Okay, so I wanted to share with you guys what's going on in my driveway at the moment. This is my younger brother's car. It's a 1978 Fiat Spyder, and my brother is 18 years old now, and he got this car when he turned 16. He purchased it from our neighbors up the street, and when he purchased it, it was not running, and of course now it's missing a wheel, but I had to push the car down the street to get into a driveway, and so my brother and my father have been working on it the past few years, and they've had it, got it up and running, you know, it's very quick, fairly quickly after getting it, but they've always been changing things, and they've taken the entire engine out of it and replaced the transmission and all that. But this is just kind of recent work, and what was interesting and why I'm bringing this up is that they had the engine block resurface, so they took the engine block out and they sent it to a machinist in Seattle who re-finished the engine block, so now it looks nice and clean as you can see here, so this is the end, I'm not too familiar with all the lingo, the mechanic lingo, so I might be making up some words that don't make sense, but this is the engine, the head of the engine block, and these areas are where the combustion reaction happens, right, the gas and oxygen mixture combusts, and there's the spark plug comes out here, and then these areas are where the valves open and close to allow the oxygen or the gas in, and so this whole thing was chemically treated and cleaned and then resurfaced, but what I wanted to point out is that these engine blocks are typically sand-casted aluminum, and you guys are familiar with sand casting, and I wanted to show you some of the surface features of this resurfaced engine block, and you can see the porosity in the sand cast, it has very high porosity, just evident just through visually inspecting it, and what's also interesting is that fiat is also notorious for having parts break down on it, and in fact, my father, I was talking with a machinist who was making these parts, or specializing in the fiat parts, and he said, yeah, the engine blocks have always been notorious for its low quality, and I think it's pretty evident of this high porosity that these are kind of poor materials, but also why this is out right now is that my brother was adjusting the timing of the car, of the engine, and the timing is important because the timing is when the spark will ignite compared to the height of the cylinder, the piston, and also the valves when they come out, and so if you want to get it at the right spot where you have the spark, right at the top of the cylinder or whatever, so you have the right compression ratio, I'm not too familiar with the exact lingo, but what happened is if the timing is off by too much, that the head of the cylinder that or piston that's going up and down will bump into the valve that's opening, which is this, the valve seat is right here, and so what happened is my brother accidentally put the timing a little bit too far off time, and the top of the cylinder bumped into this valve that would open and broke off, and so now it's inside this chamber that's bouncing around, you can see how it's deformed, that the pieces of this cylinder valve have broken off and deformed inside while the engine was running, and so if we had to take the whole thing, they had to take the whole thing apart again, this time the engine could stay in the car, just taking the head off is easier, and what they did recently is they went to, I guess, a scrap yard that had the same car and got the same part from an old, retired car, and so here's the same kind of part, you see these are the valve parts that kind of open and close, and that's what was destroyed, and so they're going to clean this up and replace the one that was damaged, which is too bad because it's nice and clean already, but anyways, just a little bit of material science, I thought I'd share what's kind of going on in our time off, but that's my brother and my father have mainly been doing that. Okay, so let's, we're going to talk about a bit more about the electrochemistry, I wanted to go over some things we learned last lecture, some things that I might have missed and clarify some things, and we'll talk a bit more about batteries. So last time I talked about the standard reduction potential, right, this list of potentials that these different species will reduce at, I failed to emphasize that these values are measured at standard state, so they're empirically determined at standard state conditions and the standard state is at 25 degrees Celsius, one atmosphere, and then also most importantly is the concentration of the species. So in all of these, these cases, with the exception for some like the metals like zinc metal, zinc metal, you can't have a concentration in solution, it's a solid, but like iron three plus iron two plus, the concentration of iron three plus is equal to the concentration of iron two plus and for standard state, I think they keep it at one molar, which is a bit high, in my opinion, but so these values are only accurate when it's at standard state, and that's why it's called the standard reduction potential, it's given by this notation, the the E with the little knot above it, that's the standard reduction potential. So if we, in just looking back at this example where we had the cell of the iron two and three plus and the zinc two plus and zinc metal, all this