 OK, I think we should get started. So there were two sort of errors. I guess you might say that. I think I made last time in reviewing what I said to you. The first one is not all that important, but your name happens to be Cotrell. And that is pretty sure that I miswrote his name on the board. It's got two T's in it. I was realizing after I finished that. The other is not so much an error, but I made a statement at the end of the hour when we were discussing this Cotrell equation that one of the things you might use it for is determining the area of an electrode, given that the area of an electrode is not typically the geometric area. And what I was picturing and saying that was that, just about for any electrode, there is some roughness factor, not necessarily a huge number. And it may be important. And you can get at it this way if you so desire. What you heard me say, because many of you work on porous electrodes, is that you could determine the area of a porous electrode that way, which I wasn't really thinking about at the point. And in fact, you're absolutely right, you can't. Because all of the diffusional type processes that happen in a porous electrode happen on a very short time scale, and therefore are hidden under that charging current that is so predominant early time. And just to show that, I was playing around with this simulator that I told you about last hour. So I set up a simulation here. We're out here the green route in the bulk. And I've put a concentration of zero product out there. That's what the green symbolizes. And here at the electrode, that's the blue thing. I'm generating a product at a constant rate. So I've generated an electrode now that has these pores in it, which is not the situation that I was talking about before, but is a situation now. And you'll recall when this simulation runs, we will see a process happening here. And we'll build in a concentration gradient as we go across this space between 1,000 molecules that I start with over here and zero molecules that I have out there. And so we'll start that simulation. And I've stopped as quickly as I can. We've only gone through 19 iterations of the diffusion gradient at this point. And recall that this means that there's greater than 40 molecules in the black there and wider zero molecules when we have these colors in between. So I get out here, we're zero to start with. But you'll notice even after just 19 steps here, this front boundary, all these boundaries actually, have washed out in terms of the structure of the electrode. So even in this short of time frame, we have totally lost the fact that we should have generated material first in the pores and then later on out here. So absolutely correct for a highly porous surface that's going to run out. And as that continues to run, just in case you thought there was a little bit of structure there, you can see that it really washes out very quickly. So even far away from the electrode, you can see there's absolutely no indication that we have a porous structure there. OK, and we can, let's see, get out of that program now. Go back to our PowerPoint, which I'll come back to in a minute. OK, last hour, we spent a lot of time deriving this control equation. And let me remind you that the important boundary condition that we're using here, we're using some pretty standard boundary conditions, which I've listed over here in terms of out in the bulk. We're always at the bulk concentration. And that the concentration is the same for the oxidized species when we start the experiment. And the one thing that I'm not writing here, but that's a given that in this particular case, the reduced species, the product, is non-existent before I start the reaction. So I'm going to start with the oxidized species around. The important thing, though, about this, the unique aspect of this equation is that we're taking a large potential step. And by that, I mean whatever size potential step is necessary so that the concentration of the oxidized species at the electrode surface is zero as soon as I start the experiment. And it stays zero. So the instant before I start the experiment, I'm here and I have bulk concentration everywhere. And as soon as I start the experiment, that drops down to zero, so I have as large a chemical gradient as I can have between the bulk and the surface in doing this. Now, you'll notice in doing that, I haven't told you anything about the mechanism, just that that condition applies. Doesn't matter whether this occurs because the system is nursed in and is therefore always obeying the equilibrium statement given by the nursed equation, and I've made a big enough potential jump that the nursed equation tells me I have nothing there, or if it's charge transfer limited, and I've simply made a big enough potential jump to get the rate constant for the charge transfer reaction to be sufficiently large, that's zero. So it doesn't really matter. It holds for all situations, and that's why I have this little bit of ambiguous statement, large potential step, because large means whatever it takes to make this condition true, independent of the mechanism. So this will hold for every chemical reaction, not your chemical reaction, as long as I'm allowed to adjust my potential step so that this condition is true. In addition, I pointed out that this is the largest current you can see, because it's got the largest chemical gradient that one can produce in an electric chemical cell. It goes from this bulk concentration down to zero. We can't do larger than that. We know that this is a diffusion limited process and that the diffusion depends on the chemical gradient. This is the biggest one you can get, so this must be the largest current. So I'm going to put a little D down here now, because this is a limiting case of a diffusion limited current. So we have our system there. Now we can start to ask ourselves what happens in other situations. The other requirement that I added in here, which is given over here, really, for this system is that it's a linear system and it's semi-infinite, linear diffusion. So what if we do something simple, like say, well, instead of using a flat, a planar electrode, that will meet this requirement, we'll use a spherical electrode. And you'll recall I gave you these sorts of operators what we need to use to solve the situation for various geometries, and I am simply not going to solve it. I'm going to write down the answer for a spherical electrode. So if we instead use a spherical electrode, we would have a limiting current as a function of time. Everything else is the same here. Then Faraday's constant area of the electrode diffusion coefficient to the 1 half power bulk concentration of oxidized species times now two terms. One that follows the Coutrelle equation plus the second term that takes into account this spherical diffusion, where r naught is the radius of the electrode, the spherical electrode. Now you'll notice that that's a pretty significant change. In that, this equation tells you that at long times, the current's going to drop to 0 when I have semi-infinite linear diffusion. This equation says that there will always be a