoretical concerns that I've been writing on the board. So hopefully, you understand the theory from what I put on the board, and you apply it to real-life problems on the exam. So much more like the homework, in other words, exam questions and the homework correlation there. Then we'll continue on. And Thursday, that's two weeks from today, will be my last lecture with you. And then you have a really, because you've taken this double set of lectures up here, you get a little treat the following week there are no lectures. And then you'll have a wonderful set of guest lectures by Professor Lewis to end the term. We've been talking about what we're going to do there, so you should enjoy that. So all of that will be on the website if you memorize that. Yeah, but just to show you what you can expect. And in particular, please try and calendar the extra lectures for next week so that we can give you the week off a little later on. OK, so we have a problem set that's due today. Let me remind you, and that can go to Tom. And there is another problem set that Tom is putting on the website as we speak. We'll be up this afternoon. Let's see, would you like to edit this manuscript for me? I'm trying to get back to it. OK, so we were talking about chronoamperometry when we last met, and I like to pick up with that. We have essentially developed all the basic chronoamperometry apparatus, that is, we're doing a simple step from a potential where nothing happens of chemical interest to a potential where something happens of chemical interest. And we have developed what happens when we make a large step, so large that the concentration of a reactant species drops to zero at the electrode surface. We've looked at what happens when we make a smaller step, but it happens to be a chemically reversible system. So the Nernst equation applies, and we saw that the solution there was very similar. And then we went and we said, well, what if we are charge transfer limited? And again, let me stress, even though we're charge transfer limited, we still have a diffusive component. So we still start off with fixed second law, and we use now, as our boundary conditions, the reaction dynamics. And in that case, we came up with a kind of complicated equation, but we saw we could back out some rate constants for the charge transfer reaction by doing that. And we looked at two cases there, the quasi-reversible case, where the forward and back reaction rates aren't too different, and the irreversible case, where you have a forward rate constant and can ignore the back rate constant, because it's sufficiently slow compared to the forward rate constant. So that's about what you can do with chronoamperometry. But after people have developed that, they said, well, how can we take this analytical technique and broaden it out? How can we tweak it? How can we modify it so that we can get even more useful information out of it? So we'd like to look at a set of expansions on chronoamperometry to start the course today. And the first one I want to look at is this idea of a chrono-culometry that you can see over here on the projector. And what I am showing you is a very short one-page paper that occurred in analytical chemistry with this title that you can see up there. You'll notice the author of this paper is one Fred Anson. And this is actually not the paper I want to start with. There's two papers that Fred Anson wrote on it. More than two papers, but I wanted to start really with this one right here. Excuse me. He didn't call a chronoamperometry to start with. He called it this long, complicated thing, potential steps, and charge pass, and things like that. And this is a one-page paper, which is what I was trying to show you. And you'll notice down here. Here's just a little note that's sent into analytical chemistry, the journal analytical chemistry. And there it is from Fred. And you'll notice here it comes right from Caltech. And if you have really good eyes, you'll see that it's in the Gates and Crone labs that the work was done. And it is a purely theoretical paper, one-page theory. Good luck on getting away with that today. And the date on this is 1962. Excuse me, you'll notice right over here. July 22, 1963, which means Professor Anson was a very unprofessor at that point. And if you notice, it was received and published in just short order. Short order, yeah. Well, it was a short paper. It has no data in it. It's just a theoretical discussion. That's why I'm saying good luck at doing something like this today. And what he starts off with is pointing out that there's this technique called chronoamperometry and that he has published, you don't have to read it, some earlier papers. And when he suggests, instead of doing chronoamperometry, you can get some more data by, instead of jumping, that you would scan the potential instead of doing a step potential. And you would do this integration of the current to get the number of coulombs past. And he published some earlier work on that with Bob Ostry-Young, who was at aerospace company in the area here. And now he's saying, you know, you could actually do better if you go back to chronoamperometry and do the integration. You'll get more information, he thinks, out of the system. Again, there's no data here. And so he starts off, and look what he starts off with. Hopefully this is an equation here that you remember from last hour. This is just a simplified form of the full kinetic statement for chronoamperometry. There's that exponential and the air function business. And he's just using k's and y's and whatnot. He's hiding all those messy alphas and f's and whatnot terms in there. But that is the equation that we solved last hour. So he starts off, and it's in the way of current. And he goes through this. And he's simply going to integrate this with respect to time, maybe not so simply because it's a messy equation, but he does that. And he comes up over here with a statement now. That is coulombs in terms of the same parameters that we're working with and makes the observation that you could plot now coulombs, not versus time to the minus 1 half, because when you integrate that becomes the time to the plus 1 half. And that should be a linear plot. And there should be an intercept over here. And he defines the term lambda, which you can see up there. And I don't know how good your eyesight is. But that is a term that's very similar to the kinetic term that we were looking at last hour that I called lambda when we went to our unitless system. Anyhow, so he has that. And then he says, well, you could really get the same thing right here by going to what he calls the diffusion limited case. That is, where the concentration he specifies right here drops down to 0. There it is, for times greater than 0 with 0 everywhere. And you write this out. And there's the Coutrelle equation. And one thing I find actually very fascinating about this is in this paper, 1963, he doesn't call it the Coutrelle equation. He says, he just describes this equation. He doesn't have that name yet. But there it is, that one you should know, even if you can't see it. And he does a simple integration of that and points out there's the results. So there's the basic chronocoolometric equation for the case where we are diffusion limited. Either we're reversible, and we make a sufficiently big step that the ratio of oxides reduced favors the product heavily. Or we're kinetically limited, and we make even a bigger step so that the rate constant can be ignored. And we'll get this situation going right here. And in passing, he has to point out that when he does the exact solution up there and this shortcut solution over here, he does run into a little discrepancy in that this is one term that says that time equals 0, the current is 0. So you have a 0 intercept on the y-axis. Whereas you have an intercept value over here. And he goes on and argues that, well, actually, if you think about the situation over here, you need a big rate constant one way or the other. And if you do that, that will change this lambda and whatnot. And so this term will dominate. And this term will be very small. And even though, technically speaking, there is a non-zero intercept. It's so small that we