this potential that we calculated the standard redox potential of the reaction, it's only accurate if we say that the concentration of iron two plus is equal to the concentration of iron three plus, and then the concentration of zinc two plus is one molar. Okay, so that that's only the case and I forgot to mention that I wanted to go over that in more detail, you know, what happens to the voltage of this cell if the concentration is not the same for each species. Okay, so what does the voltage look like in that case? And this follows what's called the Nernst equation. Okay, so this equation relates the voltage, the redox potential of the species related to its standard reduction to potential at standard state, and then how it changes concentration. So minus RTR is the gas constant T is temperature in Kelvin and is the number of electrons involved in the reaction. So for iron two plus iron three plus it's only one electron, like zinc two plus the zinc metal is two electron reaction. F is Faraday's constant relates the amount of charge in one mole of electrons. And then q r is the I want to say reaction quotient reaction quotient. That is the ratio of the products of the reaction to the, the reactants of the reaction. Okay, so it at standard state, the products and the reactants are at equal concentration. And here I've plotted the reduction potential versus the ratio of the products to the reactants of that reaction of this, this redox reaction, you see if they're equal, so it would be one over one, or it would be equal to one, that is the value of the standard reduction potential and for iron to the three oxide, it's 0.77 volts. But as we increase the amount of reactants, which would be iron, in this case would be iron three plus if we increase the concentration of iron three plus relative to iron two plus, the voltage of this of this redox voltage increases. Okay, and the opposite is true for if we increase the amount of products. Okay, this, this should be familiar with, you should be familiar with this. You've probably learned this in freshman chemistry, and even high school chemistry. This is a principle called Le Chatlier's principle. Okay, and that says, you know, if you have more product, then the reaction will be favored to go in the reverse. Or if you have more, more reactant, it'll be favorable in the forward reaction. Alright, so all we're saying is that we have more reactant versus product. And so there's going to be a higher, greater thermodynamic driving force for that reaction to proceed instead if it had more product than reactant. Okay, I mean, even if even if the amount of product is very low, or excuse me, if the amount of product is very high, there's still a positive potential, which means there's still a thermodynamic driving force for that reaction to proceed is just lower than if it was the other way around. Okay. So here's an example of that. I've taken two cells, a solution A and solution B, and they are both made up of iron two plus and iron three plus. But the difference is the difference in concentration of the three plus to two plus. So in solution A, we have a greater amount of fire and three plus solution B, solution B, we have a greater amount of iron two plus. So you can use the Nernst equation to calculate the the redox potential in solution A. Okay, so we have the standard redox potential of iron is 0.77. And the natural log of the concentration in this case is one, the concentration of the product iron two plus is one in this case and the reactant is 10. Okay, so it has a redox potential that's above its standard redox potential. And in the opposite, it's true for solution B as a redox potential that's lower. So if you compare these two together, there is a potential difference between these two cells. So if you were to take the multimeter and measure between you see a voltage and that voltage is the difference between these two. So the cell voltage is just the difference between the reduction potential of solution A and solution B. So you can always take the higher number minus the smaller number. And that will be that can indicate which direction like electrons will flow and which what species will be reduced or oxidized. So in this case, I've used the reduction potential. But in the previous example of the zinc, I had used oxidation potential, which is just the negative of reduction potential. Alright, so is the opposite here. So you have to add the oxidation potential in this case. So the question is, is this when will this reaction stop? Alright, so if we if we were to short circuit this, you know, just put a wire across this, then electrons will flow from this side, the iron two plus will oxidize will give electrons and flow into this side and reduce the iron three plus and that will continue until a certain point and that is of course, equilibrium of the entire system. But the question is when does it reach equilibrium? Okay, so we know at equilibrium, there is no net product production of product and there's no there's no net forward reaction and no net reverse reaction. Okay, so that means that the potential between the two cells at equilibrium would be zero. Alright, or Gibbs free energy would be zero, right? There's no thermodynamic driving force to go one way or the other. And so the question is, you know, when at what concentration of these two cells will that occur at when will the the reaction stop taking place? And in this example, I should have started