current, because I went on time-dependent parameter right there at this current. So for very long times, I'll still have a current, and that's because of the change in the diffusion field from something that just goes out like a plane, like we're just looking at to this thing that goes out spherically, which you saw the last hour. So we expect some differences there. So if you go back to this data, which I showed you, I believe the very first lecture of a Chrono-Am program, this is the oxidation of ferrocene. This was a potential step of 0.6 volts in the absence of any oxidizable material in acetonitrile, tetrabutyl monium perchloride supporting electrolyte. And I showed you this because I wanted you to see the non-Ferdic current that decays exponentially there. Above that, you see we've added in about a millimolar of ferrocene into this system. And the same size potential jump. And you can see a very different shape to the decay. That shape follows the Cattrell equation. And you can see two limiting situations here. Number one, at time equals 0, the Cattrell equation predicts that the current goes to infinity. That's a little hard to prove from this, but you will notice that it goes very high up. And if the Cattrell equation is right, that means that the current has gotten so large that your recording instrumentation and your tracking instrumentation, your potential stat, aren't following it. So there's some very short time in here where you've lost control of the system. And then it falls down. And we expected to follow this t to the minus 1 half dependence. And again, as I said a moment ago, we expect this was a planar electrode at long times that this will go down to 0. Now, I should say this was an approximation of a planar electrode. And so you can see this is heading down to 0. But at much longer times than I'm showing you, this is 2 milliseconds per box. So we have to go out much more than about the 10, 12 milliseconds I'm showing you there to see that that is in fact going to get back to the axis. And in fact, because it's not truly a plane, it probably will have a residual little current term that's in there. But if you take the data that I've shown there and you start, you throw out the first two or three milliseconds to get rid of the RC component, the non-ferred data component, and just plot that data. You can see, in fact, it does follow the control equation. There's the current as a function of time to the minus 1 half power. And there's the fit to the line. You can see that we have a r squared value of 0.99 plus. So that should convince you that, in fact, the control equation works. And quite obviously, if what I was interested in was the diffusion coefficient, for example, in this system, I could extract that from the slope of this line. Again, assuming I know what n is and what the concentration is in the area of the electrode. So that's one approach that we can consider. What if I have a reversible system, a nursing system, such as the thyracine system that I'm showing you there, but I don't make such a large potential jump? What if I do this and I jump into a regime where the Nernst equation tells me there's some ratio of oxidized to reduce material at the electrode surface? So what do I expect in that case? It's probably semi-intuitively obvious that what I'm going to get is a curve that looks like this curve, but it should fall off. And that is, it should be below this curve for all points in time, depending on exactly what potential I jump to. I'm going to show you that that's the case in a moment. But it's going to have that shape still. It's going to have the coutrelle to the minus 1 half dependence to it. On the other hand, I might consider a system where I have a rate constant that is the limiting beast. And in that case, again, I'm going to expect something that has overall this dependence, but not necessarily a simple t to the minus 1 half dependence now, because I have the time associated with the kinetics that has to get convoluted with this. But once more, I expect something, a curve, that falls below this curve. That is, at any given time, there is no current under these circumstances that I can get that's larger than what's shown there, independent of what the mechanism is, pre-unit area, et cetera, et cetera. So let's see how that works out. So let's start off with the situation where I do a step, but it's not quite as large a step as I have over here. So I am going to limit my step to something smaller, but I'm going to stay with the reversible system. Well, I say stay. I have to be careful. I'm going to use a reversible system, because I just argued a minute ago that this really does not need to be reversible. It just has to be a large enough step that this condition is true. It will be true for a reversible system, but it need not be a reversible system. So what I'm going to need to do, in other words, I'm going to keep my boundary, long boundary conditions the same, but this is the boundary condition that's going to have to change. So the most general statement I could put in, whether this is reversible, irreversible, EC mechanism, et cetera, would be that at the electrode, we have to conserve mass. And so what I'm interested in doing is taking this statement over here, if you will, and I'm going to exchange for it a new boundary condition, which is simply that the sum of the oxidized species that flow into the electrode area and the reduced species that flow in or out of the electrode area are constant. There's no net change in flux, if I consider all the species that are available. And then as I said, I still want to maintain the assumption that reduced species everywhere in the cell before I start the experiment is zero concentration. This is just an assumption of convenience and that the limit is x goes to infinity for the reduced species for all x and t is equal to 0. And then similar statements for the oxidized species. That is, the oxidized species before I start the experiment is equal to the bulk concentration, oxidized species. And in the limit of being far away from the electrode for all times, I have the bulk concentration. So I'm going to maintain this idea of the semi-infinite linear diffusion going on. Now, since I want this system to be reversible, these conditions are not limited by reversibility. This is a very general set of conditions. Yes. Now, this is simply saying that the flux of molecules entering. Oh, I see what you're saying. You're saying that if I end up with two molecules coming out for every molecule that goes in, yes, I'm assuming my molecule is not splitting in half or something like that. It doesn't matter whether the value of n, of course, as long as that doesn't happen. So it's not stoichiometry in terms of number of electrons, but yes, if I, for example, took a water molecule and changed it into a hydrogen molecule and an oxygen molecule, obviously, I'd have to make an adjustment here. But you would take care of that in your mass balance in your actual statement. That is, I suppose, write down my, legally correct here, my generic statement that specifically what we're talking about here is oxidized molecule plus n electrons going to