 OK, we can get started. All right. There is a new problem set for your, I don't know, edification was one that was thinking, or your enjoyment, I think, is the right word here. Perhaps edification. It's on the website or we'll be shortly and it will be due next Thursday. It's a little bit shorter actually, I think, than the last one I gave you. And to the best of my knowledge, it has no fundamental flaws in it like the last one I gave you. That will really cause you to just decide you have no idea what you're doing. So you can take a look at that. Let me remind you, I will not be on campus tomorrow or Monday. So if you would like to chat with me about the problem set or anything else, I am available on Tuesday and Wednesday. But I have to run back to Princeton and think about who we'd like to see in our graduate class next year. Is your office in North here? Yes, I'm now in Fred Anson's office up on the third floor, room 312. So that's where you could find me, hopefully. A little bit more accessible. We were talking about mass transport limited situation last hour. And I now want to focus on that in some detail and not just wave my hands and actually work out the problem. And the specific problem I want to work out is when the mode of mass transport is diffusion. Because that is the most general control potential experiment. We have a quiet solution, and we'll just let things diffuse up to the electrode and ask what happens. Now I'm going to work this problem out in some degree of a specificity. And I've discovered in the world that there are two kinds of people. There are people that just love mathematical equations and really understand the world once they've seen the math behind it. And if you're that kind of person, then we'll be moving in that direction today. And then there are other people that are sort of, and it turns out there are even chemists like that, that are math-phobic or challenged or something. It's actually phobic, I guess, because they want to run the other way when they see an equation. And for those of you who might feel like that, this is one of those things where I just think it's sort of like medicine. It's good for you. So if you will just grin and bear and nod, like you think you know what's happening here, it's really not that bad, actually. It's the sort of thing like there are just certain derivations that everybody in the world should see once. Well, maybe not, but you're going to. Just so you understand, so you can say to someone someday, I know where that comes from. I saw it back in Chem 117, and you just can just bluff at that point. This is not the sort of thing where I'm going to ask you to derive it on your own or work it out for a different set of boundary conditions probably, or things like that. So this is for your edification, and probably not your enjoyment, depending. So what are we going to do? We have a situation where we have, and by the way, I should tell you, where are we going with this? Why am I doing this? Because probably the primary experiment that one does, that's a controlled potential experiment, is a chrono-amperometry experiment. We do a potential jump, and you watch what happens as species react at the electrode surface. And so if all goes well today, by the end of the period, we will have, at least on the board, done that experiment. Now we start a little bit further away from that. We're going to start off asking ourselves exactly how does diffusion work. So to start that discussion, we need to think about this general idea of a flux. And I'll use j to symbolize a flux. And excuse me, but I just remembered, I can't just start on a random board here. I'm sorry. We'll start with j over here to symbolize a flux. And let me draw a little picture first. I want to do everything in one dimension. Actually, there's a real wonderful thing here. We'll work out everything in one dimension. Then I'll turn to you and say, but of course you own an electrochemical cell. That's really not one dimension. It's sort of a three-dimensional thing. And we have to expand this. And we'll talk about that for a minute. And then I'll say, wouldn't life be nice, though, if we could work in just one dimension? And you'll say yes. And then I'll say, so here's how we can set up ourselves so that we can be pretty certain we're always working in one dimension. And my little assumption here will turn out to be perfect. So I have a tube here. It has one dimension. It has some material that can move down this tube in that dimension in either direction. And I'm going to have, say, some little particles or molecules at that point in the tube. And maybe at some other point in this tube, I'll have the same molecules, but perhaps a different number of them because it's really random. And the way these molecules move in this tube is by a random walk. And this is going to be a one-dimensional random walk. So they can take steps that way or take steps this way. And it's going to be totally random, which way they do that. So that's a pretty good picture. I think you're all probably familiar with it of what molecules tend to do. There's this lovely thought that all the molecules in the air in this room might end up in that corner over there through these random processes. And then what will we do? It turns out that's kind of low probability, but this model lets that happen at least just to make life exciting. OK, so we have molecules over here. We have molecules over here. And you'll notice that since they need not, at any given point in time, have exactly the same number of molecules at every place in this tube that I have some kind of a concentration gradient in this tube. And in this particular situation that I set up, you'll notice that there's a higher concentration over here than there is over here. So we have a concentration gradient in this particular example going like that. And so we would predict that this random diffusion is now perturbed by that concentration gradient. That is that in general, there's going to be a net diffusion that is going to balance out this difference in concentration. So I expect molecules on average to move in this direction, even though they're randomly moving. And the way I might monitor that is I might put a little plane in here. And I might simply count the number of molecules that move to the left or the right through that plane. And if I made that plane of unit area, so I guess it's no longer a plane, it's a bounded plane. So I have square centimeter or something like that. And I carried out that counting process. Then the number I counted would be the flux. So the flux is just the number of molecules through a unit area per given time. So we're talking about something that might have units on it of moles per centimeter squared, if that's going to be our unit area per second, for our flux. Now, we are going to want to deal with our standard chemical reaction that can't