could call it a zero intercept. So this equation makes sense. So there's the magic equation right there. That is the equation that, if we wanted to give everybody a name, should have become the Anson equation. But we missed out on that one. And then he does something even more important, actually. And that's over here in this little highlighted area over there. He points out, why would you want to do this? And he says that if you make a plot of charge versus now t to the 1 half, the square root of time, you expect a linear plot. And that you expect a intercept again. That's going to be zero based on this. But there's going to be situations where you don't get zero. And he points out the specific case here. One thing that would be interesting is if you happen to have molecules absorbed on the electrode surface. So they can't diffuse. And remember, this is the diffusion equation over here. So if they either absorbed when they get oxidized or desorbed when they're reduced, if you're looking at a reduction reaction, then you expect a non-zero intercept, because it does not have the same time dependence. And this would be a great way of looking at surface processes, chrono-colometry. And he leaves it there. There's no data to support that. That's that, end of publication. And then he comes back. I believe we're now two years later. Now he's publishing a paper by himself. It isn't with Austria anymore. Still out of Caltech here. He now has the name for this technique, chrono-amperometry. He's showing some data now. So here's an example of a step you can see where you have dropping mercury electrode. It is the old days still. And you have some cadmium ions around that you're reducing. And then he shows what happens. If you have no cadmium ions around, you get a different situation here. And the first thing to point out about this is when you think about chrono-amperometry, of course, we get this huge current going up and coming down. But here you'll notice we started zero. Obviously at time zero, there are zero coulombs that have passed. And now we have a curve that's never undefined anywhere. That's a big plus to start with. And in addition, all this data over here, which I have argued before is the good data because it has the biggest time rate of change in it, is available to use. That is, he does a step here in the absence of cadmium ions. And there is, you can see there for a charging current right in there. But all this data beyond that, that's good stuff. So now we can get a much better idea of the early-time dynamics. And it goes a little further and writes down an equation that probably is the most important equation. And that's right here, equation two in this paper. See if I can raise that up a little bit so you can see that better. And what he says is the total charge is this charge that we got from last paper, which is just the diffusion-limited current for a large step, plus something he hasn't noted before, double-layer charging. That's obviously right there. So you should add in a number of coulombs for the double-layer charging, he says. And then finally, he adds a third term over here. Now that's the term he refers to, he and Austrian refer to, in the prior paper, which is this surface absorption. So he said there is another quantity of charge that is time-independent. So it happens that time equals 0. And that's equal to the number of electrons that takes to oxidize or reduce a surface-confined molecule times Faraday's constant to get that into moles times the area of the electron times gamma, which is called the surface excess. It's the moles per square centimeter of this material that's absorbed on the electrode surface. And so he's pointing out in this equation that you always expect a non-zero intercept because you always have charging, double-layer charging. But in addition, you might have this second component which would give you even a bigger non-zero intercept. So in other words, if you go like he did and take that data and plot it versus the square root of time, so there's charge versus square root of time for the data I showed you in figure one, you get these linear plots just like he said. And you'll notice this is done by hand. This is wonderful. These are real pencil dots. And that when you do it in the presence of the cadmium ions, you get this non-zero intercept. And you get the same non-zero intercept in the absence of cadmium ions. Of course, here you have an intercept that goes as the diffusion coefficient and concentration of the cadmium ions. Here you're not oxidizing or reducing anything, so you have a zero slope. But you get exactly the same intercept. So that must be the double-layer charging intercept. So one can back out the double-layer effect by simply looking at that. Now this, of course, assumes that there is no surface absorption. He is sort of that in this case because he does it with and without, and he sees no difference. So there must be no third term, no surface absorption that's present. But then he goes on and points out, you could get yourself in a situation pretty easily where you couldn't simply subtract out a molecule to see whether or not there was surface absorption because it might be a complicated process. And you might only have surface absorption present when your redox active species, not only when your redox active species was present, but when it wasn't present. And then you would be fooled by this. And so he comes up with a variation on the experiment and says we should do double-step chronoculometry. So now we're going to change our waveform for the first time. First thing we're going to do is we're going to step to some potential here, and we're going to sit there for some time period tau. And then at time period tau, we'll step to a second potential and monitor the number of coulombs as a function of time. And he has set this up based on something that was already out there, which is double-step chrono amperometry, which we're going to look at in a few minutes. But he points out that the way you want to do this for chronoculometry is the potential you first step to and the potential you step to second. That second potential should be the initial potential. So you step to one potential, then you step back to where you started. And if you do that, this is again with the can of my end, that plot. And if you're having trouble copying all this down, don't worry about it, because I've given you this plot on your homework problem set. And you're going to get to play with it and do the fit to the Anson equation. And I'm going to show you what you better get as a result. Here's the answer to the homework problem. He makes his t1 half-clots for the forward step and the backward step. And he notes that he gets non-zero intercepts there. And he argues if the deviation from 0 at this axis is the same in both directions, then that's just the charging current. And if it's different, then one of them must be the charging current and the other must be the sum of the charging current plus whatever surface absorbed. And so you can get all three of those parameters by doing a double-step experiment. And so when you have the situation now where you have surface absorbed material, you can get a handle on all of this. And certainly in chrono-amperometry, you can't do that. So I'm just going to leave that. Let's see, I'm going to blank that out on you guys. I'm going to leave that for a moment. And we've run through that now and looked at the data and seen it qualitatively. And it would be worth just taking a moment here and writing some equations on the board so you can see a little more quantitatively. And besides, you guys should be getting good at this by this point. We do everything the same as before. In fact, we'll start off with a shortcut here. We'll start off with Anson's statement that if we integrate from time equals 0 to time equals t of the current that we get in a chrono-amperometric experiment, that would be the charge given as a function of time. And we can see that there's a factor of 2 that pops out there, n, f, a. This is just the integration of the control equation. Diffusion coefficient to the 1 half power, assuming it is not time-dependent. Volt concentration of oxidized. Time to the 