with nine and one to make the total 10. But the total concentration of iron is actually 11. So that's kind of my mistake. But that's fine. So the total concentration of iron, in this case is 11. So when this cell has a concentration of 5.5 millimolar iron three plus and 5.5 iron two plus, and that's the same as this cell, that's when the voltage between the two will be zero. Okay, and so that's when the reaction will stop. The electrons current will stop flowing between the two. There's no potential to drive the current. So another question is, you know, in this example, it's fairly simple, the concentrations are the same in both, right? The total amount of iron in both cells is 11 millimolar. What if one of the cells had a much higher concentration of iron, for example, let's say solution a had a 100 millimolar iron three plus and one millimolar iron two plus, but solution B was kept the same. All right, then what would the final concentration be? Now, before this lecture, I had prepared the slide, but unfortunately, I lost it and it didn't save. And I'm wondering we could go through it. But actually, this lecture this morning's lecture was kind of long. So I think I'll skip that. But the answer is, you know, for if you have to think of as you take electrons away from solution B, all right, for every like one millimole of electrons you take away, right, you're you're creating one millimole of iron three plus and then the same for this solution, you're you're giving it one millimole of electrons, you're you're creating one millimole of iron two plus. So if this was 100 millimolars, the the most this could change in concentration at the maximum, let's say it took all the electrons away from these irons, then the most it could be would be, it would be nine or excuse me, 10, it only has 10 electron millimolars of electrons to give. So this would be become 90 instead of 100. So that would be the maximum, but it's not going to reach that state. Because remember, the reduction potential of these solutions depends on the ratio of product to reactant. Okay, so actually, before it depletes all those electrons that ratio is going to create such a potential that the it's going to equilibrate and turn to zero. It's a bit it's a lot more clear when I had that extra slide, I'll see if I can upload it later, it'll make more sense. But let's move on. So this is something I was thinking about last night when I was making these slides is how how is this different than having just iron and two plus and three plus in solution, right? I mean, what's what's stopping from stopping this iron two plus to from giving electrons to this iron three plus and reducing it, right? It's kind of kind of similar to this. And if the concentrations are different, right? I'm embarrassed to admit I spent a long time thinking about this. And I was driving me nuts thinking why why? It didn't make sense. And the reason is, well, of course, in solution, you are having this electron transfer between iron two plus and iron three plus all the time, right? It's fine. It's natural to have some exchange. But the point is that I was missing is that if you remove an electron from this iron atom and give it to this iron atom, the concentration of iron three plus and iron two plus does not change, right? You just removed it from one iron and gave it to another. And it took me about an hour to figure that out. Like, oh, I was a little bit embarrassed about that. Anyways, so it is different. This solution, you could say is that equilibrium, it has equilibrium potential. This system, this system is not at equilibrium. Okay, until the voltage read zero. But if you're just looking at individual solutions, you could say, yeah, that solution is at equilibrium. But the system is not. And that's, you know, particularly introducing the salt bridge will put the give you this this potential between the two electrodes. In a side note, I was at a museum in Germany is related is related. I was in Germany with my wife when I was an undergrad, and we were vacationing and I went, I think it was in Dresden, we went to some science museum, and there was this really cool demo that talked about the like corrosion potential between different metals, or the activity between different metals, where they had a metal like a one type of metal like aluminum and had another type of metal copper and they're like plates. And it had a voltmeter just like in this setup. Okay, and the idea was you would put your hands on these plates, and then the voltmeter would read a voltage. Okay, so how does that work? It's it's very similar to, to this setup, where you have, you know, say you have the copper plate on one side, the aluminum plate on another, and they're connected to the voltmeter that we're looking at. And then you your your own body is making the salt bridge to connect the two. And that's when the voltage reads so your body is the electrolyte to for this to make this to measure this potential difference. I thought it was really cool because there's all these different metals and you put your hands on different things and make the voltmeter go to different voltages. Anyways, so pretty cool demo you can you can try out if you have a voltmeter and see if it works. Okay, so here's a method of determining the the standard reduction potential. This is called Potentiometric Titration. And there's other types of titrate titration is when you when you when you add a species to a solution like