seem to get away with. And that is an oxidized molecule plus an electron. Do you think by now I could write this reaction out? Going to reduce molecule. So just keep it up there. And so let me put a little j-knot down there so you can see that what I've decided to monitor is my oxidized molecule when I get to the electrochemistry problem. And the way I'm going to do this is much like I do chemical kinetics, where when I look at a rate of reaction, I have the option of either looking at the appearance of some molecule or the disappearance of some molecule. And typically, we look at the disappearance because the molecule that's disappearing is the reactant, and that's what we have control over. And we're going to do the same thing here. We're going to start off with our region, say, of high concentration and ask what happens as it goes to lower concentration. So let me put a negative sign out there to take that into account. And let me suggest to you that it would make sense if that flux was in proportional to the change in concentration of our ox species with time. So I have a constant proportionality here, d-knot. And I have a concentration that can vary as a function of time and position. And I'm saying that changes with time, that will give me my flux. Excuse me, not time. That was wonderful. Position. I've been talking about time all term here. That changes with position. I'm looking at different points along here. That gives me the flux. And we know that our units of concentration over here will be something like, well, actually, I have to be a little bit careful here. Do we know what our units will be? Our units of concentration over here will be moles not per liter. Electrochemists have decided to redefine the world, but moles per cubic centimeter. So 1,000 times smaller volume than chemists normally work with, why have electrochemists gone and done that? That's because the world likes to use units of centimeters squared per second for this quantity over here. So to get the units to work out, you see, we need to be in moles per cubic centimeter for this system. So we have a statement here. In fact, this statement is so important. It defines a flux that has a name. This is Fick's first law. It simply tells us what a flux is. And this is not specific, by the way, to electrochemistry. Anybody that's worked out a transport problem has run into this equation. And you probably can guess if Fick had a first law, there's a reason for that, such as he had a second law also. And it's actually a second law that's the more useful one for what we want to work out here. So how does that come about? So what we're going to do is we're going to take his first law minus j0 is equal to a diffusion coefficient d0 times this change in concentration with position. And we will differentiate that with respect to position a second time. So if we do that, we're now looking at the change in j0 with position. And we will assume, which is pretty good for most electrochemical problems, but not all, it's assumed that d0 is not sensitive to not a function of either position or time. It's just a constant, in other words. Now, the reason I bring that up is it's a mathematical requirement, obviously, over here, because I would like to take this d0 out of my second derivative. But I bring it up explicitly because one of the examples I have to show you is an electrochemical case, not today, but down the road, where in fact, this is wrong. So it's important to understand that it is an assumption. And even though it's an assumption, it's the only case I am aware of an electrochemistry where it happens to be wrong. So it's a pretty good assumption most of the time, but it is an assumption. So we have the second derivative there. And let me make sure I wrote that out fine. Yes, I did. And we're going to just leave that equation there for a moment. Be totally lost as to where we're going, in other words. And I'm going to draw another picture for you of our tube, one dimensional diffusion, and point out that if I were to drop two planes into this tube, say one over here and one over here, and they're separated by a small distance, dx, that one way I might define the concentration and how the concentration is changing in this volume element that I just drew up here is I might look at the flux going through the first plane, which we'll put at position x, and compare it to the flux coming through the second plane, which is at position x plus dx. And further, I might say that that might change with time. At one point, I might look at that, and I might see certain fluxes going through these two planes. And I might look at that a little bit later on and see a different flux going through those two planes. So in general, I could say that if I was interested in the concentration in here and how that concentration changed with time, the time dependence on that concentration, that I could write that as the derivative of concentration with time given that that concentration may also have a positional dependence to it, but I'm only interested in the time dependence right now, is equal to the difference between the flux going through that first plane, whatever my observation time is, minus the flux going through the second plane, think sloppy here, divided, if you will, by the volume element, which is one dimension, so that's simple, right? Change in the number of molecules over here, going through this plane, minus the molecules over here, going through this plane, and then to turn it into a concentration I divide by my one dimensional volume element, which is just dx. And then I could point out to you that looking at just this term right here, that if I take, if I look at that term, that term is really just equal to the flux, not out there, x, plus the change in flux with position times the change in position. What I'm saying is I take the flux here, and I add or subtract from it the change across here times how long that volume element is. So putting that together, and I have the statement then, and you'll probably want to give me another second to get out of the way, I think, that the change in concentration with time as a function of position and time is equal to the negative of the partial derivative of the flux with respect to x. What I need to do then is take the derivative of that with respect to x and multiply it by dx, my volume element. And so that is just equal to minus j x of t dx. That is, if you look at this statement over here and put this in, then this term, and this term obviously cancel out, and all I'm left with is this nonsense. We produce over there, and that simplifies down to this derivative right here. And of course, from first law, the derivative with respect to x of fix first law, I have a statement for that derivative already. So if I discombine this equation and this equation over here, I have a statement now that tells me how the concentration