reduce molecule. So assuming that that's the balanced chemical reaction, my statement is true. But you're absolutely correct. I do have to make a correction if there's some stoichiometry in here that's not unity. And I'm writing this arrow right now as reversible, because I wrote the word reversible over here. But in fact, again, this set of conditions in the green box does not require reversibility. The condition for reversibility that I have to add is that the Nernst equation holds. That is that the ratio of oxidized species at the electrode's surface to reduce species at the electrode's surface for all times is equal to what the Nernst equation says it should be, the exponential of nf divided by RT times the electrode potential minus the standard redox potential. So that is the special condition here that we're going to want to apply. But let's start off with the general set of conditions. So let's just assume that the general conditions are where we're going to be. If we do that and we play the same game that we played last hour of Laplace transforming our statement of fixed law, then we come up with the statement that the transform value of the oxidized species, and I am going to note oxidized and reduced now because I'm going to consider the general case where I might have both of them around. For any position in this cell as a function of s, our transform variable is equal to the bulk concentration of oxidized species divided by s plus an undefined function in s at this time exponential of minus s over d0 to the 1 half power. And likewise, for the reduced species, the transform concentration is equal to minus this function a of s times the ratio of the diffusion coefficients oxidized over reduced to the 1 half power exponential again of minus s over d0 r to the 1 half. So this is playing the whole game of taking a fixed second law, transforming it, and applying these conditions to it. And that is the general solution. That is, these two equations should hold for all possible solutions to the chronoamperometric experiment. Now, for the special case that I want to consider, the case of reversibility, we now have to add in the boundary condition that the Nernst equation holds. That is that the concentration of the oxidized species at the electrode surface is equal to theta times the concentration of the reduced species at the electrode surface for all times. Where theta is just the Nernst equation here. I can now take that equation and transform it. That's one of the easier things to transform. So if I take that into my s-coordinate system, that just says that the transform concentration is a function of o and s is equal to theta, a constant, times the concentration of reduced species transformed, taking it o and s. Now, something that's going to become very important in the future is that reason I can get away with this transform, you'll notice, is that the Nernst equation has no time dependence in it. So theta is just a constant. So it makes the transform particularly easy, and it's become problematic as we go on. Given that, then I've now defined this function a of s as equal to minus the bulk concentration of the oxidized species divided by s times 1 plus this ratio diffusion coefficients to the 1 half power times theta. Brackets and things there. So now I can, of course, plug this back into this, and then you'll recall that the whole trick here, the way the game is played, is I then figure out my current as a function of s, which is done by simply realizing that the change in concentration with time within a few constants like n, f, and a is equal to the current. And then once I have that, I can back transform that into current as a function of time. And so speeding up this whole operation, if you do all those steps, and this is just algebra that I'm talking about right here, then we come up with the fact that the current as a function of time is equal to n, f, a diffusion coefficient to the 1 half power bulk concentration of oxidized species divided by pi times time to the 1 half times 1 plus this ratio diffusion coefficients times theta. And you'll, am I saying it out of the way right now? And you will notice that theta given by the Ernst equation over there trials between essentially 0 and 1. That is, I can get other ratios. But at some point, if these two concentrations become too different, even though this equation would say I can get any ratio concentrations I want here, I in fact don't have enough for one species around to maintain this potential. So theta is always going to be a positive number. And it's going to be a number somewhere around 1, let's say. By that, I mean with an order of magnitude of 1. And I'll use my standard assumption that the diffusion coefficient for the oxidized species and the diffusion coefficient for the reduced species are about the same number. So this is about 1. So I'm adding a number on here to 1. And so this term down here is always larger than the large potential step. And therefore, the current is always smaller. But you'll notice that this functional form of this equation is identical to the control equation. So my statement that I made a moment ago that when we do a potential step that's smaller than the diffusion limit potential step that is controlled by the Ernst equation, we'll get a curve that looks just like this curve. It has the same functional form, but the current will be smaller for all times, with the exception of time equals 0, where it goes off to infinity. So if you wish, I could then just rewrite this current as equal to the coutrelle current, i sub d, divided by this 1 plus the ratio of oxidized to reduced to 1 half times theta term. OK, before we consider other cases, so we've now considered two important cases that we're going to run into. Let me point out from a practical point of view some of the pluses and minuses of this. Once again, assuming I know what the redox potential of the system is and whatnot and how much material I've put in, then I can use this to calculate a diffusion coefficient. It's probably one of the better ways of getting a diffusion coefficient for a system. And once again, if I wanted to, I could use it to get at the area all of the things being equal. Assuming the area is not for a porous electric, but one that's going to respond this time scale. In general, I would not do it this way, because I can get it from the coutrelle equation with a big step and not have to introduce these other terms over here. And so this would just be adding inaccuracy to the system. So although in theory, I could get it that way, experimentally it doesn't make a lot of sense. In addition, what you're going to have to do to use this equation experimentally is make the assumption that d-ox and d-rad are approximately the same. In general, a good assumption, but not always. And so that gives you another opportunity to mess up. You'll notice, by the way, since in this particular equation they go as the square root, if you're a little bit off, they're sort of evens out for you. So the math helps you out a little bit there. So we have those are the pluses. On the negative side, don't forget about this, that at time equals 0, this