drop by drop, and you're measuring the volume added to the solution. Okay, so like there's you're probably familiar with a acid based titration, right, you might have a solution that has an indicator that changes color when it reaches a certain pH, and then you titrate with an acid and you're measuring the volume added to that solution, and you're monitoring it and then all of a sudden it hits that that that pH and it changes color. So that's an acid based titration. In this type of titration, you have a species of redox active ions in solution. So in this example, we have a 10 to plus an iron two plus just 100% 10 to plus 100% iron two plus. And we're titrating it with this this ion. This is and correct me if I'm wrong. I think it's cerium. I believe cerium. I could be saying that wrong cerium four plus. So cerium four plus has a very high reduction potential 1.6 volts. So that means if it comes in contact with any of these ions, it's going to easily oxidize those ions. So those ions are going to oxidize they're going to, they're going to iron the cesium four plus is going to steal an electron from those ions. Okay, so if we start with this solution of two plus ions, and what you do is that you measure the redox potential. So the potential between the platinum and a reference electrode such as standard hydrogen electrode. So you're measuring the whatever reaction is happening at the surface of this electrode, as you titrate it. Okay, so this this is very much at equilibrium, you add a little bit of cerium, you let it mix up together. And so it goes to equilibrium and you measure the voltage. And so you take these little different points as you add the cerium and you measure the volume of cerium and that's relative to the concentration of the cerium. Okay, and you'll get a plot like this. And this plot very much is the Nernst equation for each species. So all I did as I modeled the Nernst equation for 10, two plus to four plus and then I also modeled the Nernst equation two plus the three plus and I just kind of added them together. And the at the inflection point is the standard reduction potential. And remember standard reduction potential is when the concentration of the product is equal to the concentration of the reactant. Alright, so at this potential and or that this volume added, you have reduced half, excuse me, you have oxidized half of the 10 ions in solution. Alright, and then you continue and at this point, you've oxidized all of the 10 ions. And then you start oxidizing the iron ions and then this at the halfway point as the standard redox potential is a half of the iron ions have been oxidized. And one question is, you know, when you drop the cerium in, why, why is it that only the tin oxidizes and not the iron? Okay, because I mean, iron is also lower in reduction potential than the cerium. So you know, both of these are fair game to being oxidized, right? But but if you were you're going to measure it, you'd only see the tin oxidize. So in reality, when when we drop this liquid in and the cerium reacts with the liquid, it could very well, it's very possible that it does oxidize the iron two plus the iron three plus. However, iron three plus also has a higher reduction pension potential than tin four plus. Okay, so that means the iron three plus is oxidizing any tin two plus that's present as well. So any iron, iron two plus that gets oxidized goes forward and oxidizes tin two plus and becomes reduced again. Anyways, okay, so that's pen potential metric titration. And again, the important of this type of technique is that it's at equilibrium. Each data point we collect is at equilibrium. Now we'll talk about cyclic photometry and I introduced this before. But what's different with cyclical photometry is that we're not we are no longer at equilibrium. Alright, so in this type of test, you might have a three electrode cell like this where you have the working electrode and that's that's going to be the reaction that you're you're interested in. So you're measuring the voltage of that reaction against a reference electrode. And then you're applying a current between that and the counter electrode. So the purpose of the counter electrode is only only to provide electrons or take away electrons, but we're not really interested in it, what whatever redox reactions happening on that surface, because we're not measuring the voltage. Now I said that we're supplying a current but I mean, really, we're controlling the voltage. So this type of test, we have we control the voltage. And we set it so that it it linearly increases at a certain rate, and then decreases a certain rate between, you know, arbitrary points that we select. It's important that these voltages are within like the electrolyte range. So if you're working with water, and if you push the voltage too far, then the water or the electrolyte will start breaking down as well. And you'll you'll get other things that you're not interested in. Okay. So here's a I should this should be a video if I click, or a, let's see. Yeah, this happened in the morning too. Oh, there we go. So this is the test. Okay. So we start out at a negative potential. Okay, let's say we're starting with iron ferrous chloride. It's 100% iron two plus. And we've set the potential to negative four. So again, we're in this test, we're applying a potential rather than just measuring it at equilibrium. So it might it's not necessarily at equal. This is below the equilibrium