changes with time relates to the change with position. So there you have it. Guess what? That would be fix second law. OK, now it should almost be immediately obvious that this is a useful equation for us. That is, I have been arguing all term that this term over here is proportional to the current for a diffusion-limited experiment, and in fact, for all experiments that have a diffusive component in them. So this is, within a few constants, the current. This tells us how a molecule behaves when it is diffusing down a gradient, exactly the situation that I set up at the beginning of the lecture. And so this is the governing equation. All we have to do is solve this differential equation, and we're all done. No big deal. Now here's the point where I go and throw in the complication and say, oh, but I just worked this out for a one-dimensional case. And the real world is not a one-dimensional case, but we might have a variety of different geometries that are not one-dimensional in our system. And so what we really should do is say something like, more generally, that the change in concentration with time and position differentiate with respect to time is equal to the fusion coefficient times an operator squared, the second derivative of position operator squared, operating on a concentration. So that would be the more general statement to make, where quite simply, of course, for the linear case, my operator is just d by dx. If instead I wanted to go into three dimensions, seems like a pretty reasonable way of describing an electrochemical cell, then my operator becomes the partial with respect to x squared plus the partial with respect to y squared plus partial with respect to z squared. So second derivative of the three different dimensions, that hopefully makes sense. But now let's think about real-life electrochemistry for a moment. What's going to control this geometry? The thing that controls this geometry is the shape of the electrode. It's not really the shape of the cell, because the flux that I'm interested in is the flux immediately adjacent to the electrode. So I don't care how many molecules are going through some arbitrary plane 2 and 1 half miles away from my electrode. I want to know what the plane looks like right before that molecule bounces into the electrode. That's the flux of interest. I'm going to want to evaluate this thing at x equals 0 eventually, where I define 0 as where the electrode is. So for example, if I have a spherical electrode, instead of having a plane of molecules coming up against the electrode, like this would suggest, then I have some other complicated flow or flux up to that electrode. So in the case of a spherical electrode, we have, for example, that this is the second derivative with respect to r squared, where r has a vector in x, y, and z directions, plus 2 over r times the first derivative, turns out, with respect to r. So life gets a little more complicated if I have a spherical electrode. Another electrogeometry that is very common is cylindrical. All of you that are like me that like to say, here's a random piece of wire. I will use it as an electrode, make a cylindrical electrode, as long as you don't do anything too fancy with the wire. But if you just stick a piece of wire in an electrolyte, you have a cylindrical electrode. And again, you have this dependence on r and a first-order term, which now goes as 1 over r. And you could, of course, come up with other geometries of electrodes, and you can make your life as complicated as you want, because you're going to have to change your operator to accommodate that. So if you want to cut your electrodes into little stars, we'll have to come up with a star operator and whatnot. That we'll take that into account. This equation, it turns out, as far as my mathematical skills go, is complicated enough. I'll leave it right there. I really don't want to have to deal with these other terms if I can avoid it. So most electrochemists feel that way. Now, it's true. If you are using, say, a dropping mercury electrode and you have certain time constraints and whatnot that are going to make that drop of mercury look like a drop, look like a sphere, then you're sort of stuck with working out the problem this way. However, if you are working with an electrode that you can make look planar, whether or not it is planar, then you can move yourself to this situation, obviously. And if you can be so far away from that plane that the plane looks like a dot, very small plane, I guess, then you can be in this situation. So the requirement for this situation is a semi-infinite linear diffusion. And every single problem I think that you will see at least worked out for the first cut in electrochemistry makes the assumption of semi-infinite linear diffusion. And then if that fails, because you can't set up your cell, so you meet that requirement, then you start adding in other terms to make life more complicated. But the idea is simply going to be is we want to look at the problem in an electrochemical cell that is sufficiently large that out in the bulk, essentially we have bulk electrolyte, and we're far enough away that we see that all going in a parallel manner to the electrode surface. Another way of putting this is the area of the electrode compared to the area of the cell is such that we can ignore edge effects. There are wacko diffusion things happening at edges. And to the extent that we can say, well, yes, there's something funny happening there, but the amount of edges small compared to this plane that everything's going to hit against, we can ignore the edges and make our life much similar. So that's really what semi-infinite linear diffusion is saying. So we'll start there, and then from time to time we'll have to throw that one out, depending on whether or not we can get ourselves in that situation. Just to show you, actually, what's really happening here with these different operators, I found this very nice little website this morning out of Johns Hopkins, where they let you, it's a great little site, you might want to, just a lot of fun. I was playing with it this morning. Tom's not going to be playing with it, caught me actually. You might want to write down that web address if you like to play with these sorts of things. Right up here, JohnsHopkinsUniversity.edu and the rest right there. We're offline right now, but I believe it's going to work just fine. What this site gives you is a possibility, I'll show you in a moment, of first of all, establishing a medium that has different regions of diffusivity in it, high, low, or zero. An electrochemical cell, typically, the diffusivity, which is control light things by viscosity, temperature, things like that, is constant. So I'm keeping that the same everywhere. It's green everywhere. And the next thing it lets you do is define a chemical gradient that's over in this square over here. So what