 OK, well, let's start. I just list here sort of what we did last time, so I remind you. We talked about electron transfer homogeneous solution where we relate the second order rate to the first order rate by a association constant that I actually didn't talk about at all, but it could be, but I will pet. And then we talked about the potential energy surface that the electron transfer rate happens on. And we talked about adiabatic and diabetic surfaces, the diabetic surfaces being the non-interacting surfaces and the adiabatic being the actual surfaces. And we then got the Marcus expression for the rate for adiabatic electron transfer in which the pre-factor is a nuclear pre-factor of something of the order of 10 to the 13th. And from this, I mis-wrote it, of course. This is delta G0 plus lambda squared over 4 lambda RT. And from this dependence, you get a normal and inverted region, so if you plot the log of the rate constant versus the driving force, you find in one half of the curve the rate goes up with driving force as you increase driving force in the other half. It goes down, and that's the inverted region. And then we talked about non-adiabatic electron transfer that assumes that you don't always cross from reactants to products with unity probability. And so then you end up with a pre-factor that's basically an electronic frequency on which you transfer from one surface, one diabetic surface to the other that can be written that way. And we talked then about the reorganization energy is the inner sphere and the outer sphere, where the inner sphere is contributed by each react and contributes its own inner sphere reorganization energy, whereas the outer sphere is a composite term depending on both of them. We wrote the cross relations, and we ended with the distance dependence saying that HAB was a function of r, and the normal expression given is an exponential function of r. That's not required. There's nothing required about that as the functional dependence. The only thing you can say is the HAB comes about due to overlap between the tails of wave functions. You have the wave function on one side and the wave function on the other. And tails of wave functions tend to be exponential. So their overlap will tend to be exponential. It doesn't have to be. And so it's written like this. It's written as beta over 2 simply because HAB ends up as squared in the expression. And so then the distance dependence is a minus beta r rather than, so it's just written that way for simplicity. And there's been a lot of work determining what the beta should be. Certainly, Harry Gray's group has done a lot of work in proteins determining that, and in other case. Now what I want to do is I want to take a digression to absorption spectra. And the reason I want to do that is I want to develop functional forms I can then use to do electron transfer on a metal electrode. So we're going to talk about absorb of charge transfer transitions. And we'll draw our normal surfaces. And the charge transfer transition we're going to look at is shown as vertical, EOP. And we know that EOP, in our notation, is equal to the reorganization energy for a zero driving force reaction. The question I want to ask, though, is what does a spectra look like? This just tells you a stick figure that at one energy equal to EOP, you should have an absorption. But we know that's not true, that there's some distribution. And so I want to know what is the absorption spectra for a whole bunch of different energies drawn like so. And if I can describe the probability of transition for the different energies, I will get the whole spectra. And I'm going to do this totally classically. And what I'm going to say is I'm going to assume that the probability for transition from this position or from this position or from this position are all exactly the same. The only thing that's going to govern the spectra is how many molecules there are at this position versus this position. And so you'll have more at this position because it's a low energy position. So this will be the most intense part of the band, whereas this will have fewer molecules and this will be a less intense part of the band. And so we'll develop that. So this is our ground state. And E of the ground state is going to be lambda x squared. And this is our other state, which I guess I could call 2, where it's not quite an excited state in our thing, but we're thinking of it as the upper state is going to be, we draw this again as 0 and 1 for our coordinate. It will be x minus 1 squared. OK. And I know that E op is going to equal E2 minus E ground state. OK. And this, of course, is going to be lambda x squared, pardon me, lambda times x minus 1 squared minus lambda x squared. And you can solve this for x and you get x equals E op plus lambda minus E op plus lambda over 2 lambda. That's a minus E op. OK. And since what I want to say is that the probability of excitation is just depending on how many there are, the number of molecules there, which I'll call say n of i, will be n of i at some value x of i, well, x, call it, is going to be equals to some n, 0, E to the minus E at that point i over RT. And this is E to the minus. This E of i is this ground state energy. And so that's going to be lambda x squared over RT. And we have x squared here. And we have 0, E, E to the minus lambda, and then I guess I'll write it lambda minus E op, squared over 4 lambda squared