equation is undefined. So we can't use the very early time. In addition to that, if we have a large Rc time constant for our system, large compared to the time constant for this process, that is, compared to the square root of the diffusion coefficient, if you will, we are not going to be able to see the ferdaic process under the non-ferdaic process. Something else that should be pointed out that is not necessarily containing these equations, that is, the assumption we've made here that we're diffusion limited. There is no other process going on. That is, if some convection sets in, for example, then all the assumptions in here are incorrect. And so we are assuming there's no convection, which means that we have to do our best, first of all, to make sure that the experiment is such that there are no gradients that are set up beyond the concentration gradient that we want, for example, a thermal gradient across the cell. So you don't go and set up your electrochemical cell next to a hot plate because your lab mate is cooking something, right, and generate a couple hundred degrees across your cell. And likewise, you need to be fairly vibration-free for the same reason. There's two things there. There's one you can control. There's the control aspect of, in the middle of your potential step, somebody goes and slams the lab door, or you have an earthquake since we're in California, or something like that, and we have all kinds of vibrations that mess up your data. But the other one is, you will get some other sort of a convective process occurring in your cell. If you simply wait long enough, you cannot avoid having vibrations, for example. You can have a vibration-free or relatively free environment for a short period of time, but it's not going to happen for a long period of time. So we saw on the one hand, because even under the best of circumstances, you have this RC time constant that's going to knock out the first couple milliseconds of your data. So we have no data for the first couple milliseconds that we can rely on without the convolution. Out here somewhere, we're also going to have the convection-type situation setting in. And I wrote down in my notes that under the absolutely best condition you could imagine, and none of us working under that condition, you might get up to something like 100 seconds for a convection-free environment. But a much more reasonable time is less than a second. Most of the buildings we work in have lots of vibrations in them. And assuming you're not going to go and put a chemical cell on a vibration-free table and things like that, like a good laser spectroscopist would do, you are working in a regime, let's say, at the most a half a second before other things start to take place. And in fact, that's one of the reasons why this current here, that should go down to zero. It's not going to go down to zero, because there is going to be a convective component that comes in before we could reach zero, as well as the fact that it's not really a plane or electrode that I'm dealing with right there. Yes? In the semi-infinite, you're basically assuming nothing comes in from the solid. No edge effects, correct. And there is in the spherical one, you basically have that. So I would guess in the semi-infinite, you should get a constant current term for stuff that's coming in from the side, too. And so that would also affect you. Absolutely. And that's why I'm saying my electrode, although planar, is not infinitely planar, right? So there is an edge here. And of course, the question is, what's the ratio of edge to area? And one can try and make that large, but you run into a problem when you do that. And how do you make that large? You make the electrode large. And then relatively, the area is large compared to the electrode. But as you do that, you expose more of the electrode to these other convective effects. So on the one hand, you like a small electrode so that you don't generate strange gradients across it. And on the other hand, you like to minimize your edge effects. And that is certainly another reason why this doesn't go down to zero. Now, in addition to gradients that you might sample because of vibrations and whatnot, one should be aware that even if you went to all the trouble of vibration isolating your electrochemical cell, you still would have a gradient issue in that as the electrode gets bigger, you do not drop your potential evenly across the electrode. That is, any geometric imperfections in the electrode, and there will be some at some length scale, will cause an imperfection in the electric field. And so you will have an electric gradient that moves across your electrode. And these gradients actually can be rather large variations on a typical scale. So even if you go to, say, an electrode that's 1 square centimeter, not particularly large, it's fairly impressive the size gradients that one can run into. Where this is an interesting issue is in automobiles over the last two years, we have come across these wonderful rear view mirrors that automatically darken when the lights get behind. You have a little solar cell and it detects some headlights coming up behind you. And when that cell sees that, it tells the smart mirror to darken. And the smart mirror is an electrochemical cell. And you really don't want to have a liquid in your mirror because that would evaporate, leak out, things like that over time. So it's a solid state electrochemical cell in which you have a redox species that is reversible and changes color from a fairly clear state to usually a blue sort of state when you do this. So things like bronzes, like tungsten oxide bronzes are material that's used for this. There are some conducting polymers that have a nice color change between a golden color and an intense blue color that are used for this. Impression blue is another material that can be used for this, going from a yellow to a blue color. Anyhow, you have a fairly high resistance in such cells because you don't have a nice solution electrolyte. And you have a mating problem between the electrolyte and the electrodes, which doesn't help in terms of the electric field. And as a result, when they first are making these cells, when the mirror changed color, it didn't change that nice homogeneous color that one expects. You had a blue part of your mirror over here and a regular part over here and some third color in between because the distribution of the electric field across this large area now, your rear view mirror, was not homogeneous. And they had to work very hard to get a homogeneous electric field. And if you want to spend a lot of money and take apart your rear view mirror that does this, you'll see that they've made electrical contacts all the way around the perimeter of the field. They don't have an electrode like you would make where you have one wire coming off one end and then a big potential drop as a result across the mirror or the electrode. But they're connecting at 100 points around the mirror in order to try and even out that electric field. So that is another issue to keep in mind. So small electrode, but there will be edge effects. Even having said that, there's the