1 half is the big change. Now it's up on the top. And the pi to the 1 half, of course, just comes along for the right. So the only change is the factor 2 here, which we really don't care about. And the fact that this thing moves to the top of the equation. And again, this is for the situation that Cotrell described. And by the way, in that second paper, when he writes down the Cotrell equation and then shows the integrated form, he now calls it the Cotrell equation. So between 1963 and 1965, the Cotrell equation was born. The equation pre-existed, but it got its name, I guess, in that time frame, at least as far as Fred was concerned. I'm not sure. I don't know. That's an interesting thing. I went an obvious question, right? So who is this guy Cotrell that we did this equation after? So I figured, well, we're historically right in the time frame looking at these ants in papers where we should be able to figure this out, because this is happening in parallel, more or less. And so I looked to see, when he writes down the Cotrell equation, who does he cite? And he references it back to the standard electrochemical textbook of the time equivalent of Barton Faulkner, which was Dele Hayes' book. And I haven't had a chance yet to look in Dele Hayes' book to see where Cotrell comes from. I'm clear. So Mr. Cotrell, we still have to do some historical research on him, but he gets the equation, nonetheless, Cotrell equation. This is the integrated Cotrell equation. Not referring to the integrated Cotrell equation, but to who is Cotrell. So we have that statement. And in fact, we now have to modify that and say that, in general, for a reversible reaction, big potential step, that we not only have this term, but we have to add another term, which is the coulombs that go into double layer charging. And so the intercept tells us about the double layer effect. And the slope tells us about the actual solution dynamics. And you see, so we've deconvoluted now this charging, this non-ferdaic current, from the ferdaic process. And that is a big plus here. And then Fred went on to explain that you could even push this further by realizing, so here's step one. Here's the advance number two. And then number three, even though he realized this very early on, that the total charge was equal to this integrated Cotrell IC charge, this term up here, plus the double layer charging, plus a term, and ferridae's constant area of the electrode, gamma, or gamma now, is given in moles per square centimeter of a material that is electroactive and is absorbed on the surface of the electrode. And so you could back this out. Because the key point is, this is time dependence, and we know the time dependence. And these two terms are not time dependent. And we can deconvolut these two terms by doing a double step experiment. Right? It's gone. Exactly. So it's not time dependent. Yeah, so it adds to the intercept, but not to the slope. Right. So that's exactly what Fred realized. It adds to the intercept, but not the slope. And of course, there is an assumption here. There's an important assumption, and you just brought it out. And that is that we're not talking about a kinetically limited case. If it turns out, say, the oxidation of the surface material has a rate constant that occurs on the time scale of the chronochoologram, it's going to have a time dependence now. So the big assumption here, big assumption, but it turns out it happens quite often, no kinetic limitation. That is, as soon as we do the jump, all that current flows. And on the time scale that we're looking at things, it's happened. And of course, in that suggests, well, there are other processes that have kinetic limitations. And if we wanted to work out that problem, how would we work that out? And the simple answer of that is, well, we know what the kinetic limitation looks like for the chrono-amperometric solution. So we could just take that solution again, and we can integrate it with respect to time. And we have the kinetic limitations. Kinetic limited, it's limiting. If we do that, we see that the coulombs as a function of time are equal to this NfAk forward times the bulk oxidation, both concentrates the oxides species. And by the way, I'm going to do the thing where I'm assuming that the reduced species concentration is 0 to start with so that we don't have to add in those terms. But you can look them up if you wish. This whole thing becomes divided by this h squared that we introduced last hour. And all of that times an exponential of h squared times t error function of ht to the 1 half power plus another term in here, 2ht to the 1 half, or pi to the 1 half minus 1, where h, again, is just this ratio of rate constants to diffusion coefficients k forward or diffusion coefficient for the oxides molecule to the 1 half power plus k backwards over the diffusion coefficient for the reduced molecule to the 1 half power. And by the time you get a statement like that, you simply say, well, there really is no advantage of using this over using the chrono-amperometric solution. In fact, this is a little worse because I've thrown in another term here. So my ability to analyze this, especially in the absence of a computer, remember it's 1960-something, is somewhat limited. And typically what one would do is say, well, let's consider a situation where h times t to the 1 half is a large number. And it turns out if you look at the sensitivity here, you would like it to be greater than 5. And if you do that, then you can simplify your life and say that the charge as a function of time is equal to nfa k forward, both concentration of oxidized times 2 t to the 1 half over h pi to the 1 half minus 1 over h squared. And so you can see, if you get yourself in this regime, and of course, you either do that by picking your chemical system so that h is fairly large. That is, if you have the opportunity to study any chemical system you want, you can find systems which have fairly big rate constants. Or if you want to study a specific chemical system, and hence are stuck with whatever its kinetics are, you can, of course, go to a time regime here where this is required. So we're going to longer time now. And in doing that, sort of losing the advantage, I will point out, of chronocoolometry over chronoamperometry. Because again, at long time, the curve flattens out. And your data quality isn't quite as good. But assuming you can find the right region, then your equation simplifies dramatically. And from this, we can again make one of these q versus t to the 1 half plots. And from the intercept now, we're going to back out h. And then from h, we get all our kinetic parameters, assuming that we know some of them. So in other words, if we can measure h as a function of time, then different times we can get this out. Remembering again, there is a hidden potential dependence and time dependence in these rate constants that I can take advantage of, this taffle business. OK. Well, excuse me, though, there are potential dependent. Right. The time dependence is explicit in this equation. And then we get a hidden potential dependence in these h's. Yes. OK, so we can back that out. OK, so that is a fairly complete treatment of chronocoolometry. But again, the real power is this thing right here, as well as an easy way of kind of getting a handle on this. All the messiness in chronoamperometry about the rc time constant disappears when you do this. You just read it right off the graph. That comes out of the potentiostat. You can do this, but again, it's comparable in messiness and effort to doing it chronoamperometrically. And so why take your chronoamperometric data and integrate it if it won't gain you anything? There may be reasons to do that, such as noise. Or to get rid of, you could subtract out this component. If you have this going on and this, you might, by doing a double-step experiment, get rid of this component, or if you wish, this component. So there may be reasons to do it. But you have to have a good reason to do it. It's not an automatic do chronocoolometry over chronoamperometry. Now, having said that, then let's go back to chronoamperometry and see how far we can push that. So instead of integrating it to push in that direction, let's