writing below, halfway on the board, that's where lambda squared RT, OK? And that all equals minus E op squared over 4 lambda RT, OK? And this is a Gaussian function. It's sort of reminiscent of the Gaussian function we got for Marcus's theory. And if I plot basically the epsilon, this is now going to give me the epsilon. It's going to be proportional to epsilon versus the energy of E op. It's going to look something like that, where it maximizes that lambda, OK? And I can now generalize this fairly directly, is if I want to look at for non-thermal neutral reaction, where the dotted line shows the thermal neutral reaction. Well, I haven't drawn it well enough, too. But that distance is lambda up there. Pardon me, let me try that again, because I'm going to draw it the opposite direction, sorry. Because I want to go up. It's not a thermal electron transfer reaction, OK? And we know that this is lambda, this is thermal neutral, and this distance is delta G zero, OK? And so if this distance is delta G zero, this distance is delta G zero up there. And this distance, overall, I've called E op. So I can rewrite this in terms of lambda and delta G zero. And I get n zero is equal to, pardon me, n i, I guess, is n zero, e to the minus lambda plus party. It's minus E op, where what I do to get a negative sign in here that has screwed it up. I think I didn't quite solve this correctly. Let's go back to solve this. This has to be equal to lambda x squared minus 2 lambda x minus lambda minus lambda squared. These cancel. You bring this over. It'll be E op minus lambda. Do I have the sign wrong or do I? What do I have wrong here? So this is going to become E op plus lambda equals minus 2 lambda x, and x will equal minus E op. Pardon me, this is not minus x, a plus lambda, isn't it? I actually solved them correctly, or it doesn't work. Over minus 2 lambda. So this should be lambda minus E op. Isn't that what I had? Over 2 lambda? Because I don't want to get it into minus E. And I say E op. Hm. And I say that E op is equal to lambda. Oh, the question is what's the sign on delta? The g, I guess. That's OK. It's going to be plus delta g 0. And then I should get, so I claim that I'm going to get, although what I want to get is not that. It's going to be lambda minus, oh, oh, oh, oh. That's the problem. That's the problem, I think. Yeah, OK. OK, I guess I wanted to do it too simply. I think what I've got to do is I've got to rewrite these. So let me rewrite these. This is going to be E2, E1, E, E ground state, and E2. And so E ground state is going to be equal to lambda x squared. And E2 is going to be lambda times x minus 1 squared plus delta g 0, right? Now, if you solve it, so E op is going to equal E2 minus E ground state equals lambda times x minus 1 squared plus delta g 0 minus lambda x squared 2. And this will be lambda x squared minus 2 lambda x plus lambda plus delta g 0 minus lambda squared x squared. That cancels that. E op minus delta g 0 minus lambda all of minus 2 lambda equals x. So I got the same sign for lambda and delta g, which is what I need. OK, so then where I was going with this before I got sidetracked is that n is equal to n 0 e to the minus lambda times the squared. So it'll be E op minus delta g 0 minus lambda squared over 4 lambda rt. And if I write it in a more standard way, I change the sign in this such it's squared, it doesn't really matter. So it will be delta g 0 plus lambda minus E op squared over 4 lambda rt. OK, now that has nothing to do with anything really we're interested in unless you're interested in calculating charge transfer spectra, which is of some interest relative to this, but that's not really why I did it. What I want to do it is I want to reconsider electron transfer so I can write it basically in terms of this sort of distribution function of species. So I want to go back and now reconsider thermal electron transfer again. I have my parabolas. And this, as we recall, was the iron 2 plus of nucleus A, the iron 3 plus of nucleus B, whereas this is the iron 3 plus of nucleus A and iron 2 plus of nucleus B. And this we call reactants. And this we call products. OK, now I want to go back and reconsider if I consider each of the species alone what is happening in the electron transfer. This is considering as a super molecule. Now let me consider the iron A nucleus and the iron B nucleus separately. So the iron A nucleus, this is for the A, looks so like that. And it has, so this is going to be Fe 2 plus. And this is going to be Fe 3 plus plus an electron. OK, and the electron transfer, when I go from one species to the other, this is going to go from some place that's going to go from an iron 2 to an iron 3. So this is A. B, on the other hand, is the opposite. OK, and this is going to, again, be an iron 2 plus and an iron 3 plus. But this starts up there and somehow that's going to go down here. OK, so this is sort of a vertical energy transfer. It's going to go vertically because the speed at which the electron transfers fast compared to the motion of the nuclei. So the nuclei, you can't move when the electron transfers. And so the actual transferring part is vertical. But you might ask, what is the motion here? We know that the electron transfer moves along here, gets to the intersection point and then crosses over and comes back down. And what I say is that's going to be the motion here. It's going to move along here. It's going to reorganize