fit to the Cotrell equation and it's a great fit. So in other words, the edge effect here, this is an electrode that's about a 10th of a square centimeter. The edge effects are negligible. OK, well, before we leave this idea of the reversal inertia system, just one last comment to make. And that is quite obviously the current here is smaller than the current in the Cotrell case because the concentration gradient is smaller. Of course, I've worked out the concentrations here for you. You can plug back in and then you'd have to reverse transform it. If you did that, hopefully it won't surprise you, discover that the concentration, the oxidized species, at the electrode surface is equal to the bulk concentration times this term essentially. In this case, it's a negative sign, 1 minus 1 over this 1 plus d ox over d red to the 1 half theta. And likewise, the reduced species at the electrode is equal to the bulk concentration of the oxidized species times the ratio of diffusion coefficients divided by this 1 plus d ox over d red to the 1 half power theta. So in other words, the gradient you're establishing here is diminished by what the Nernst equation limits you to, as well as the diffusion coefficients. And again, in handling this, we are typically, even though I've written out everything explicitly here, we are typically assuming that this ratio is a number close to 1, like it's more complicated if that's not the case. OK, let's now take a step to the next more sophisticated situation. Instead of instantaneous kinetics, which is what essentially I've been telling you here, let's assume that we have ray constants that we can observe. So let's assume that we are no longer diffusion limited, but have a charge transfer component. Yes? Yes, c star. No, c star. c star r is 0. We're assuming that we start with no r in the system. So this is all related back to the bulk concentration of the oxidized species, which is your reactant. And I would like to erase some board now. So do I have permission to erase something over here? OK, we'll leave ctrl up, and I'll make a slight change in my chemical reaction. And that is the size of the arrow. So I no longer have the reversible case, but I'm assuming there is some difference in the kinetics. And I happen to have drawn this as k forward being larger than k back, but it could be different. But now that comes into play. So how's that going to affect our situation? Well, to start with, let's assume that k forward and k reverse are similar in size. And I'm drawing a distinction here between the words similar and identical, obviously. If they're identical, then we're nernstien, and I have nothing to work out. So they're not the same number, but they're the same order of magnitude. And that would be the situation. When I had that situation, that's the quasi-reversible case. Assuming they weren't trivially slow, I mean, you could have them identical and really slow, and then you wouldn't be nernstien. Nernstien does have the assumption that things are happening so fast that we can't see it on our time scale. But they can be very different. Yes, very different and very fast. They would be nernstien, yes. Here I have two rate constants that I can observe. They're in the time regime I can reserve. They're not identical, though. That is, the nernst equation will not, at any time, give me necessarily the right result. I want to be able to use a kinetic statement to get to the correct result. And so how do I handle this? Exactly the same way. I start off with fixed second law and say that's going to be controlling. And then I go and say I'm going to have a situation where I am going to conserve mass of the same set of boundary conditions that I just erased over here. And that will lead me to this solution right here. So we're starting off with this general solution. So I don't have to do that again. That's right at that point. Now I have to pick a new boundary condition, and this is what we're just discussing really. My boundary condition now is going to be that the current density is now dependent on time. So I have a current density that still changes as a function of the diffusion gradient evaluated at 0. But that now is going to be equal to the forward rate constant times the concentration of oxidized species at the electrode surface minus the back rate constant times the concentration of reduced species evaluated at the electrode surface. So just a standard first order kinetics. I am assuming it's first order here. That's happening, which if it's something just simply hitting the electrode, it ought to be. I am hiding in that statement something that you already know. And that is that these two rate constants now are potential dependent, as well as temperature dependent. So in fact, when I write kf, what I'm really am telling you is that there is heterogeneous charge transfer rate constant, k0. And we take that times the exponential of minus alpha little f, electro potential minus the standard redox potential. In other words, the over potential. And we have our value of the rate constant, where again that little f is going to be nf over rt, where the big f is Faraday's constant. And likewise for the back reaction, I have the same self-exchange rate constant because that's the other half of the couple exponential of 1 minus alpha little f e electrode minus the thermodynamic redox potential. OK, in doing that, I've done something really important here that at least took me a long time to understand. And that is, Tafel was probably the only person in the history of the world to find a reaction that was purely charge transfer limited, where nothing else mattered. And all you had to do was look at the charge transfer dynamics at the interface. For the rest of us, we have to understand that even when we're looking at a charge transfer limited reaction, there is a diffusion component. There's a mass transport component. So in other words, what I haven't written here is I have not written, well, I have I guess, but no, I haven't. I haven't written down fixed second law. So I'm saying, look, it's diffusion limited and it's charge transfer limited. And even for a reaction that is charge transfer limited, the rate that things come up to electrode matters. It's really coupled together. So this becomes a boundary condition on my diffusion limit. And that will, for all intents and purposes, always be the case. Well, if we take that as our boundary condition and we take this as our general solution to fixed law, then we come up, and this is going to be wonderful in terms of the photography I just realized, let me just reproduce those equations over here. Our solution is this general solution. Nothing has changed. OK, there is your time out. So we have our general solution right here. I'm missing an equal sign here also, I think. Minus and a minus sign. Thank you here. And on the next one also, right? See, my board copying ability is next to nil. OK, I think I got it. Thank you. So what that's going to give us then, when we plug in this new boundary condition, I keep on pointing the top of the board here, I mean both the top of the board and this section over here, is a solution for A of