I've done is I've said, OK, we will have over here an electrode. And I'm going to consider now my ox molecule out on solution, and we're going to make the red molecule. And we're going to follow the red molecule. So I define the concentration of the red molecule over here as being zero to start with, and over here as being a constant. And that's what this color means. I'll show you in a minute. And then we let this simulation run. And there's a little counter that's supposed to be there. It just appeared, but OK. That's not too critical, I think. Let's see if it runs. It does run. OK. And we'll just stop it there for a second. So what's happening here? You'll notice a scale on the right-hand side. If there's zero molecules around, it's white. If there's greater than 40 molecules around, it's black and various numbers in between. So we started off making that stuff at the electrode. It turned black instantly. So we made a lot of molecules there. And as time goes on, you'll notice that there's this wave moving out from the electrode. So we're far away from the electrode. There's still no molecules out there. We go a little closer to the electrode, and we see 0.01 molecules per unit area, and then one molecule per unit area. And eventually, we get up to this very high concentration per unit area. And as we allow this thing, of course, to continue, eventually we would fill up the whole cell with these products. But given that this is a large distance, typically, in a cell, you're never going to get to this situation. That is you're not going to be able to maintain a gradient across the cell unless the cell is very, very small. If the cell is very, very small, then I don't have this situation that I want over here of a semi-infinite linear diffusion. So I'm assuming that there's a large spacing over here, and that gradient falls apart before I get to the other side. So that would be the case for linear diffusion, certainly. And again, if those electrodes are very far apart, so I have a big cell dimension for semi-infinite linear diffusion, you can see how that works. Now we can go back to that, and we'll put in a concentration field here. So again, I'm going to go and set a concentration that will be far away from my electrode over here as being at 0 of the product. I'm not going to set it. Let's try that again. Rectangle constant equals 0. This is fun, but it's sort of hard to do the graphics. Oh, good. OK, so there we have it far away at 0. And now I'm going to take an electrode where I again have a material generated. So I'm going to generate it at a constant here, high concentration. But instead of using a planar electrode, I'm going to go over and generate a spherical electrode. And now we can run that simulation again. OK, so the first thing that happens is probably more or less what you would guess would happen. And that, as you'll notice, first of all, the material is not diffusing out from the electrode in a planar manner anymore. But it's just diffusing out in this spherical manner that's given by the shape of the electrode. So just changing the shape of the electrode changes the concentration gradient across the cell. We don't see anything, though, too funny. Now, if I go out radially from where I initially put that electrode, then I look at slices that more or less look like the linear case before I see this drop off with distance. But you have to go out radially. Now, again, if you design your cell so that there's a short distance between the boundary, that is the edge of the beaker, and your electrode, things can get pretty complicated in this case. So now I have a very hard-to-define geometry there because I'm hitting the edges here, and that changes the shape. So it's not that I have it spherical the whole time anymore. It changes over time. Now, this would be taken care of, of course, by using this operator and taking into account a set of boundary conditions, if you wanted, that described what the cell looked like. But in general, we're not going to want to do that. So quite simply, when we change the shape of the electrode, we change this diffusion field, and life gets exciting. So we'll leave that there. You can go and you can play with all kinds of other diffusion fields if you want in there. And we call kinds of pretty patterns. So that's one thing that we need to be concerned with. Am I still working on these boards over here? Are they photographed? The ones on them are not photographed. So let me put one more thing up there before we tell it to smile. We have a differential equation here, and we have to solve that differential equation. And you all know that if we are going to do that, we need some more information. So we need some initial conditions, some boundary conditions, and whatnot. So let's assume we're not going to have to deal with these things to start, but just with the semi-infinite linear diffusion. What are we going to use then potentially for our boundary conditions and our initial conditions? The obvious initial condition, assuming semi-infinite linear diffusion. And this is going to be nice. So this is why I want to assume this. I guess I should write in big letters, assume. If I assume that, that means I never run into the situation I ran into at the end of those simulations. That is way out in solution. The concentration doesn't change. It doesn't matter what I do. That's easy to set up in an electric chemical experiment, but you can work hard and mess that up if you wanted to. So in the limit of x going to infinity, the concentration of the oxidized species is equal to the bulk concentration. I use that asterisk to indicate that. Or I'll also let it be equal to zero. It doesn't matter what value of t I put in there. That is, either I start off with all of my molecules, say, in the oxidized state. Or they don't have to all be in the oxidized state, but some concentration in the oxidized state. And I'm going to just make sure I run my experiment for a short enough period of time that when I look far away from the electrode, that's the concentration that I see. Given the currents that one normally uses in an electric chemistry experiment, it's hard to avoid this situation. That is, typically we don't run enough current that we could change a large number of molecules and make a change in that bulk concentration. It would take so many hours to do that, in other words, that any assumption about a concentration gradient that existed in my cell would have broken down long before I saw a change in this bulk concentration. The other possibility is, I might say, well, I'm just going to put the reduced molecule in at time equals zero. And so there will be no oxidized molecules around. And I'm going to run my experiment for a sufficiently short period of time that if I look far away from the electrode, I won't see any oxidized molecules. Now, both those statements, I point out, you do not mean that if