take it and let's do a double-step experiment. So we're going to do two-step chronoamperometry. So the idea will be, as a function of time, we'll start off at some potential E initial. At some magic point in time, which we usually call T0, we do a step up to some potential where something exciting is happening. Before we call that the final potential, so let me continue with that notation, even though it's not the final potential anymore. And then we're going to step at some later time to a potential, I'm going to use a lower case r here because I don't want you to confuse it with the redox potential, to E sub r, the reverse potential. So I have three potentials I could deal with, one where nothing is happening, one where I'm carrying out an oxidation reduction, and one where presumably I'm going to carry out the reverse of that reaction. And so presumably the way I'm going to want to set this up is I'm going to have something like the redox potential for the system, somewhere in the middle there. So I start here and everything is in the oxidized state because we're going to use our generic equation. So down here everything is oxidized. I jump to a potential up here where reduction takes place. So I'm making red, not red, but red as a reduction. And then I jump back across the redox potential here. So I'm now taking the reduced species and oxidizing it, reduced species that I made up here. Now I might actually choose to jump back to the same potential I started with. That would do it. But in most cases, that's exactly what you do, by the way. But I'm just drawing it this way to show you that you could be more creative than that if you wish. There is no requirement that you jump back to the initial potential. But for most experiments, that makes the most sense. It meets the criteria that I said there. Now the first question is, why do that? That is, after all, if I had a reaction like this, I've already argued I can characterize it completely using a single step. So why do this? Well, in fact, if that's all you have going on, then a single step was all you need. However, and this is probably why electrochemistry is interesting to at least a slightly larger group of people than the half dozen electrochemists that currently walk the earth. And that is the following. It's often the situation that you have an EC mechanism. And this is just one EC mechanism that I might write out, but that's the one we're going to look at. I start with a molecule. It's reduced. And that reduced species is chemically reactive. It does not involve a charged transfer. It just changes into a product. That is, a bond is made or broken. And we get a product. So if all that's happening over here is this, then that's really exciting to people that are interested in charged transfer. But it's not exciting to the rest of the chemical world that likes to make and break bonds. But what this statement is saying, so this would be, by the way, an electrochemical step that is reversible, so another R down there followed by a chemical step that is irreversible, so an I down there, so an ERCI mechanism. And you could play with those R's and I's and come up with a whole bunch of different mechanisms you wish. But what this says is I can measure in theory a rate constant using electrochemistry that's got nothing to do with charged transfer. And people like to do that. So the idea would quite simply be that if this rate constant was 0, then the forward step, the current decay and the forward step, and the current decay on the backward step would be the mirror image of each other, be totally reversible the way I've defined this. To the extent that this rate constant gobbles up, reduce species, then when I jump back, I don't have those reduced species to reoxidize. And so I perturb that back step, and I don't have the mirror image anymore. So before we get involved in the wonderful mathematics here, let's just look at this chemically. This is an example that I took out of Giliotti's book. And we have this azo-benzene molecule right here. A horrible thing for an inorganic chemist to put on the board, but there it is, the alive organic thing. And it is well-established in the presence of acid that this thing undergoes a reversible 2-electron charge transfer. That is the presence of a very small amount of acid. And you end up with this hydra-azobenzene product. And you can go back and forth there. However, if you have enough acid around, then this product right here is reactive. And you cleave this nitrogen-nitrogen bond. And then the thing goes back together, a horrible free radical type of reaction. Things that I think even organic chemists don't totally appreciate. And you end up with a whole slew of products here as a result of this thing coming apart and going back together. Another product, of course, is this one. If it goes back together the way it came apart, it's a product, but not too interesting one. And so if you were to do a double-step experiment, you'd start out here. You'd do a big jump. You'd go off to infinity because we're doing chrono-amperometry now. You come back down. And if this dwell time out at this potential, which we'll call tau in a moment, even though it has a different nomenclature right there, if that dwell time is very short, there'll be no time for this backwards rate to happen. And so I'll jump back and I'll get the dashed line there, which is supposed to be just the mirror image of this. I could just superimpose those two curves on top of each other. And in this particular case, because this is not a simple first-order process but is dependent on the proton concentration, in fact, I can control this. I get a k observed here. And if I want that k observed to be very slow, I just don't have a lot of protons around. And so I get that dashed line. But if I add more acid in, then this k observed becomes faster. And so when I jump up, stay here for a while, same period of time, and jump back, I get this current over here, which is diminished compared to the no-acid curve, because I have used up some of this stuff. And so when I jump back, I can't reoxidize what I reduced. Another way that this might have been done, a lot of the systems where you don't have this dependence, you have the situation where you could change the dwell time here. And if it was very long compared to this rate constant, you'd see a lot of stuff disappearing. And if it was very short, compared to the typical rate for this reaction, you could reproduce the forward-going step. So that's the situation that we want to work on. And so how do we handle this? Well, now unlike the situation for single-step chronoamperometry, our potential has a little more sophisticated time dependence. We essentially had a time-independent potential before we said it was either uninteresting or interesting. So all the potential we were looking at before was the potential up here. But now we have a time dependence to our potential. I'll start that over here. So we could say that the potential for this experiment, if we want to do this analytically, is equal to the initial potential plus some function that you'll call s of tau, where tau is going to be my dwell time that is time-dependent. That is the difference times the, excuse me, not initial here, but final. The difference between e reverse and e final is what we're doing there. And then for this particular case of two-step chronoamperometry, this is a very general statement I've just made. I simply would say that s of tau for time equals tau, excuse me, time equals less than tau, is equal to 0. It's just going to be the initial potential. So we're going to do two-step up here, and s of tau for time greater than or equal to tau is equal to 1. So I've put in a simple step function there. Now why have I bothered to write it out this way when I've done that simple? I've just given you a general solution. That is, you can define s now anything way you want. We're not limited to two-step. We can do all kinds of things and come up with solutions. So we're using that general situation. There's two-step, the initial step and reverse. So in other words, for time less than tau, this term falls out so you step to the final potential. And that longer times than tau, so I should