potential of the iron, but iron has already been reduced completely. So you're not going to see any current. Yeah, the y axis is current going through and x axis is the applied potential. So we're not at equilibrium at negative point four. And this is voltage versus the standard reduction potential. So at zero volts, that's just going to be for iron 0.77 volts versus SHG. So yeah, so we're below the equilibrium potential. But again, all the iron has already been reduced. So it's no, it's can't be reduced anymore. And we sweep the voltage, we increase the voltage, and then we start to reach the reduction potential. And that's when we see current and the current corresponds to the reaction, the electrons are being delivered to the iron or taken away from the iron two plus and it's oxidizing at that surface. Okay, and then we reach a maximum point. The reason why we reach a maximum limit in this case is that the concentration of iron two plus at the surface of this platinum electrode is decreasing as as we're as we're oxidizing the iron two plus the concentration decreases. Okay, and eventually we reach a limit where there's no longer any, any iron two plus available, but there's still there's still plenty of iron two plus in the bulk of the electrolyte, it's just not at the surface of the electrode. And so what happens is we've reached a diffusion limit. So we're we're waiting for the ions and solutions to diffuse to the surface and then also the iron three plus in the against the surface diffuse away from the surface. Alright, so that's why the current starts to go down. Even though there's plenty of iron two plus still in the solution, the current starts to go down because of the diffusion limit. So it's it's a kinetic limitation. And this peak current also can depend on you know, the size of this electrode if you have a higher surface area, of course, you're going to have more, more higher current than a smaller like a wire, platinum wire. Okay, so we're continuing to push this, this voltage, let's see if I can, okay, go up. And then we reach the the final voltage and reverse it. Okay. So we've reached our end voltage, which is what we this point here, and then we start to sweep the voltage down. Okay, so at this point, we still have positive current. See, it's above zero. If we have positive current, that means that electrons are flowing away from this electrode, they're flowing out of that lecture, we're oxidizing. So we're still oxidizing the iron because there's still plenty of iron available until we reach zero. And at that point, then we switch from oxidation to reduction. Okay, so now the electrons are flowing to the platinum electrode here, and reducing the iron three plus species. And then the same thing happens that you you have a diffusion limit of the iron three plus available at the surface, and then it continues to go. Okay, so that's that's an interesting to think about. And this kind of the fact that, you know, we're at we're at a positive potential versus sh e. So remember, this is zero, it should be really just 0.77 actually, we're at 0.77 volts above sh e. And we, we have both, you know, we can have positive current or negative current. And, you know, for the longest time, I forget when I learned this, but I always thought, you know, if you have a battery, and you discharge the battery, right, you're getting current out of the battery.  And that is entirely incorrect.  Alright, if you want to recharge a battery, let's say your lithium ion battery discharges at three or four volts, to recharge it, you actually want to apply a higher voltage across the same terminals, you're going to increase it to four volts or five volts. Okay, so it's kind of interesting to think about, you know, even though in both cases, we have positive voltage across the electrodes. It's, it's, the current can be either positive or negative, which is kind of strange. Now, now, that's versus a reference is, of course, versus the other electrode, the potential is those change. So that's that's where that's a bit different. So here's another thing to think about. Let's say we're running our test, and we start increasing the current, and we hit this zero volts, right, they were at the standard state, if we were to stop the test, let's say let's say we're not necessarily stopped, but we hold the voltage at zero, if we were to hold the voltage at zero, what would happen? Okay, we hold the voltage at zero, we don't, we don't progress any higher or lower. What happens is the current will decay along zero, okay, until it reaches zero volts. Alright, and at that point, we know that zero volts, the versus the standard reduction potential, that is when the product is equal to the reactant. Okay, and actually, in this case, because in this setup, it doesn't make too much sense. Actually, I'll show in the next setup. But because there's a bulk solution of iron two plus, we would probably never reach might not ever reach zero, because there's still going to be plenty of iron two plus in the bulk that's diffusing. So we'll probably reach as current limit. I want to correct myself there. But in the next slide, it makes more sense. So yeah, this type of cyclic voltometry has this diffusion limit. Okay, but there's another type. If you take this electrode and you move it around, for example, this is called a spinning disk electrode. So if you it's basically a like a platinum electrode, that's a disk, and it spins in solution. And as