I look right adjacent to the electrode that I don't see a concentration change. In fact, in general, I'll see a massive concentration change right at the electrode. It's far away from the electrode that this is important. So out in solution, that might define what's happening. Close by the electrode, we might use, excuse me, that's an initial condition, not out in solution. That's initial, right? For all times, but it starts at equals zero. Now, in terms of a boundary condition or a set of boundary conditions, there's conditions that have a spatial dependence to them. Nope. You know what? I have totally messed this up, and I apologize for that. I did write down a boundary condition here. So I'm going to put that underneath. Far away, it's going to be what I said. OK, the initial condition now would be independent of what is going to happen far away from the electrode. When I'm looking at the starting material, which all define as ox, at any position in the cell, at time equals zero, excuse me, that'll be the bulk concentration, again, or zero. Either I put some stuff in that cell, and it has a concentration, or I don't put any stuff in the cell, and it doesn't have a concentration associated with it. It doesn't matter initially where I look in the cell. This is before I started the experiment. So it doesn't matter what the value of x is. So we have what's happening out in solution here. We have an initial condition, and now we have a situation when we look at what's happening at the electrode in terms of a boundary condition. And the most general statement I might make there is that the concentration of the oxidized species at the electrode, so at position zero, for any time, could be a function of the electrode potential. Since this is a controlled potential experiment, that seems like a pretty safe statement to make. And then we could ask ourselves, OK, exactly what functional forms might we use here? The one I like the most is this one. It might just be zero at all times independent of the, it's not a bad one. Makes life really simple. That is, if I have molecules in solution, and I'm not going to move to a potential where they're either oxidized or reduced, then the concentration won't change. Now, we've already talked about the situation where only a non-fair day at current flows. And one might use that to probe the interface, and then you could use a condition like this to solve this situation. If it doesn't change at the surface, it doesn't change at the bulk, either. But yeah, this is normally zero by simply setting potentials so that the molecules might be there, but they're not electroactive. Yes, for example, let's set up an electrochemical cell in acetonitrielectrolyte with some tetrabulomonium perchlorate supporting electrolyte. And let's not put anything else in. And let's say we're going to limit our potentials to potentials that neither oxidize nor reduce the acetonitrile. Now, I have molecules around. I have concentrations. But as long as I limit myself to those potentials, this is a true statement, that there's molecules around. They're just not electroactive. Something perhaps a little more exciting, since most of the time we want to have electroactive molecules. Let's move it over here. So we are continuing with our boundary conditions. We may say that we're under Nernstien control, which we've already agreed is something that could happen. And so the Nernst equation would always define what the concentration of oxidized and reduced species would be at the electrode interface. And so all I need to do to cast that in this format is rewrite the Nernst equation in terms of concentration instead of potential. So for that case, the Nernst equation tells us that we will have a ratio of oxidized to reduced, which is equal to the exponential of N Faraday's constant over RT times the difference between the equilibrium potential and the standard redox potential for the system. And I've done a change of sign here. And I've put this thing over to balance that just so you can see the subtlety there. So this would be for reversible system. Now this is Nernst always assumes equilibrium. So we're at least at equilibrium right at the electrode surface. So in other words, what I'm saying is this potential here that I've labeled equilibrium is the potential of the electrode. That's the assumption that makes this one work. Nernstien equilibrium. And again, we get there simply by making sure that the charge transfer processes are fast compared to the diffusion processes. And we're in that situation. Another possibility, because we're not always in that situation, is that we might just simply conserve mass. And we might say that the number of oxidized molecules that come up to the electrode, so there's the flux of oxidized molecules at the electrode surface, plus the flux of reduced molecules at the electrode surface, better equals 0. That is, a molecule goes into the electrode and it bounces off the electrode. And it either comes out as an oxidized molecule or a reduced molecule. We don't lose any molecules in the system. That's just mass conservation. And you can see a condition like that is going to be true independent of whether your mass transport limited or charge transfer limited or whatever limited. So this is nice in a specific case where you know you are mass transport limited, that you have very fast kinetics. But this is a more general statement that can be used when you don't know what the situation is. Now, having laid out that problem, we need to take a brief interlude here and do a little bit of math. So the problem we have now is that we have a fairly messy differential equation right here. We have some boundary conditions that can make it even messier. And we need to solve that differential equation. And so what we would like to do is turn it into an algebraic equation instead of have to do the calculus. And the way we can pull that stunt off for this particular form of equation is use something called the Laplace transform. So just to remind you, in case you're really not on top of your mathematics for chemist things, what we're going to do is the transform is carried out specifically on the time parameter. So we're going to turn our time parameter into a transformed s parameter. And the transform, the Laplace transform operating on some function, f of t, is defined as the integral from 0 to infinity e to the minus s times t times the initial function integrated with respect to time. Those of you who like other sorts of French transforms like Fourier and whatnot will notice this is a fairly similar transform to that same sort of idea, and this is going to give us a new function which we will call f bar, and it will now be a function in s. This happens to be a linear transform. That is, if I carry out this transformation on the function a times f of t plus b times g of t, then that is the same as a times f bar of