itself to some distance. Then it's going to transfer. Likewise, this guy is going to reorganize himself to some other distance. And then he's going to transfer. And I'll call this delta G vertical A. And this is delta G vertical B. And when you view it this way, you have to say that when the electron transfers, that delta G vertical A has to equal delta G vertical B. And the probability of doing the transfer at this distance, I'm going to say is just dependent on the number of molecules that have this distance. So it's going to be very much like an absorption that I did here. I'm just going to say how many molecules have this energy of absorption. I'm going to be dependent on that. And so now I can write all that, hopefully. So the probability for the a nucleus to be at xi is going to be the total number of a's, which I'm going to throw away in a minute. But 4 pi lambda rt, yeah, I didn't go and evaluate n0. But if you think about n0, there's something about the integral of it. The integral of all the n's has to be equal to the total number of n's. And if you work that out, you find n0 comes out to the square root of 4 pi lambda rt. And this is going to be e to the minus. So what was it? It was the delta g squared plus lambda minus eop. So it's going to be minus delta g vertical a. And it's going to have the lambda in it. And then it has to have the delta g, which is the distance between the minima. And that free energy distance is the minus q e0. It's the cell potential between them. So that's going to be minus q e0. And to make it come out right, it would be nice if I don't do a self-exchange reaction. So I don't get rid of the driving force. So I'll assume that this is the qa, and this will be minus q e0 of b, where they may, in fact, be different. And this is all squared over 4 lambda rt. And pb of xj, we'll call it concentration b, all over the square root of 4 pi lambda rt. It might be better. Times e to the minus q e0 b. This was a plus lambda minus delta g v b squared over 4 lambda rt. And the probability of a transfer to a particular position i and j is going to be p, how do I write them? p of xi, pb of xj, where they're different distances, they've reorganized to different amounts, as we know they will because they have different force constants. And the total probability of the transfer is going to be somehow the integral of pij times the integral over all delta g v. So that's going to give us a b, the square root of 4 pi lambda rt. Now we've got to be a little bit careful, I guess. And I haven't been really careful. The reorganization energies for this couple and this couple, if I'm assuming the driving forces are different, the reorganization energies could be different. So just to keep them all separate, we should call them lambda a, lambda b, lambda a, lambda a, lambda b. So that we have everything distinctly marked as to who he belongs to. So I have two of these square roots times an integral e to the minus delta g vertical plus lambda a minus, what is it? I've gotten it wrong. It's minus q e a 0 plus lambda minus delta g vertical. This is for xi squared over the 4. This is a for lambda a rt e to the minus minus q e b 0 plus lambda b minus delta gv. I guess I called this b. I guess I called this a. I didn't distinguish them with the x's for lambda b rt e delta gv. And as I say, to have the electron transfer go so that you can serve energy, delta gv a and delta gv b have to be the same. So I can get rid of their distinguishing marks. They are forced to be the same. And now we have the integral of two Gaussians. And you can go show on Mathematica, or if you're really good at calculus, you can show that this, the integral of two Gaussians gives you a Gaussian. And the Gaussian comes out to just sort of what you would expect, because we know the answer to this is going to be a, or almost the answer, times b all times 4 pi lambda a plus lambda b rt e to the minus minus q e a 0 minus e b 0 plus lambda a plus lambda b squared squared. Got to get the signs in the right place, rt. It just comes out to the Gaussian, which is the Gaussian we know, which is just e to the minus delta g 0 plus lambda squared over 4 lambda rt, where the lambda is the sum of the individual reorganization energy. So that somehow we can write the overall term now as distribution functions for two individuals and get them to be here. And that's in a sense what I want to do, because I want to go and do electron transfer at a, OK, this 4 pi rt. Well, when you integrate a Gaussian, the Gaussian has some width. If we write the Gaussian as e to the minus x squared over a, this width is, what is it? 2 times the square root of a, OK? So what does that mean? So when you're going to integrate the Gaussian, it's basically the width of the Gaussian is going to determine a lot about the integral, right? It's going to be sort of a height times a width. So these guys, you always get forms like this to normalize the Gaussian. So when you do this integral, you're going to get like a 4 pi rt out, and you have to replace these two separate things with normalizes separate Gaussians by one that normalizes the two Gaussians. And it just works out magically. That's how it works. But the 4 pi, the lambda rt are basically normalization functions for the Gaussian, OK? There's a difference between them. And OK, you're asking, did I get it correct? Let's see, that's hard to see. This one is minus, and this one is