probably make this explicit on here and point out that tau is this dwell time right here. OK, now since you're all pros at this solving fixed-second law business, you know that once you've got that figured out, what you're going to do is come up with a set of boundary conditions. And so we might say, well, initially the bulk concentration at time zero everywhere in the cell is equal to the bulk concentration of the oxidized species. And again, to make our life simple, we might say the concentration of the reduced species under the same condition is equal to zero. And then we can say for the forward step, first step, we have initial concentration at the electrode for all time, which is equal to the initial, I'll call it prime now, concentration out in solution. And that r concentration likewise takes on a value where I can do this one of two ways. If the potential step is large, then c0 prime is equal to zero. That's just my semi-infinite linear diffusion condition. And if the final step is more modest, then given that I've defined this as a reversible reaction, it's governed by the Nernst equation. So then in that case, we have c0 prime is equal to theta times r prime. And I'll put a prime on the theta also, showing that it is the theta that goes with the forward step. Because remember, theta is the Nernst equation, and it's got an e minus e redox in it. So it depends what potential I step to. And I'm stepping to two potentials in this. So we put a prime there to indicate the first potential that we step to. And then we have the reverse step. And we could, again, say that the concentration of oxidized species at the electrode surface for all times is equal to another number, which is c double prime. And likewise, the reduced species is equal to cr double prime. And again, if we make a step that's big enough so that we drive one of these species to zero, in particular, cr double prime, we could note that. But the more general statement I'll just write down will be for the reversible system that c0 double prime equals theta double prime times cr double prime. So Nernst equation holds, in other words. And then we could stipulate these sort of normal things like, in the limit of being far away from the electrode, this is our infinite diffusion, mass transport limit of situation, being infinitely far away from the electrode. The concentration of the oxidized species is equal to the bulk concentration. And the reduced species is equal to zero. And likewise, we could say that there has to be mass conservation so that the flux of oxidized molecules up to the electrode has got to equal the negative of the flux of the reduced molecules away from the electrode. And now you're all ready to go and do your Laplace thing, Laplace transform, fix second law, plug in Laplace transform all your boundary conditions, plug them in, solve for a current now since you have all the concentrations in Laplace space. So you get current in Laplace space, and then you back transform, and you're done. And we really don't have to do that totally. We have to do that maybe a little bit, but we already have part of the answer, because if I just look at this situation right here, the first step, if I stop it right there, we've solved that one already. And so I can just write down that answer since we did all the work the other day. And so the current I that we're going to get only go from the initial to what I called F. I've got to stop calling that final, but just F. I sub F of t, we already know for the situation I've written there, it's just NFA, diffusion coefficient to the 1 half power, bulk concentration of oxidized divided by pi times time to the 1 half power with this other term that takes into account the Inner's equation. So that's simply what we've written down before. Thank you. Yes, I need a theta prime now. And that would be for the situation, again, where t is less than tau, of course, greater than 0. And now I can consider the situation where t is greater than or equal to tau. And if one runs through that, then we come up with the statement that I, I'll call that I sub little r of t, is equal to minus NFA, diffusion coefficient to the 1 half power, bulk concentration of oxidized divided by pi to the 1 half. And now there's some, there's a new solution here, because I have a new equation that I'm Laplace transforming and back transforming. And I'm not going to work it out, I'm just going to tell you that that comes out to be 1 over t minus tau, I want to do the 1 half power minus 1 over t to the 1 half if we run through that. Now, the way I have worked this out so far is k equals 0. This is the system, the solution for the system, ox plus n electrons, reversibly goes to red for a two-step experiment. So for my more general statement, I'm assuming k chemical is equal to 0 here. So in terms of my picture right here, I've worked out the situation where I don't have a lot of acid around and the reverse step is just the mirror image, the dashed line there, or the forward step. Oh, there should be, oh, excuse me, you know what? I wrote that down for the mass transport limited case. For mass transport limited, that is where the concentrations dropped to 0. However, since you asked for it, ready for some writer's cramp here, the more general case is that this second current is equal to again minus nfA d0 to the 1 half, both concentration divided by pi to the 1 half times 1 over 1. You know what I'm going to have to do here? I'm going to get writer's cramp too much. Let me define this ratio of ox to reduced. And we take the square root of that. Let me define that. It's this Greek letter right here. Squiggle, yeah, like that squiggle, theta prime, double prime, no, excuse me, theta single prime, minus 1 for 1 plus, same squiggle, theta double prime times over 1 over tau minus t minus tau to the 1 half, minus 1 over 1 plus squiggle, theta single prime to the 1 half. Now hopefully that equation, when you look at it, will have not a lot of physical significance to you. It's a mess, right? But if you go and you plot it out, you get exactly what I told you that is for the forward going step, we're going to jump up, we're going to come down, we're going to be decaying towards 0. And at some point, we're going to do our, what we call tau, right? That's our point tau right there. We do our back step, and we get the exact mirror image within the quality of my drawing ability of the forward. That's what these equations are saying. Now, one thing you could do, I'm not sure why you'd want to do this, but if you had this data and you had this reaction, again, all of the parameters are the normal suspects, the area, the number of electrons, concentration, whatnot. So you could solve this for D, again, if you wanted a diffusion coefficient, or if you knew D, you could solve it for one of the other parameters there. Now, why would you do that? I'm not quite sure, because you'd get the same information from a single step. However, what about the situation where you suspect that a reaction is reversible but aren't certain? That is, maybe this isn't happening, but maybe I shouldn't have two arrows here like this. I might, in theory, use this to determine whether or not I truly had a reversible reaction. And how might I do that? Well, I could go and solve this whole thing here and this big mess here and see if all these equations held. That is, take my experimental data and does it fall on top of this theoretical curve that I've drawn right here. But an easier way to do this, which would get around the issue of I don't know my electrode area or I don't know the number of electrons, maybe that transfers in this step, I don't know the value of that. I just want to know if it's reversible to start with, is I could go into one of these dimensionless schemes. So I could look at the ratio of the i sub f to the i sub r. That would be unitless. And I could compare that in this curve at the same time into the curve. That is, let me pick this point right here arbitrarily in the curve. And I'll call that a time t prime. I could then go the same distance into this curve, t double prime. And if I picked off the currents at the same time past the jump, if this was purely a reversible process, then that should