s plus b times g bar of s. And now I could do something really nasty like, say, go home and come up with your 10 most favorite functional forms and figure out what Laplace does to them. But let me just pick some random ones and show you what it does. So we'll start off with f of t. And we're going to turn it into f bar of s. So the first thing we might run through this little machine is some constant a. And if we do that, we'll find that what comes out the other side is a divided by s. So constant just introduces this new variable. We might try something like e to the minus at, where a again is a constant, and we'll find that that comes out as 1 over s plus a. We might try running just a simple transforming t through this that comes out as 1 over s squared. And then just quite arbitrary, I'll pick out of the air the function f of t equals pi t to the 1 half. Excuse me, t to the minus 1 half, yes. And if you happen to pick that one, you'd see that that is 1 over s to the 1 half. And why do I arbitrarily and randomly pick that one? Because that's the one we're going to run into in a few minutes. But before we run into that, let me point out another attribute of these transforms. And that is that if we take the transform of a derivative as a differential equation, we will move into the realm of algebra as I suggested. That is, if I ask Laplace to take this derivative and transform it, then what I get back is my variable f times f of s minus f, not f bar, but f evaluated at 0. So I've taken this differential equation and turned it into an algebraic equation. And likewise, if I did the same thing on the second derivative with respect to time, then I come up with s squared f bar of s minus s f bar evaluated at 0, excuse me, f evaluated at 0, minus df of t dt evaluated at 0. So I went with a little bit of a derivative there, but it's only evaluated at one point. And then the final important aspect of all of this is that if you take an inverse transform of a Laplace function, you get back what you started with. So I can do the opposite, and that gives me back my starting function, all which is to say, so I can take my statement of fixed second law, I can Laplace transform it to turn it from this messy differential equation into, I guess, a messy algebraic equation. And then when I'm done doing the algebra, I can back transform and I have a solution to the differential equation. OK, there's your math lesson. That was good. You are already now to do chrono-amperometry with that. We have this captured already. But you do not have, you have to get out of the way here, because you don't have this captured, right? And I would like to erase it. I guess I could, well, I can come over here. You don't have this either. And I just walked in front of it. Very good. I was bound to screw up sooner or later. There it is. You want to try again? Yeah, it has to wait. I have to wait. All right, so can I, I'm going to walk across your shot here for a second, since I've already messed it up. OK, we'll work over here. OK, so we're going to do chrono-amperometry. The chrono-amperometry experiment, I'm not going to erase. It's the simplest way form that one could think about applying to a cell for potential control. We're going to start at some potential, you not, and we're going to, at some point in time, jump to another potential. And the point that we make that jump at in terms of time we will define as our zero time. Presumably, we would like to set this up so that between this potential and this potential some chemistry happens. OK, what sounds pretty obvious, but I find over and over again is violated by people in the laboratory, is the first potential you pick over here should be a potential where nothing is happening. That is, you wouldn't want to start off with five amps of current going through your cell and jump to some other potential. OK, the idea is that, remember, I just erased it all, but that at time equals zero our bulk concentrations are well known and constant everywhere. That's the condition that I want to imply here. And that later on, still far away from the electrode, the bulk concentration hasn't changed. So the condition for E0 is you find a potential where no current is flowing after the non-ferdaic component has been subtracted out, as there is no redox process. And then if you're interested in exploring a redox process, obviously you would jump to a point over here where you've picked up enough potential that the process may occur. If you're interested in a non-ferdaic process, then that's not a requirement. I want to add a second requirement to start with. And that is, let's pick our final potential such that we guarantee that this process is diffusion limited. That is, let's make a big enough jump so that we are certain that we will convert every molecule that hits the electrode to product. So if we're under control of the Nernst equation, that means that we'll have a big change in delta E over there. And therefore that ratio will change a huge amount and will be all that product. If we happen to be in a charge transfer limited situation, we'll make the jump big enough so that the rate constant, which remember is potential dependent, is big enough that once again we will convert all our material from reacting to product. So that might be a lot more. We might have to apply an over potential that's a lot more than what the Nernst equation tells us to. But that's the assumption we're going to start with here because that makes for the simplest possible problem. And what do we expect then when that happens? Well, when we do that, if we monitor current, which is the idea here, we expect that the current will jump up and then fall off. So at the point we make that jump at time equals 0, the current jumps up and it's going to fall off following some functional form here. And we've already discussed the fact that actually the equation I'm solving here is only going to take care of part of it because I will be ignoring the non-ferdaic part now. This is just the part that has to do with things moving in solution. And so there is obviously hidden in this a non-ferdaic part that we will ignore. OK, so how do you go about handling this problem? Well, we're interested in current and we know that that current already is proportional to the change in concentration of the oxidized molecule. We'll assume that we're doing ox to red. That's at a function of time evaluated at x equals 0, the electrode surface. And we have this beautiful fixed second law now that tells us that the concentration of this molecule everywhere and for all time must obey these equations. That is for all times where we're diffusion limited. And so for the situation I've set up there, what are our boundary conditions? Well, we will go to the semi-infinite boundary condition. So we will say that in the limit of being far away from the electrode, the concentration of ox is