plus, because you're going the opposite direction. And so that this one should have been plus, and this one should, and now is it written so delta G0 is correct? I think it is. I think it's written as delta G0 is correct, OK? So now, this is, turns out, this is the probability of electron transfer, or it's more like a rate because it has these rate constants, has these concentrations in there, but it doesn't have any frequency factor. So it needs some sort of frequency factor to actually fix it up. And we know what the frequency factor is, so the rate is actually going to be 4 pi squared HAB squared over H, all times my friend, where I've divided out the concentrations. So this is 4 pi squared HAB squared over H times 1 over the square root of 4 pi lambda RT, e to the minus delta G0 plus, OK, which is all what we know. And this is sort of our frequency factor in some way that we have to add. OK, so now we want to go on to electrodes. Now we're finally ready to do electrodes, and hopefully we'll make use of this. So now let's think about our metal electrode. Here's our metal electrode, OK? And the metal electrode has a whole bunch of energy levels in it. It's filled, it has a whole bunch of, and they're filled up to some level here, so they're all filled, where this is the Fermi level. Actually, I promised I'd write the energies as a scripty whereas potentials as a non-script to try to. What you will see is there is a great propensity to become confused between potentials and energies. And now we're going to think of some energy level, EI, and that's going to have a potential of minus QEI, OK? And when we put this in a solution, the solution is going to have some potential. And when it comes to equilibrium, you will have the potential of the solution equal to the Fermi level of the metal. They will come to equilibrium, we know that. And there's going to be some other position of the QE0. If the concentration of the species in solution, the oxidant and the reductant are equal, these will be the same place, but they don't have to be. And we want to ask about the electron transfer like so, OK? And we're going to assume that this redox couple is in a solution that has a high ionic strength, so it's basically outside of the drop of the potential in the solution, so we don't have to correct for that. That could be corrected for, but we don't want to correct for it, so we'll assume it's all outside of that. And what we're doing is we're going to transfer an electron to or from this band structure. And I guess what we are going to do is assume that Marcus theory works. We're going to assume that Marcus theory is the correct description for the microscopic description of electron transfer. And now we want to write a rate constant. So I'm going to say my rate constant, k, is going to be my 4 pi squared HAB squared, all times some integral of a distribution function for the bands of the electrode that have some energy epsilon, and times a distribution function for the redox species. How come I can't keep track of that eraser? Integrated over all energies. And the question is, what are we going to write for the various distribution functions? Now, the first thing we, I guess, need to ask is what is going to be the reorganization energy of the metal electrode? Does the metal electrode reorganize? We're going to put one more electron in, or take one electron out of the metal electrode. But the metal electrode has a band structure, and that band structure is dispersed over all the atoms in the electrode. So there's basically 10 to the 23rd atoms that we have this electronic structure separated over. And so the amount is going to affect any individual atom is minuscule. And so you aren't going to expect that it be any bond distance changes. You're going to expect that it be any bond distance changes in the metal electrode. So we will assume that the reorganization energy for the electrode is zero, that there's no. So the only thing we need to then say is, what's the probability of getting an electron out of there, and what's the number of states? And so we're going to assume that there's a density of states here, rho. That's a function of energy. That tells us how many states you have per E v, and it's going to be a function of that. So we're going to have a rho of E times the Fermi function, which tells us the probability that an electron will be occupying the state. Remember, these states aren't really all filled, and these are really all empty. It's some sort of Fermi function that basically goes from one that these are filled to zero up here. So we need to write some sort of Fermi function. That's a function of the energy 2. And that Fermi function we have to worry about is whether we're transferring an electron out of the electrode, and that's going to be the Fermi function for the unoccupied states, or if you're transferring it into the electrode, it's going to be the Fermi function for the unoccupied state. So for our Fermi function, for our filled states, which I'll call f, which is going to be 1 over 1 plus exp to the E i minus E f over RT, where this is the energy of the Fermi level and the energy of the ith level that we're considering, or in terms of potential, which is what I, it's going to be minus Q times E i minus E f over RT. And the Fermi function for the empty bands is going to be just equal to 1 minus the Fermi function for the filled bands, which ends up as E to the minus Q