equal one. If it's purely reversible, then in fact, are the mere image of each other. I have to throw a negative sign in there, I guess, because one of those currents goes the wrong way, but I get one. And of course, I could do this for a series of t primes and t double primes, just to make sure that it wasn't being fooled. And I could come up with a statement about whether or not this was a reversible system. Or another way of putting this is that more generally. May I erase this board over here? We're OK on that one? Let's get a little more space here. And it really didn't matter the way I wrote that, because it was going to be equal to one. But I happen to have written down this next part as the inverse of that. My apology. I'll stay with that. So the ratio of t r to t f, negative sign to get rid of the change in the axis system, is equal to, in general, t f over t r minus tau to the 1 half minus the ratio of t f to t r to the 1 half. And now I've changed notation here, because there is no assumption about the relationship between t r and t f, other than t r f refers to this step and t r refers to this step. t r is actually measured from 0 in this. Now I'm going to change that all the way across. Maybe I should put that. I think my computer got tired of looking at things. Well, you can notice if you come over there, everything's measured from 0. Maybe you can't notice on that screen scale. But let me change that now and say, let's keep the relationship, so we'll add a limit here, that t r minus tau equals t f. That is, I now instituted my t prime and t double prime. As I'm saying, t r equals t double prime over there and t f equals t prime. But again, I have to measure it from 0, right? This one measured from 0. That is, instead of just saying, OK, I'm going to pick this point here and then I'm going to close my eyes and pick that point over there, some random point on this curve, I'm going to say, I'm going to go a certain number of seconds after this jump and a certain number of seconds after this jump. And that's the ratio of currents I'll look at. And then if you do all that, life becomes very nice. And the ratio of currents is equal to just 1 minus the quantity 1 minus tau over t r to the 1 half power. And so you see, I can now make a working curve that is this ratio versus, say, this quantity over here. And one can come along then without using any equations, if one so desires, and get that ratio and read off the appropriate time and tau values. And if that fits, then it's reversible. It's a more sophisticated, more sensitive check on reversibility than what we've looked at so far. OK, what if it's not reversible? Actually, this is the case of a not reversible case. And now I'm not talking about irreversible. Remember, irreversible means that we have a slow charge transfer step. And now I want to talk about the situation where k takes on a value, chemical k. Things are beeping all over the place here. And now we have a non-reversible situation because we're going to consume the reduced species in a chemical reaction. And then if I was going to do that, you can see I could start writing out some new boundary conditions. So for example, for times less than tau, I might say that the flux of the oxidized molecule at the electrode is equal to the rate of a forward reaction, chemical reaction, minus the rate of a backwards chemical reaction. I'm making this more general. I have a forward and a backwards in here now. And likewise, I could write similar statements for t greater than tau that the flux at the electrode was equal now to I'll even allow the rate constant to change, if you wish, and have them double prime. This is the most general statement I could write. Probably some of these numbers become the same. And I could go and I could solve for that system, for example. And when one does that, you do the same thing, the Laplace transform, et cetera, et cetera, you have to Laplace transform these statements, life gets messy. When one does that, the general form of the curve is you can see it over there, but the mathematical statement gets really ugly and is probably best hidden as follows. The current for the forward step is equal to, divided by NFA, just to get it there, is equal to the bulk concentration, fusion coefficient to the 1 half power, time some function phi, which has to do with the rate constants, the time, and the tau that I pick divided by pi t minus tau to the 1 half minus 1 over pi t to the 1 half. That is the equation that I'm showing you over there. And this, again, would be first the situation where I have ox now plus n going to reduce reversibly and then a chemical rate constant that is non-zero and irreversible the way this is solved that takes me to the product. Now, that is too nasty. The nastiness comes in into that little phi. It's slowly hidden right there, which is a confluent hypergeometric series. You should be impressed I was even able to say that. There is a series solution to this thing. And one can work out working curves. And here's a good case where you might as well just use the working curve, because life is getting too complicated here. You really don't want to have to solve this explicitly. The working curves are worked out, and you could just look at a ratio of I ox to I red in this system and come up with a rate constant. So let me show you now how this actually works. Now, for the last couple of days, probably if you're a chemist, this has been fairly new stuff to you. And you do not like Laplace transforms. And the idea of mass transport limitations is less than appealing to you. On the other hand, if you are a chemical engineer, you have said finally something that is easy. And straightforward, I understand mass transport, and I understand limitations. So this is a little break for the chemist. I am going to apologize to the chemical engineers in the crowd. You're going to have to learn some chemistry for a minute here. But I am an inorganic chemist. So let me show you a problem that we were interested in several years ago. Before I show you the problem, let me show you the molecule. This is the molecule right here, tungsten tetracharbonyl bipyridine, a beautiful red colored molecule. And one might ask, why would we be interested in it? We're interested in it because it's sort of, if you will, a poor cousin of a molecule that is a very famous inorganic chemistry, which is ruthenium tris bipyridine, which probably every inorganic transition metal chemist has heard about or handled or admired or something like that. This is the molecule that was discovered a long time ago, could potentially absorb sunlight and split water. And it does it to a small extent. And people have worked on that system very hard and long and wanted to understand it. And they understand a lot of things about it, but not everything. And one of the things that was not well understood it, thank god, because Brustel has a job as a result. One of the things that actually is still a problem, pretty well understood at this point, was less understood when we did this, was that sometimes the molecule, when you made the excited state, did charge transfer to water, and you ended up with hydrogen and oxygen. And sometimes it fell apart. And so the question is, what were the dynamics that gave rise to it falling apart? So this particular molecule, intense red color, as I said, just like ruthenium tris bipyridine, well not just like, a slightly different spectrum. But the intense red color is due to a metal to ligand charge transfer band that moves an electron from the tungsten and excited state out to this bipyridine ligand, just like on ruthenium tris bipyridine. But the lifetime of this molecule is much shorter than for the ruthenium tris bipyridine system. And as a result, what one observes is nothing. That is, it does not spill water, it doesn't fall apart. It just sits there. So you can do nice spectroscopy on it, and people like Brustel like it for that reason. But it doesn't do any of the exciting stuff. So he shouldn't like it as much. But what we know because of studies on this spectroscopic studies is that there are 60 orbitals. We know that actually from freshman chemistry. 