always equal to the bulk concentration. We'll also take an initial requirement that at time equals 0, everywhere in the cell I have the bulk requirement. So those will be my boundary and initial conditions. We want to solve this now so we can take this equation and we can Laplace transform it as well as Laplace transform our boundary conditions. And I'm free to erase over here now. Thank you. OK, so our statement becomes then that s times the Laplace transform of concentration with respect to s and t minus the bulk concentration, which does not have a dependence to it, is equal to the fusion coefficient times the second derivative of the concentration that's been transformed. We can take now the boundary. So there's my Fick's second law, Fick after Laplace, OK? For my boundary condition, excuse me, I continue to go on playing with this and I will come up with C bar of x and s is equal to the bulk concentration divided by s plus some function that I don't know yet in s e to the minus s over diffusion coefficient all to the 1 half power times x plus some other unknown function in s exponential of s over the diffusion coefficient to the 1 half power times x. So the difference between these two terms, besides the a and b functions, as you'll notice, there's a negative sign right there. So I've just run through and done all the Laplace transforms for you to get the second law. OK, and what we're going to now do is use our boundary conditions to figure out what these functions have to be. Yeah, x sub s. No, but C bar over, excuse me, very sloppy. That was a x and s, right? So in other words, C was a function of x and t. C bar is a function of, thank you, of x and s. There's that board penmanship. Taking this, I have to run. I still have to do that, right? Yeah. Everything's in, yes, I was sloppy here, but all my concentrations are in terms of oxidized. Like you said, not typically, but when I transform something, I sort of drop the subscript because it's now not really a concentration in the normal sense of the word. OK, we can now go over to our condition for semi-infinite linear diffusion. And now that becomes a question of what happens as x goes to infinity now of C bar, x and s. And if we carry out the transform there, then that is just the bulk concentration. We saw over there, divided by s. If you go back to your table of functions, you notice if you take a constant and you Laplace transform it, it's the constant divided by s. So I'm just Laplace transforming this whole condition. And now I will argue, based on that statement, if this statement is going to hold and this statement is going to hold, then that implies that B of s has got to be equal to 0 under all conditions. That is, for large values of x, this term over here automatically goes to 0 because of the negative sign that I have right here. This term blows up, but I have this constant that comes out of the fixed law. And so I'm going to make those two statements true. We've got to get rid of this function right there. And the way to do that obviously is to get rid of B of s. We also know that right at the electrode surface, we have a 0 concentration for all times after the jump. I think that is another one of our conditions. Remember I said we were going to set up this jump so that this reaction happened as quickly as needed to, so you would never be charged transfer limited. That is, the molecule hits the surface and it's gone. So as soon as we do the jump, we turn all of our oxidized molecules and reduce molecules right at the electrode surface. And so from that, we have that C bar taken at s equals 0. s is equal to 0 also. It's doing the transform. And therefore, what do we have? We have C bar. Let's be consistent here since I've dropped my little knots. Of x and s is equal to C star divided by s minus C star divided by s e to the minus s over do not to the 1 half power. That is, I have, based on that assumption, I come to the conclusion that A of s is equal to this. You OK with that? Now, we now have our inverse concentration dependence totally under control. And we have a choice to make now. We could back transform this into concentration and then relate concentration to current. Or we can transform our current into this s parameter and go all the way to current and then in s and then back transform into the time domain. So what I'm saying is what we are after is current as a function of time. And let me just write that as n divided by nfa. Since we know that that parameter now is equal to the derivative of the concentration with respect to time, right? This we've already seen and agreed upon for a diffusion-limited case. And I should point out explicitly, we're talking about the concentration at the electrode surface. And so that is the same thing as saying that the transform of the current in s divided by nfa is equal to the diffusion coefficient times the transform of the concentration differentiated with respect to x and evaluated at the interface. So I've just written fixed first law, in other words here, in this Laplace transform. Diffusion coefficient is left out, above here. This is just a kinetic rate statement. That just says the current is equal to the rate that molecules are oxidized or reduced at the surface, right? This statement is true for all cases of electrochemistry. Yes. Yes, yes, yes. C star and C star not. I was sloppy about that, yes. These are this. Doesn't matter if I put that. They are not. In fact, I'll get rid of it, so I'll be consistent. C star is the bulk concentration, whether or not it indicates it's the oxidized species. All we have is the oxidized species in the cell at long distances from the electrode. Other questions on this before I jump to the next point? We're almost there. Hang on for 10 more seconds, and we will have a statement. So what do we have down here? This says that if we simply take the derivative of C bar with respect to x, we're home free, right? And we have a statement of what C bar is. So all I have to do is differentiate this with respect to x, and this is not a bad one to do, given that I don't have this term and I know what this term is. And we are in good shape. So if I do that, take that derivative. So let's see. Let me call this equation one. Let me call this equation two. And let me call this, there's no room over there, equation three. So what I'm suggesting is we can take equation two and we can solve that by simply taking the derivative with respect to x of equation three, and then we're done. If we do that, then we have i bar of s is equal to nfa, just moving this to the other side of the equation, diffusion coefficient to the 1 half power, or the oxidized species times. And I am going to write explicitly now the bulk concentration of the oxidized species divided by s to the 1 half down here, equation two, probably. Well, yes and no. Yes, if you go back to what I wrote at the beginning of the hour, but