E i minus E f over RT divided by 1 plus. OK? So now we have the distribution functions for the distribution functions for the oxidant. Of course, we've sort of written a bunch of times by 4. And we're going to write it as exp to the minus lambda plus delta g0 squared over 4 lambda RT, which we've written before. It's this function where the E0 is for the redox species in solution, and the vertical energy is basically the Q E i for this band. OK? So now I can write the whole rate constant. k equals 4 pi. I guess there's actually, there's a 1 over the 4 pi lambda RT, which I left out, is 4 pi over h, hab squared, all over the square root of 4 pi lambda RT times the integral of rho of epsilon over. OK, I guess we now have to decide which one we're talking about, and I'm going to talk about the transfer from the electrode to the redox couple. You could write either one. They come out clearly somewhat different, but basically the same thing. OK? So now what do we have? What do we have? This is going to be the E. Now the question is where? The E i dependence you see here, but it's actually also in here because delta G0 is going, we are going from this state to this state. So it's minus Q E0 minus a minus Q E i. So this is minus Q E0 minus E i. So we have a dependence of E i in the delta G0, we have a dependence of E i here in the Fermi function, and you could even think that we might have a dependent in rho. In general, what we're going to consider is that rho is, the density of states is constant enough. But let's look at what we have a second. What we have this part is a Gaussian. So I'm going to draw him as a Gaussian. And where does he peak? Well, he's going to peak, well, maybe I should write him out. My Gaussian looks like exp of minus lambda minus Q E0 minus E i squared over 4 lambda RT. And the Gaussian peaks where this is 0, so lambda minus Q E0 minus E i is equal to 0. That's where it peaks. So I can say Q E i is equal to, let's see if I can solve it so I get it actually right. This has to come across. So it's going to be minus Q E0 plus lambda. So if I'm plotting this as my energy E i up here, this point is going to be minus Q E0 plus lambda. You don't think it's right? When I multiply through here, and then you say I have to take this to the other side and make this minus, I'm sure that's what I want to do. Oh, yes, yes, yes. OK, yes, that's what I want to do. It's the problem of the potentials in energy. If I look at energies, these are energies now. These are energy terms. This is in terms of potential. If I'm plotting it in terms of energy, so I want it to be minus Q E i going up. So high energy is over here. Low energy is over here. High negative potential is up here. So on, and so forth. So I get, in terms of energy, this is minus Q E0 plus lambda. And that's where my peak is. My Fermi function, however, goes through a transition. And it goes through a transition where this is equal to zero. And so that's where minus Q E i minus E f is equal to zero. And so minus Q E i will equal minus Q E f. And so this goes through a transition like this, where this is minus Q E f. Now what we want is the overlap between these. So it's clear how I've drawn it now that there's very little overlap, and the rate you're going to get is very small. As I move the Fermi level up, the overlap becomes more and more, and I get a bigger and bigger rate. Now what does it mean to be moving the Fermi level up? The Fermi level is the level that is in the electrode. And when I'm not at equilibrium, if I apply a potential to the electrode, I can push this up. And then I have current flow precisely because the forward rate is faster than the backward rate. And so by pushing the Fermi level up, I actually push the rate up. In other words, the higher I push this Fermi level, the bigger the rate I get to my solution level. And that's because I'm in a sense populating this much more. And clearly when I push the Fermi level up here, so this is a full population, I'm getting as fast a rate as I can get. And so once I get up here, if I push the potential up higher, I never get any increase in rate. So what do I expect in terms of rate constant versus minus QEF? When I'm down here, I have a very low rate. And something happens, but probably not too much. Then I come over into the tail, and I really start catching some of the density. And the rate starts to go up. And then I get past the peak, and it flattens out. OK? Now, I think the question that I wanted to address is why that happens and what can we understand about the electrodes versus normal solution, normal redox species and solution. Here's my electrode again. With all my levels, I draw some sort of Fermi function where this is 1 and this is 0. I draw my minus QE0. And this is energy in that direction. And I want to draw my Gaussian here on my electrode. And he's displaced from QE0 by plus lambda. So I draw him where this distance is lambda. OK? And basically, if I think of it in terms of a solution, where in solution I have one redox couple transferring electron to another redox couple from his half cell potential to his half cell potential. And so I think of it as that I have rates that transfer, say, from this level to that guy, from this level to that guy, and from this level to that guy. This guy will have a very small rate because the reaction is uphill. He has plenty of population, but the reaction is uphill, and so there's not much