60 orbitals, so I shot five d orbitals, and six electrons in those d orbitals in there. And that is a low spin complex. So the electrons are sitting in three of those orbitals like this. The orbitals are split, so there's two higher energy ones. Now in fact, I told Brustel he'd have to sit very quietly while I did this. I totally corrupted the MO scheme right here. This molecule, just for those of you who like energetic chemistry, is obviously does not have octahedral symmetry to it. And I'm using an octahedral d-orbal splitting scheme here. These orbitals are split a little bit as are these. But for my purposes, that splitting is small compared to everything else. And therefore, we'll ignore it. What we can't ignore is there is a bipyridine pi star orbital that happens to fall right in between those. And so the red color that one observes this confidence due to a transition of an electron from this filled set of orbitals up to that bipyridine orbital. The T2EG orbitals down here are primarily non-bonding orbitals. Slightly pi bonding, but we'll call them non-bonding. The pi star orbital is, assuming the bipyridine orbital is anti-bonding, but with respect to the pyridine, these are nice strong bonds here still. It's these orbitals between the carbons that are somewhat anti-bonding. And these EG orbitals that are empty are sigma orbitals, and they're sigma anti-bonding orbitals. Everybody likes this? I have real-life inorganic chemists looking at me strangely. What's happening? That's pyridine that doesn't look right. Oh, the pyridines? Yeah, well, talk to, you know, that will get me off. Talk to ChemDraw about that. ChemDraw that does wonderful organic structures and just cannot seem to handle inorganic things. I was going to play with it for another half hour to try and get the symmetry back, but forget it. So we can't do photochemistry on this, because this thing doesn't do photochemistry, because it's got a very short lifetime. But we can do electrochemistry as a result on it. And so what we're going to do is we are going to simply take the molecule. There's the electronic structure I showed you a moment ago. And we're going to reduce the molecule and pop an electron into the next open orbital right there and ask what happens. So the idea is, here's the molecule. We reduce it by one electron. We make the anion when we do that. We happen to have thrown in another molecule there that, in a lot of it, that's a ligand. It's a triphenylphosphine. Just for the fun of it, all this electrochemistry is done in a cedarnite trial, which is a ligand also. But we wanted to have a ligand that we know would be fine there. Those of you, by the way, that are synthetic chemists, what we're going to end up doing here is we're taking off a CO in the end and putting on, in this case, the triphenylphosphine. But we could equally well put on the cedarnite trial. We do this. So this is a great way of removing a CO ligand, which is usually a tough job. It doesn't very selectively. You could do this synthetically if you wanted to. Anyhow, but that's not what we're up to here. What we're up to is the double-step chronoamperometry, because we want to study this process now, not the reduction process. We don't care about that. But what happens after we put an electron up here? We want to know the kinetics of what happens. Now, what happens, as you can see, we know from other experiments, as the CO falls off. We see that spectroscopically, and by IR, and by mass spec, and all that wonderful stuff. And that's confusing. That's a big question mark, because we're putting an electron into an orbital that's supposedly isolated on the ligand, has nothing to do with the metal-to-ligand bonds, and yet a ligand falls off. And it's not the right ligand even. That is, you'd think if a ligand was going to fall off, it'd probably be the bipurity that we're playing with. Instead of CO ligand, which allegedly we're doing nothing with, is falling off. So how could you potentially understand that? Well, if somehow after this electron got into this orbital, it ended up in these orbitals up here, then you could understand it, because these orbitals are sigma-antibonding with respect to the COs. And so you put an antibonding electron in there, and you'd think you'd weaken the metal CO bond, and maybe the CO would fall off. Of course, you have a little problem with that, in that it looks the way I've drawn this, like this bipyridine orbital is significantly lower in energy than these orbitals over there. So how do you manufacture the energy you need to get the electron up there? So one possibility is, and the one we went after was these orbitals aren't quite as far apart as I've drawn them there. They're pretty close together. And if you get the electron into this orbital, you can get some of it into this orbital. If Bruce was giving this talk, he'd start talking about fexy states and things like that, but we won't say that. We'll just say that there's a thermal equilibrium, and we could get the electrons in there. So what did we do? We took our bipyridine ligand, as ugly as it looked on the last transparency. Whoa, where'd it go? There it went. We took this ligand, and we put substituents out here. And the purpose of those substituents was to move around the energy of the pi star orbital. And then we looked to see as we moved that energy around whether or not the CO fell off. And you find that when you move it to certain places that are, in general, lower in energy, which we observed spectroscopically, the CO doesn't fall off. And when you raise the energy closer to the EG state, the CO does fall off. So how are you going to prove to the world quantitatively that you know what you're doing? The answer is we did double-step chronoamperometry. So here's some real-life data. Again, we have triphenylphosphine around just to make our life simple. And by the way, this is going to be a good news, bad news story. It's going to show you the advantages and the disadvantages of double-step chronoamperometry. So you get something here. That's real-life data. And you can see it kind of shoots off to infinity when you do your downward step, your first step, your reduction. And then you come back and it recovers. And you'll notice it's not symmetric. That is, the number of molecules I initially reduce is not equal to the number of molecules that I oxidize back. And that all makes sense because we say that the CO is falling off. And now we no longer are doing that. And so we go and we do this for a series of different towels. And this is one way of boxing in the rate constant. You don't know what the rate constant is. You have to pick a towel that matches with the rate constant. If it's too fast or too slow, we won't be in the regime where this works. So we pick a bunch of different towels. And then we can take time off. This is just one of these plots. This happens to be a towel that goes 40. And we pick a bunch of time points off of here. And over here, the symmetric points. So we can use this sort of simplified analysis. We get our unitless time parameter right here, our ratio of anodic and cathodic currents right there off of this particular tau equals 40 data set. And then you go to the literature. In fact, you can even find this in Bard in chapter 12. And you pull out the set of working curves. And this may be a little hard to see here. But you have the ratio of the anodic to cathodic currents right there. And you have k times tau as the horizontal axis down there, if you have really good eyesight. And then each of these different curves is set up for a different t minus tau over t value. So you know the t minus tau over t value, because you've got to pick that. And so you find the right curve that you're on here. You find your ratio of currents. And you just go down to the bottom axis here and read k, which is the first order rate constant times tau. You know what tau is. And so you back out k. And you do that now as a function of all of your different substituted bipyridines. And you find out there's a very nice quantitative relationship between the rate constant for CO loss and how close that bipyridine pi star level gets up to the EG star levels. And you're convinced, except the data doesn't exactly fit. Could be experimental error. But it's a little bit