the no is that we have this sign convention, which gives us two minus signs. The one I wrote at the beginning, plus the one I wrote the other day, and so no. That's in case the math is not sufficiently confusing. We want to make sure that no one knows what's happening. So all we need to do now is back transform this. So we take the inverse Laplace of this thing, and that gives us the statement that i of t is equal to nfa, diffusion coefficient to the 1 half power, bulk concentration divided by pi times time up to the 1 half power. That's why he put that function on that random table of functions. Now remember, the situation that we did this for is for a potential step that's big enough that whatever's happening, the molecule becomes instantaneously reduced when it hits the electrode surface. So this is a limiting current or a limiting condition. And again, that condition is that at the electrode surface for all times the concentration of oxes is zero. That is the equation that describes the chrono-amperometric experiment. That is, I said we would make a potential jump, and we would evaluate that as a function of time, and we would see something that looks like this. And what this is telling us is that the current will fall off as a function of t to the minus 1 half power. I'm going to do that. So this equation is what you're looking for when you do chrono-amperometry. It is of sufficient importance that it actually has a name. This is the Cattrell equation. So in other words, if I take a reversible system to make my life simple and I do a potential jump and I jump far away from the redox potential, then I expect the Cattrell equation to be behaved. And it works quite well. What could I learn from this if I did such an experiment? Well, assuming I knew a potential where there was no electrochemistry and I knew the potential to jump to where there would be mass transport-limited electrochemistry. And assuming I know what concentration I put into the cell, and I know what the chemical equation is, so I know n, then I'm going to get the diffusion coefficient for the species out of this. And this is probably one of the better ways of getting a diffusion coefficient electrochemically. On the other hand, if I happen to know the diffusion coefficient and the concentration, I could back out the area of an electrode. There's a lot of interest here, for example, on electrodes where the geometric area and the actual area are not the same because they're made of centered nanoparticles and thus the actual area is huge compared to what you see when you simply look at the thing. If you wanted to evaluate that area, the Cattrell equation would be a nice way to do it. You could put a known reversible redox standard in there of given concentration. Do your jump. You know the diffusion coefficient and come up with an area. Yes? If you're roughening up the surface of the electrode, aren't you starting from your diffusion? Well, again, the caveat behind here is that semi-infinite linear diffusion holds. So how will I know that or not? And the answer is that if this equation holds, then it's OK. So in other words, if I can make a plot of current versus t to the minus 1 half, and if that plot is linear, I'm good. Now, when it isn't linear, then you're right. And we have to go to some other boundary condition. And that may or may not be worth doing, depending on how desperate you are. But if you're in the case, it's supposed to be so thick that you only made the diffusion. If the diffusion layer is sufficiently variegated, then you get into the situation that we were just describing where, even when you're far away, it doesn't look like a flat surface. But there is a pretty wide range where you can have a pretty rough surface. And if you're far enough away, it looks OK. And then the equation will hold. And it will not just get the planar area, you'll get the actual area. I have a way of getting around that also, but you'll have to come back probably next hour to hear about that one. But yes, that, in fact, that's a very good point. It's an important point. I'm glad you made it. When you use this technique, your best data is sort of in this region. If you go out too far, the currents are so small that you can't read them and you don't get high quality data. There's a lot of error in reading the data. If you go to too short a time, you have this huge non-ferdaic current, which you're going to have to deconvolute and probably can't do it successfully, at least not analytically, because you're subtracting two big numbers here from each other. And that's going to cause a problem. So you're in this intermediate area. And although chronoampereometry has a lot of utility, more than just this equation, probably its biggest drawback is that where your highest currents are, you can't use the data. And it forces you more towards this region of insensitivity where you have good data. So and, of course, actually in the case if you have a really, really high surface area, then you'll have a high capacitance that will increase your RC constant and that will shove it out. So you might get yourself in a situation where you're left with a little tail out here and the quality of the data isn't good enough to do a good fit and come up with good parameters over here. Absolutely the case. However, it turns out that a gentleman who recently retired became emeritus from this institute told us how to get around that problem. So Professor Anson had to develop another technique, which we'll look at very closely related to this. Finally, I don't know why you would want to do this, but I will point out if you happen to know the diffusion coefficient in the area of your electrode and whatnot, you could use this to determine a concentration. Mathematically, it works. From a pragmatic point of view, you probably would never want to do this. First of all, you have the same issues of picking out intermediate currents, making sure you have good data. Second of all, it's a fairly blind experiment. That is, aside from asking yourself whether or not that curve fits the Cotrell equation, you don't know anything else. So you really don't know what's happening. You can just say, oh, it appears to fit the Cotrell equation, or it doesn't so far. And it is going to turn out in general when I finish this whole derivation next time we meet that the changes you see in this curve are fairly subtle as you change your kinetic scenarios. So you can't just look at it and say, oh, yes, that's a charge transfer limited chronoamperogram, or that's a Nernstien chronoamperogram. It does not jump out at you. You have to do a lot of analysis of the data to get to that point. And so one usually doesn't want to get involved with that, since there are other techniques for getting at that. This is a pretty good jumping off point. So are there any other questions about this? OK, then we will finish it up next class.