reaction. This guy will have a pretty good rate because he's pretty much thermonutrile. So the Gaussian is reasonable size, but his Fermi function is about half because it's about at the transition. And this guy will have a little rate again because he has very little population because the Fermi function is 0, but his driving force is good. And so he'd have a good rate there. On the other hand, now if I push the Fermi function up, this guy has a very good rate. This guy still doesn't. Well, let me go push the Fermi function way up. So now this guy has a really good rate because he has some favorable driving force, and that favorable driving force is going to give you a favorable term here. That's a negative term that's going to be lambda minus that favorable energy transfer that makes this small. So this term here is very, very favorable. And now we're driving the reaction as fast as we can go because this is a really good rate. When we push the rate up even further, now I've got to draw it again because I don't have room to do that. This is my Qe is 0. My Fermi function now is going to look like this. And I ask about the rate here. What is that rate? Is that rate going to be big or little? Big, little. Well, we got both in a minute. It has a really big driving force. So what has happened here? This has become a very big negative number. When that becomes a very big negative number, it becomes larger than lambda, and it becomes inverted. So the rate goes down. So this guy has a slow rate, but there's always a guy down here now who has just the optimal rate. His delta G is just enough to wipe out the reorganization energy making it 0, so this rate is now fast. So you still get a fast overall rate. So in this case, it was this one that was fast. As you push it up, the top one is no longer fast, but you still have somebody in there who's fast. So that on a metal electrode, when you push the potential at a metal electrode up, the rate goes up and up, but ultimately limits. You never see the inverted region, because you have lots of levels, and you always have a level then once you fill your potentials high enough that is an optimum, even though you have some that are not optimum. Now I want to draw the other half of the picture. Now we want to ask about going the other direction. Now, where's the equations to go, the other direction? OK. OK, let's write our equations again. We have the rate, which I wrote the rate. I call it from the band structure to the oxidized molecule. And I wrote it is 4 pi squared HAB squared all over the square root of 4 lambda pi lambda RT. Togs an integral of rho e to the minus lambda minus q E0 minus E. Well, just leave it as E, because that's the E of the electrode. It could be E. I could write it as E applied or something, but one is squared over 4 lambda RT divided by 1 plus E to the minus q EI minus E of the electrode all over RT. The DEI is EI. This is EI, isn't it? That's EI, is that right? Everybody agree? And then I can go back and I can write the other one from the reduced to the band structure. That's all easy. And this part is easy. And the distribution function is E to the minus lambda. And since it's going the opposite direction, I've got to change the sign in delta G. So this is going to be plus q E0 minus EI squared over 4 lambda RT, OK? So we had our picture over here, where the Fermi function changes at EF, and this changes at E, q E0 plus lambda. And I'm going to ask now where this changes. That's because I want to draw the same picture now for this guy. And this is going to be minus q EI increases, becomes more negative in that way or positive energy. And this is going to peak where lambda plus q E0 minus EI equals 0, or minus q EI will equal minus q E0 minus lambda. So this is at minus q E0 minus lambda. And this Fermi function makes the same transition at the same place. Only instead of coming down, it goes up. So now you have something that goes where this is minus q E, as I'm calling it now, rather than EF, OK? So to draw my picture on my electrode, so now to make here my levels, here's minus q E. We're going to draw the equilibrium. We'll assume that ox equals red. So this will be minus q E0, which equals minus q E of the solution, right? And then you're going to get one guy who looks like this, one guy who looks like this, this distance is going to be 2 lambda. So this is going to be a lambda negative in energy, and this will be a lambda positive in energy. And when we transfer from some level minus q EI here, we're going to be transferring into this one, and this guy we're transferring out of, OK? Now that's the garusher model. You draw distribution functions for the species on the electrode, and you look at the rate of electron transfer. It all is basically a sort of Marcus theory that is redone for electrodes. OK, what more do I need to do there? I guess that's as much as we need to do in pictures. So now what I want to do is I want to get the standard rate constant for it, the rate constant that applies when you have a nice clean board, have a driving force set equal. Well, OK, I'll tell you. OK, so we're going to get the standard rate constant, which is going to be k 0. And that's going to be, so we have k equals 4 pi squared h A b squared over the square root of 4 pi lambda r t times, well, assume the density of states is it's constant, so we don't have the energy dependence in it. We'll pull it out of the integral, and now