off. I never published this data. By the way, in this particular case, you get a k times tau of 0.86. And you back out a k of 22.5 per second for that, if that's interest to you. That's with the unsubstituted ligand on it. So here's the problem. It looks like it fits an EC type mechanism. Everything I said there is just fine. The data qualitatively looks like it should for an EC type mechanism. I can fit it to those curves. It falls right on that plot. Everything looks fine. But when I go and change my tau's, and when I go and change my ligands, more of the tau's, I don't always get the same rate constant. Keep the same system, change my tau. I should always get the same rate constant, obviously. And I don't. It changes some. Is that experimental error or something else? Should we just go and publish this and say we're done? Well, it turns out in doing other experiments. Yeah? If you change your t on one experiment, you get the same. If you change, well, there is a range. If you're really close to the pulse, you could argue you've got some charging going on. So you don't use that data. There's a range right in here that looks pretty good. And then this was the real tip off. There's a range where you go a very long time where it doesn't look so good. And the problem there is on the one hand you could say something's wrong. On the other hand, you could say, well, the data isn't very sensitive out here. Maybe I'm just getting an error. And when I calculate for the rate constant out here and over here, maybe even though they're different, that's experimental error. What do you do? This is all the data tells you. When we did other experiments and we found out what's happening is this molecule that you make over here with the triphenylphosphine in this case on it is obviously got a different redox potential than the starting molecule. And it's such that it can reduce this molecule. When we do our jump, we jump into a region where we make reduced species here. And it can then go back and reduce this molecule. And so I actually have a nice catalytic mechanism. So actually, one can use it synthetically, but it turned out analytically it wasn't very useful. Now the problem with that is you say, OK, we'll just go and we'll find another mechanism in the literature that has this catalytic step in it. And in fact, you can go into, oh, I didn't put it on here maybe, huh, I lost it. You can go into, and Bard has this, there's a whole table in Bard that has all different kinds of mechanisms, but it does not have that mechanism in it. And as you look at these mechanisms, as they get chemically more complicated, this thing gets chemically more complicated. So sitting there and working it out unless you have a real love for mathematics becomes a big challenge. And even if you have a love, it's going to take a lot of time. And you have to work it out for all possible kinetic cases. Just because we know it's catalytic doesn't mean we know exactly the various arrows that we should draw there and how many rate constants and whatnot. So you have to work out a whole bunch of different ones and figure it out. And so at that point you say, it's not worth doing. We have the qualitative result that as you move pi star closer to the sigma star levels, it seems to work. But you can't quantitate it because you can't get rate constants that you believe in. As a result, I never published this data in there. But that brings up the big problem. From a chrono-amperometric curve or a chronocoolometric curve, you can't just look at it and get any sort of pattern recognition and say, oh yes, this is what's happening. In addition, one experiment, say jumping to one time or jumping to one potential, does not fill all of space up. You don't know the answer. You're sampling a little bit of data when you do that. Now what you should do to do this right is if you sat there and you took, this is a single step, if you took chrono-amperograms and you jumped to 1,000 different potentials and you monitored them for various periods of time and whatnot, then you would get a shape, a surface that looks like that. This happens to be the surface that we calculated for the completely reversible case. So this is real, not real data, but real calculations, not a random drawing. But you'll notice, if you look at the curves in the parallel to the current potential plane here, you're getting the shape that we developed for the mass transport-limited system before when we start talking about mass transport, those are those S-shaped curves. And if you look in the current time plane, you see those are the Coutrelle equation right there from various points. So that's the whole thing. So you could sit there, if you had a lot of patients, you could sit there and you can map out that whole surface and you still have two problems. Number one is nobody has that much patience. And number two is even if you had that, if you looked at that and instead of being nursed in like I've drawn there, it had some slight rate constants associated with it, it wouldn't pop out at you, that surface would look just about the same. And so these techniques are only useful after I know the mechanism. And I want to get a rate constant or a diffusion coefficient. But they're not useful for coming up with the mechanism, these chrono techniques. They're good for quantitating it but not coming up with it because you don't have any pattern recognition capability. So what do you do? Well, this would give you all the data and you could fit it into some kind of mathematical model and you would have pattern recognition with an appropriate computer now because presumably if you had all the data, then it would only fit one mechanistic model that you could crunch out in your computer. But a lot of work again and still you don't really know what you have until after the fact. So the suggestion is if you were to take this surface and you were to slice it with a plane and pick your plane so it's not parallel to your axes because if you're parallel to this plane right here, that's just a chronoamprogram and if you're parallel to this plane, that's just a volt hemogram but go off axis and slice the plane and look at the projection on that plane. And that projection now has less information than that whole set of data there but it has all the components projected onto it of the current time and potential dependence. And so you limit your amount of data which gives you pattern recognition capability and at the same time, you're not throwing away all that much data. Some, yes, but not all of it because you're sampling throughout space. That's the idea of cyclical tammetry. We're going to slice that off axis and we're going to get a two-dimensional shape out of that three-dimensional structure that will give us pattern recognition capability. So we can determine mechanism from that and given that, we can decide how we want to quantitate that either by cyclical tammetry or one of the chrono techniques to get at the detailed rate constants. Yes. Yeah, I'm finished. Yeah. I'm going to something like KB. Are you done? I am good. You want more? KF, KB. K forward and K back for a chemical process. This is a more general set of equations. I wrote this. Yeah, I wrote that one out for the possibility that you have something like this. K back, chem, K forward, chem. I just was giving you a general set. But it's only solved over here for what I had drawn a moment ago. We don't need this photographed. CO loss occurs before the phosphine comes in, yes. In other words, the rate constant for this process is independent of the incoming ligand. We don't have a bad, no, no, yeah, the CO falls off and then we actually put the phosphine in that case because we wanted to be able to limit that concentration and make sure it was bimolecular. The initial experiments were done in acetonitrile electrolyte and so the acetonitrile was the ligand but you had no control over concentration then, okay? Very good, okay, so we will see you next Tuesday. Now by the way, another homework is coming up tonight. Homework is due here. I am out of town again on tonight through until Monday evening. So if you'd like to talk to me about any of this, I am around on Tuesday, Wednesday, Thursday, Friday.