s, which is equal to minus k forward divided by the diffusion coefficient for the oxide species, the 1 half power, times bulk concentration of the oxidized species divided by s times h plus s to the 1 half. Where h is defined as k forward divided by the diffusion coefficient to the 1 half power plus k backwards divided by the diffusion coefficient for reduced species to the 1 half power. And again, in doing this, I'm also hiding the fact in here that I have this potential dependence. So it's all there, but it's implicit. So I can go and I can plug that in. If I do that and I back transform it, OK? And to make my life simple to start with and yours, let's just assume that the reduced species at time 0 is non-existent. Then I'm going to come up with a statement that the current is equal to n, Faraday's constant, times the area, times the forward rate constant with its potential dependence, times the bulk concentration of oxidized species, exponential of h squared times time, air functioned, and it's multiplied by the air function, of h to the 1 half. So now you'll notice we have a different functional form here. May not really pop out at you what that functional form is, but it certainly isn't a simple to the minus 1 half dependence associated with it. If, by the way, you wanted to have some reduced species around, if it's not true that it's equal to 0, then the statement becomes I of t equals n f a times k forward bulk concentration of the oxidized species minus k backwards, bulk concentration of the reduced species, which now exists, times the exponential as written above, times the air function as written above. So you see life gets even more exciting. OK, now what do we do with this? We realize that there is a potential dependence that's hidden in this thing, and it shows up here, it shows up here, it shows up in these h's because the h's depend on the rate constants also and their potential dependence. So there's a very intricate play of potential that's occurring here. However, once I say I'm going to do a chrono-amperometry experiment and I'm therefore going to jump from potential x to potential y, I've fixed my potential. So for a given experiment, the potential jump is a constant and therefore k forward, k back, and h are constant. So although there's a general potential dependence, there is no potential dependence on a specific chrono-amperometric experiment once you've said this is a potential jump that I'm going to do. Now you'll notice once I've done that, what's going to happen? If I make a potential jump, I'll get a maximum current somewhere and that current will fall off. And that maximum current is equal to NFA, forward rate constant, times the bulk concentration. I can't get a current larger than that in this system. It is falling off, but not exponentially. It's a yes, something that looks sort of exponential. Yeah, right. So we're going to have a curve here that falls off faster than the curve I've been showing you on the PowerPoint. So these currents will always be below this. And unlike the Coutreux behavior or the Nernstien behavior, I now have a maximum current. It doesn't shoot off to infinity, but I have some value for i when I carry this out. So the curve is obviously lower at time equals 0, but then it continues to be lower because of this error function term, as well as this rate constant term as time goes on. And there's a question, yes? The C-I star of reduced species. Right, because up here for this equation, there is no reduced species. But I'm saying now, if you want to throw some in, we can rewrite it. And all it's going to do is add another term. So is the argument of the exponent supposed to be positive or negative? The argument of the exponent is positive. I did that right, yes. Just double-double check, yes, because the negative term comes in right here when I go back, so it's positive over there. But this one is going down. So again, you have something that is going down, but it doesn't exactly look like that. Now by i, you're probably not going to notice much of a difference other than for the short times, you don't have as high a current. But of course, you still have your non-fair day at current convoluted on top of that. So whether you can pick that up or not is a bit of an issue. So how might we handle this in terms of getting some information out of it? One way you could do this is you could say, well, the current really can be thought of as a function of things that have to do with charge transfer, like the rate constants. And a function of things that have to do with mass transport or diffusion, which gets back to this idea that even if we're charge transfer limited, we can't ignore diffusion. Both things are happening. There's a convolution here. OK, now, did I just do something that was fair? You probably all agree on the charge transfer parameters here, things like the rate constants, the number of electrons. That all sounds fair for charge transfer. But is it fair to include H as a mass transport sort of phenomena? After all, look at H. It has diffusion coefficients in it. That certainly has to do with the transport of material. But it also has rate constants in it. Well, one way of answering this is saying, well, if you look how you're going to do the experiment, you sort of finesse it. And that if we're looking at short times, then we can always make these numbers as small as we need to by adjusting the timescale rod since we're multiplying by H. That really doesn't totally answer the question I'm getting at. The answer is more on the potential dependence of these parameters. And that is, I can always find a potential since I'm charge transfer limited, where these numbers are sufficiently small that I am dominated by the diffusion coefficients. So remember, this is not the situation where I'm doing a large potential jump. And therefore, I become mass transport limited. I'm interested in this intermediate situation where I'm doing a potential jump where the rate constants are important. So if I pick my potentials correctly, my statement over here is a true statement. And it turns out, if H times 2 to the 1 half is small, I can rewrite this as the current is equal to the standard parameters nfa k forward bulk concentration times 1 minus 2 times H to the 1 half divided by pi to the 1 half. So fair amount of math going on here. And we end up with this statement right here. And I can guarantee that this condition is met either by choosing my potential step correctly or my time correctly. And it's a potential step. That's the one you're going to want to control because you don't have that much control over time because of the square root dependence right there. And on top of which, you've got to run your experiment over a reasonable amount of time. So by picking the correct potential step, I can get this situation. And then you see, if I make a plot now of i versus t to the 1 half, then I predict a linear plot. And out of the intercept, I can extract the value of kf. And of course, I'd have to do that for several different potentials now because kf is potential dependent and I need to get that piece of information. So from this approach, I have the opportunity