we have this integral of exp to the minus lambda minus q E 0 minus E i squared over 4 lambda r t all over 1 plus exp to the minus q E i minus E all over r t d E i. And what I want to do is I want to evaluate that for when minus q E 0 equals minus q E. OK, that's the state. In other words, you apply a potential to the electrode that's equal to the standard potential of the redox couple in solution. And you can then measure the current and clear the current relates to this rate. So in order to do this, I need to evaluate this. So I should define this integral, the whole thing. I won't write it. We'll just know what that is. OK, and to do this, I'm going to introduce, I'm going to change my variable a little bit so it becomes more obvious. I'm going to say that epsilon i equal minus q E i minus E f. That's really if I take my electrode and here's my Fermi energy and here's the level i, it's this distance. It's the energy difference between the i-th level and the Fermi level. And therefore, q E i is equal to q E f minus epsilon i. No, that's not right. It must be this. Even though my notes don't say it, the notes have to be wrong. OK, so we want to plug that all into our integral. And our integral will equal the pre-factor all times 1 plus E to the epsilon i over kt. I call it epsilon i because I'm talking about the i-th level. But then, of course, I integrate it over, assuming the levels are basically continuous. So the i sort of subscript is sort of meaningless. But it's easy to keep in mind what you're talking about. So I leave it there. So this will be minus lambda. God, the signs look all wrong. Minus q E 0. And I plug this in. So this is going to be minus q plus epsilon squared over 4 lambda rt. Epsilon i. There's a q play to convert between potential and stuff. This was the i, and this is probably a minus sign. But that'll switch the limits. So I'm not worrying about any of that like this. There's something that mattered with my signs. And my signs here go back and forth. And so I'm not clear where it's lost. You see where it's I've lost? Plus, oh, that makes me very happy. Very happy. Thank you. And I'm going to evaluate it where this is equal, or the f is equal to E 0. So these are going to cancel one another. And so now I have n. We're going to need to ask where the overlap is. Now what we have, as I've drawn it now, our Fermi function occurs right where epsilon is 0. Well, this is 0. Epsilon is going up here. And our peak occurs at lambda. And this width, which we say is a square root of 4 pi lambda RT, is also controlled by lambda. So even if I make lambda big, what I do is I increase this width and stuff. So the relationship, in a sense, stays the same, independent of lambda. And what I find, if you do this, is that the overlap is always right in here. That you're enough in the tail of the Gaussian that you haven't gone to 0 here, and the increase here basically gives you all the gap. So that the only place you get any overlap is for very small epsilon. That epsilon being up here equal to lambda, the Fermi function wipes it out. So what I'm going to do is I'm going to expand this and linearize it. And then I'll be able to integrate it. So I expand it, x squared minus 2x epsilon plus epsilon squared over 4 lambda RT, 1 plus e to the epsilon over RT, that is the lambda, what I've got x's here. So I can write this out as a product of three terms. The first term is not a function of epsilon. So I can take that out. So I get out of the integral, it's going to be e to the minus lambda over 4 RT times the integral, e to the minus, I'm OK so far. I haven't gotten, screwed up the little signs yet, e to the epsilon over 2 RT, e to the minus epsilon squared over 4 lambda RT, all divided by 1 plus e to the epsilon over RT, d epsilon. And there's that piece in front, the pre-factor that I haven't written, the square root of 4 pi lambda RT. Now, what I've said is basically it's only epsilon values close to 0, so that piece I can throw out, because he's the smallest piece. And so now I get the integral e to the minus epsilon over 2 RT, 1 plus e to the epsilon over RT, d epsilon. And what I claim is you can integrate that, but just to show you can. If you divide through by an epsilon, minus epsilon over 2 i t, the numerator and denominator, you get this is 1 over e to the minus epsilon over 2 RT plus e to the plus epsilon over 2 RT, d epsilon. And everybody recognizes the denominator as the cosinch, 1 over or 2 times the cosinch of epsilon over RT, or epsilon over 2 RT, which is it. Now, I've forgotten. I think the 2 has to go in there. I don't know if I wrote that down. Now, I got a sign wrong. I divide numerator by, yeah, I know. The top is plus, and yeah, that's OK, right? So I divide the numerator and denominator by e to the, or multiply numerator and denominator by e to the minus epsilon over 2 RT, and then the numerator goes to 1, yeah. So then it works. So this is the cosinch of the, and that you can integrate, and you get, oh, now wait, I've lost a piece. You let me lose a piece there. This piece I've lost over here. So this is e to the minus lambda over 4 RT. So we have e to the minus lambda over 4 RT times all over the square root of 4 pi lambda RT times pi k RT. Difference between r and k, among friends, we won't mention. We don't have to worry about. And so I get k0 is 4 pi squared over h, hab squared, times the square root of 4 pi lambda RT times pi kT, e to the minus