it spins, it's creating a convection current in the solution. So you're always delivering fresh iron two plus to the solution. And so it no longer has that diffusion limit where the current goes back down, but still has a current limit. And this current limit depends on factors like the concentration of ions, the the size of the electrode, the the speed that's spinning at. And also what's important is the diffusion of the the redox active species, right? So you can use this type of test. And if you know all those other parameters, you can calculate the diffusion coefficient of that ion in solution. So that is kind of an interesting test to do. If you need to find diffusion coefficient of ionic species in redox ionic species in the solution. Okay. So again, these this type of cyclic voltometry, this type of test, we're taking it away from equilibrium. And driving the reaction. So in these, what I've showed is that we have a redox active species as part of the electrolyte, it's in the solution. But batteries and other electrochemical materials, the redox active solution is not, you know, it's not an aqueous ion, it's actually a part of a solid. Okay. So in this case, all the species are already present at the platinum. So depending on what the material is, you don't have to worry about diffusion of your redox active species to the surface anymore, it's already on the surface. And if it has high electrical conductivity, then it has no no problem. And then you expect a curve like this where you'll have zero, when if the entire species is reduced, and then you start oxidizing, and then you're starting to run out of the material is starting to become fully oxidized. And then at this point, see, all the redox active species on the electrode have been fully oxidized. So it should reach a current of zero. And then you can do the opposite as well. But oftentimes in electrode materials, like battery materials, there's still an influence of diffusion. So you'll still see that these peaks will shift away from zero and because of the diffusion limit. And that's because, you know, if you have a lithium ion battery, you're still relying on the lithium from solution to diffuse into the material. And so you can, you can still deplete the concentration of lithium at the surface. And so you're still limited by that. And not only that, that's the solution resistance, but also the the diffusion resistance inside the material, right? So you can, you can intercalate lithium or sodium right into the surface of the material, but it still needs to diffuse throughout the material to make room for more lithium or sodium. And so that can also limit the limit this reaction. And so if that happens, you'll see a splitting of the peaks, okay? And in general, if if you increase the scan speed, right, if you if we're if we're sweeping the voltage faster, you'll see a bigger split in the peaks. And also the current will also increase. But you'll see a bigger split because diffusion is more limited at faster scan speeds. Okay. Oh, yeah. So in this example, this is where I can talk more about, you know, if you were to take this material and bring it, you know, measure it up to this point at zero volts versus its standard reduction potential, and you were to stop the test. Okay, what would happen? Well, remember, even even if we have are applying zero volts versus the standard reduction potential, it's not at equilibrium. Even if we're applying the standard reduction potential, it's not at equilibrium yet because we have current, we still have current. If you have current, it means there's a reaction undergoing. And so that means it's not equilibrium. But over time, if you hold it at zero volts, the current will decay, the current will go down, and then reach zero current. And then once it has reached zero current, then you're at equilibrium. Okay, another thing to think about is, you know, what if we had taken this material pushed into zero volts, and then stop the test, and then do no longer control the voltage, if we if we no longer apply a current, basically, if we disconnect this wire, you know, we push it to zero volts, and we disconnect this wire, what would happen to the voltage, the voltage would go back down. Because when we had brought it to zero volts, by forcing electrons in or out of the solution, that's not at equilibrium. Remember, this cycle of photometry is not at equilibrium, we're pushing it away from equilibrium in order for the reaction to proceed. And if we had stopped it at this point, it would go back, the voltage would go back to whatever its equilibrium voltage would be would probably it depends on the number of species that had already been reduced, maybe probably somewhere. And again, if we had if we had driven it all the way to zero, held it there, the current goes down to zero, left it, then at zero, half the species are reduced and half the species are oxidized. Okay. Okay. Oh, yeah, so another question to think about is in these three electrode cells, I mentioned this before, you know, we're taking electrolyte electrons away or giving electrons to the counter electrode. So there's some sort of redox reaction occurring at the counter electrode, but like I said, we don't really care about it. But in my research, I was interested in learning what exactly it was going on at that. In part of my research, I was investigating mixing non aqueous and aqueous solvents together and doing three electrode tests. But one of