when my rate constants are in the right regime so I can control H of extracting the rate constant. Now let's make life worse or better, depending on how you want to look at it, and say, OK, we're kinetically limited, but now let's move out of the regime of crazy reversible and go to irreversible. That is, let's make one of our rate constants much larger than the other rate constant. And let's see. We'll leave that up there. Guess that is true still. This, OK, so if we do this and now have k forward much larger than k back so we're irreversible, and we run through the math again, start with these equations, put in our kinetic statement which only has the concentration now dependent on k forward times the concentration. Just first order in oxidized species, I should say the current, excuse me, dependent on k forward. Then we come to the conclusion that the current, after we've done all this wonderful Laplace stuff, is equal to nfA, bulk concentration of oxidized species, forward rate constant with its hidden potential dependence, exponent of k forward squared times the time divided by the diffusion coefficient times the error function again of k forward times the square root of time divided by the diffusion coefficient. Similar functional form, but you'll notice now that we explicitly only have to worry about the diffusion and the forward rate constant. And the key point to make here, the reason to show you this equation, because I doubt that this has any explicit meaning to any of you, is that even in the irreversible case where charge transfer is limited by the forward rate constant, the diffusion coefficient still comes to play. There still is a diffusion component in this reaction. You can't say there's no diffusion. It's just charge transfer limited. OK, how are we going to handle this? Because this is a bit of a mess. We're going to handle this by introducing a concept that used to be extremely important, but now it's useful, but not extremely important. So the concept I want to introduce is this idea of a dimensionless geometric solution to these problems. And to get there, all we do is we define a new parameter that's got a bunch of these terms together, kf t d0 to the 1 half power. Now, why do I do that? You'll notice what I've done in doing this. I've taken parameters that have a certain dimensionality to them and removed the dimensionality. That is, I've gotten rid of the units. Because k forward is going to have units of centimeters per second, you'll recall. Time would have units of seconds. And our diffusion coefficient, centimeters squared per second. And so when I multiply all that together, I now have a lambda that is unitless. So I've put all my time parameters, if you will, together and made a unitless parameter. So if I now substitute that into my expression over here on the left, I have that the current as a function of time is equal to nf a diffusion coefficient to the 1 half power bulk concentration divided by t to the 1 half power times this lambda term exponential of lambda squared, air function of lambda. So in doing this, not only have I come up with a unitless parameter over here, I've cast it so that the front part of this equation looks like the Cotrell equation. This is nothing more than a Cotrell equation. I multiply by time and divided by time to make that work out. And so this is just equal to the Cotrell result times pi to the 1 half to make it actually the Cotrell result lambda exponential lambda squared, air function of lambda. And collecting my currents up on the same side of the equation here. OK, now clearly on the right-hand side of the equation, I have dimensionless because I've designed it that way. But you'll also notice I've designed it so on the left-hand side of the equation, I have dimensionless. Doesn't matter if I measure my currents in amps or picolamps. On the left-hand side, it doesn't matter if I use hours or fortnites on the right-hand side of the equation, the equation holds nicely. OK, so now I can take this equation and I can plot it. And there's a lot of different ways one might plot this equation. But it turns out in this particular case, the way one gets a fairly useful curve is by plotting the ratio of the currents versus the log of this lambda parameter. And if you do that, you get something that looks like that. And so we have a unitless or dimensionless working curve here. So if it was 1960, and I didn't have, therefore, an Excel spreadsheet that I could plug things into, I could go and find an appropriate textbook or journal article that had this working curve in it. And it didn't matter that the author, when he was testing the curve, used a potentiostat and was looking at things that happened on a minute time scale. And I came along with my nice fancy new potentiostat and could do things on a millisecond time scale. But the curve is the same curve because it's unitless. And so I could come along and look at the ratio of the control current to whatever current I'm interested in. I get that number. I can read off of that a lambda value. Once I have a lambda value, I have a definition of lambda up here that, assuming either I know the rate constant or the diffusion coefficient, lets me calculate the other parameter. So in other words, I would do this a series of different times. I'd get a plot here and I'd extract, presumably, the rate constant given the diffusion coefficient. How would I get the diffusion coefficient? I'd do a very large potential jump, which takes us to the pure control solution and calculate the diffusion coefficient from that data. So I need that large potential jump because I need this current anyway for all time. I get the diffusion coefficient out. I get this current, so I can get this ratio. I do smaller jumps. And that allows me to back out, using this equation, my rate constant. So now I have a way of getting at kinetic parameters. And the convenience of the unitless working curve is quite simply the author of the curve did not have to guess exactly what time scale I wanted to work in and what concentrations I wanted to work in, et cetera. There was one curve for all things, and I could read it right off. Now it turns out, since we all have Excel spreadsheets today, and if you download the special analysis packet I discovered this morning, you can even get the error function, an Excel spreadsheet. You really don't have to do that. All I have to do is say, OK, here's the equation that we're using, and you can go over your Excel spreadsheet and you can kick out your numbers just as easily. So we don't have that reason for doing this any longer. However, it is still a useful exercise in that this shows you what's happening. I'm rationing my current against a diffusion current, and I can see now that lambda, in fact, is the reduced kinetic parameter. And I can see from this over here once more, just to kind of beat a dead horse, that, in fact, it has both a diffusional component in it and a charge transfer component in it. So in terms of just understanding what's happening, this is useful. And in terms of understanding also that the way I'm changing that curve is by changing this rate constant, by changing the