lambda over 4 RT. And that turns out to be 2 pi squared over h, hab squared times the square root of pi k RT over lambda, e to the minus lambda over RT, 4 RT. So I actually can explicitly get a rate for the reaction. So that's the standard rate. And I could do that for the other reaction, too, but it clearly is just a repeat of all of that. So we won't do any of that. The last question to ask is about reorganization energies. We clearly say there's a reorganization energy, and we basically attribute it mostly to the redox species, but there's some solvent problem, too. I mean, in the homogeneous reaction, we had the solvent polarized around the two species. And when we moved the electron, we had a repolarized solvent. Well, here you have an electrode that's a metal electrode, and you have some ion that approaches up to it. And the solvent will be polarized in some way around this. And if you bring up an ion to a metal electrode, you can think of it as that you have an image charge behind the electrode at the same distance behind that the charge is in front. And so you can work that out. And where do I have it? Somewhere, or if you've worked it out, or should I say somewhere, Rudy Marcus worked it out. And so if we say the anosphere for a metal electrode is going to be delta q squared over 4 pi epsilon 0, all times 1 over 2a1 minus 1 over 2d times 1 over dop minus 1 over ds, where a1 is the radius of the ion, and d is the distance from the ion to its image charge, which is, if we define this distance as r, d is equal to 2r. It's quite persistently in the literature written as 2d, although clearly d is sort of a hypothetical distance. The dop is, again, the optical dielectric constant, which is usually taken as a square of the refractive index. And ds is the static dielectric constant. So if I say that this can approach the electrode reasonably intimately so that d is equal to 2a1, then I get, at close contact, times 1 over 4a1, I think so, which I could rewrite as 1 over half times delta q squared over 4. Now if you recall, for the solution, anosphere for the solution, or for the homogeneous in solution, we wrote down as delta q squared over 4 pi epsilon 0 times 1 over 2a1 plus 1 over 2a2 minus 1 over r times 1 over dop minus 1 over ds. And this was for two ions in solution, a1, a2, radius, and the distance between them was r. And if we do close contact and we assume the ions are the same, we say a1 equals a2 and r equals 2aA, then we can work this out, and this is delta q squared over 4 pi epsilon 0 times, these two add up as two of these. This is 1 over 2a that subtracts, so it's a 1 over 2a times 1 over dop minus 1 over ds. And so that lambda outer for solution is equal to, or one half of lambda out of solution is equal to lambda outer at the metal electrode, OK? Likewise, we can say lambda in has a relationship. Lambda in at the metal electrode is one half lambda out in solution, because for the metal electrode, you have one ion that's reorganizing, whereas in solution, you have two ions reorganizing. So overall, lambda for the metal electrode is usually approximately one half lambda of the solution value. Now we have a little bit to say about SAMS, and then we're all done. OK, so now we've talked about the reorganization. Now you can write relationships between the rate constant in solution and the rate constant at the standard rate constant at the electrode if you want, and that's often done. But I'll leave that out, because I want to make one remark about SAMS. Now at the electrode, oh wait, oh yeah, OK, well two remarks. These rate constants I've written at electrode are all for a molecule basically at a distance from the electrode. I haven't talked really about the problem of diffusion toward the electrode or any of that. So if you want to worry about that, that has to be worried about somewhat separately. But there's a large class of large cottage industry of work that takes a metal electrode, generally a gold electrode, and puts a self-assembled monolayer on. These are usually alkane thiols, so they have sulfurs that will bond to the metal electrode. And what they have is one in, I don't know, 10 or 100 of these capped with a redox-active species. And they can very well characterize the SAM, and they can say quite well that it's nice and densely packed and it's all stacked at a particular angle. And then by changing the number of alkane units in, they can measure the rate constant for electron transfer. And they can get a distance dependence of electron transfer. And so it's one way of getting this beta value. And for this work, the betas you get for saturated alkanes are about 1.1, if you want, angstroms to the minus 1. And if you, as you might expect, if you make these aromatic and you make them planar, the beta goes way down because you can transfer distances much further. And if you make them so that you, well, then there's other work that has measured beta in solutions that don't have full chains, and then the beta gets worse. And so there's a whole industry of measuring and characterizing beta that's been done, and much of it on electrodes has been very helpful. OK, I think that's where we should end. Maybe I just should give an advertisement there. You can now do all of this stuff on semiconductor electrodes. And there are many differences and special things about that in semiconductor electrodes. But you'll have to take Nate's course to get any of that.