potential, hidden over here, you can see that affecting here also. Current goes up, and lambda goes up, lambda goes up, because the rate constant goes up, the rate constant goes up, because of the taffle type of kinetics that we're dealing with. You can, no, what you would, the value for lambda is just constructed theoretically, right? It doesn't, all I'm saying is I can put in, you put in any numbers you want for lambda here, right? And generate a plot of ratio of current to lambda. That's not a real plot, no, it's a working curve, right? It's a working curve. Now, you're going to come along, and you're going to do some experiment, okay? And you're going to get this ratio from your experiment, right? And you're just simply going to read off this plot that somebody else has prepared. It's in the literature, it's in bar or whatever. You would have read, oh, my value of lambda is that. Given that, you would use this equation and calculate something of fundamental interest, okay? But this is just theoretically constructed. This is, yeah, by putting in whatever numbers you want for lambda. And actually, it is somewhat useful in terms of teaching yourself, and I know this because I was playing with this this morning, to actually go and start sticking in numbers and see what kind of numbers. For example, you know that this ratio can't go over one, right? Because I've already argued, hopefully, convincingly that the control current is always the largest current. And so what kind of values for lambda gets you in the right range here, and then what does that mean in terms of this equation? As I said, probably, from a practical point of view today, you would just go to this equation and your Excel spreadsheet, okay? Which you can't maybe tell from this equation, hopefully you can tell from what I just argued, is you can't just do one potential jump and get out a rate constant. You can't even really just do two. One of them is the Cotrell jump that'll get you the diffusion coefficient and the denominator over here, and then you need a series of these eyes, right? Because you have a time dependence over here that you have to be able to evaluate. At different rate constants, that is over potential dependent. So it doesn't do much good for you to walk up to me and say, oh, I had this great reaction, I measured the rate constant, and it's a 10 to the minus two centimeters per second when I do a potential jump of 0.6 volts. What does that tell me? It doesn't tell me anything, right? So I need to do several of these. Okay, let's see. Now, I think we have gotten through most of the situations that I wanted to hit. Okay, so what do we do next? Well, the good news is, I don't have anything else to say exactly about chrono-amperometry. We've gone all the way from the reversible, mass transport limited case to the reversible case to the totally irreversible case. However, there was a problem that was brought up last hour, and that's the next thing that we have to think about, and that is that you have all this data in this region of the curve in chrono-amperometry and you are totally blinded by the non-Ferdaic response. And come back here. Wake up, wake up. That was perfect timing about that, totally blinded thing. Whereas, of course, where you'd like to be working is out here, where you have no RC component. And in this particular case, you have a reasonable current, but bigger numbers would be better, and there's obviously more sensitivity in defining this curve up in this region than out here. So how do you get around that issue? One of the ways of getting around that issue was suggested by Anson and Ostryan. Professor Anson, you know he was working with Professor Ostryan, Robert Ostryan, who just passed away this past year, and they suggested instead of looking at the current as a function of time, that one should look at the number of coulombs, the amount of charge passed as a number of time. And you can see very simply without doing anything what that's going to do, that's gonna take this curve and as time goes on, there are more coulombs passed. So the numbers get bigger as time goes on. So out here you'll have big numbers now where you have good data, and down here you'll have small numbers, which is fine. That is, in the region where you can look at the data, you'll have your most sensitivity. So it's a simple idea, a very clever idea, works quite well. In addition, what they were particularly interested in was not so much the RC time constant, because you can get a handle on that, but other processes that might be happening at short time, which would be totally buried in that current spike that we're looking at. And specifically the process of interest was chemisorption on the electrode surface. If you were oxidizing something and it was sticking to the surface or maybe sticking first and then being oxidized, that all happens very early in the time and you can't see it here, okay? So what you would like to be able to come up with is a technique that'll let you see that. And the way of handling that is simply to go back to these equations and integrate them with respect to time, and now we have coulombs versus time and we're all done. And really, the only one we have to do this for is the original Cotrell equation. That is, one would probably not use things like chrono-amperometry to get at specific rate constants because we're gonna run into a sensitivity issue. So that is, there is no advantage of chrono-amperometry over chrono-culometry and looking at an equation like this. The advantage is in looking at this equation over here. So we just have to deal with the simple case. You don't advance to chrono-amperometry over chrono-culometry, you don't advance to chrono-culometry over chrono-amperometry. They're equivalent. If you have a question like this, it doesn't matter which one you use. To the extent that they're good, they're good. In the extent that they have a sensitivity issue, you're stuck with that, okay? Because you're not gonna be in a region. This automatically pushes for chrono-amperometry the timeline out, right? So you do tend to get beyond the non-phrodetic response because you have a rate constant here that slows things down for you. So your data out here is quite good now and there's more of it, okay? So switching over to chrono-culometry doesn't help you. On the other hand, if you're interested in the early time data and that's this equation right here, then you have an advantage, right? So, and we'll simply get that by integrating. If you had no advantage, if you're using this, you really would want to do it because you're inserting another step which introduces more experimental error and whatnot. That is where the thing we detect coming out of our electrochemical cell is current and somehow we're gonna have to change that into coulombs whether that's done by a numerical integration or a electronic integration, but it introduces more components and hence more errors. So you only want to do this if it's gonna help you. And where this is gonna help you is under the condition where this thing shoots off to infinity and you lose all information. Okay, so we will pick up with that topic next hour.