the problems I wanted I wanted to keep the concentration of water in my non aqueous solvent consistent. And the problem is, in a three electrode test, if you have your counter electrode, providing or taking electrons, it means that something is breaking down at that side, which meant that the water was breaking down, I wanted to keep that consistent. So in this experiment, I just had a completely aqueous cell. So it's all water and I had a electrolyte, I had my material on carbon cloth electrode. Okay, and what's different about this cell is that, you know, typically, right, I have my voltage against the reference electrode, which is in this case, with silver silver chloride is a different type of reference, and the material, and then the current between that and the counter electron electrode, but also on another channel, I was measuring the voltage between platinum and reference, which is kind of unusual. This is the typical CV curve of my material, which again, which Prussian blue, and this is in water. So it's, you have two different species of iron, and that's why you get two different peaks on CV, because you have there are two different chemical, chemical environments, what's surrounding them is ones is carbon, the other ones nitrogen. So that changes the energy of the iron. So that changes the reduction potential of the iron for each one. And so what I did, this is a little bit confusing, but I'll go through it. So on the first graph is the measuring in blue, this is the cyclic voltometry scan. All right, so we're just controlling the voltage, we're pushing it away from equilibrium, and, and it's sweeping it. And then the red curve is the current of the cell. Okay, the current between the carbon, my, my active material and the counter electrode. The second graph is the voltage in blue, again, from the counter electrode to the reference electrode during the same, during the same time, during the test. Okay, and what you'll see is, during high peaks of current, when we, when we need to supply a lot of electrodes or receive electrodes, you see the voltage against the counter electrode is negative. And then when we switch the current, it goes to the positive. And so if you see this plateau here, I measured that plateau, that potential, these two plateaus for the counter electrode, that corresponds to the, the, the breaking down of water, the electrolysis of water. Okay, so that's what is happening on this counter electrode during our test, is that we're breaking down water and either giving away electrons or receiving electrons. So if we look at that in more detail, so during the oxidation of my, my material, I need to take electrons away from the material. So this is reaction A on the counter electrode, on the counter electrode, the water is receiving electrons at this negative potential, and breaking down into hydrogen gas and hydroxide ions. Okay, and then on the reverse during reduction of my material, I'm giving it electrons. So that means that the counter electrode needs to supply electrons. So we have positive potential, where we break down water into oxygen gas and hydrogen ions, and we give away electrons. Alright, so that these this is, you know, the where these electrons are coming from in the circuit, right? But again, we're not too interested in this. And as long as the cell is a large volume, you know, any changes in pH should be rather insignificant. Okay, unless you're going to run the test for, you know, a very long time, or if the cell sizes are small, then it could influence it. One more thing, you know, so I drew these lines, the lines represent the standard reduction potential, or the reduction potential of water, either reducing or oxidizing. But you notice that the actual voltage is either lower or higher than those lines, corresponding to the amount of current that is needed to supply. And so this this either over potential under potential, we call it over potential. And it's an additional thermodynamic, you know, driving force to increase the rate of the reaction. I like to think about it in related to how phase diagrams work for like, for example, the iron carbon phase diagram, the steel phase diagram, you guys are familiar with this. You know, if you're going from austenite to pearlite, for example, or austenite to alpha ferrite, right, you at high temperature, the y axis is temperature, and you start in austenite and you lower the temperature, and you reach that line, right, that line is the equilibrium line. And then in order to transform austenite into pearlite, you need to go take bring the temperature below that line. And so that's that's called under cooling is the amount of temperature below that phase equilibrium, they have to bring it to increase the thermodynamic driving force for this transformation to occur. And you guys know if you bring it if you quench the material, if you bring it down faster and faster, more and more, that's a higher thermodynamic driving force for that transformation to occur. The same idea can be applied to these voltages is that, you know, this line represents the phase equilibrium between, you know, the breaking down of water. And we need to bring the voltage lower than that to increase the thermodynamic driving force for this reaction to occur, so we can either supply or receive more or less electrons. Okay, so that's called over potential. And the amount of over potential is related to the resistance in the cell. For example, diffusion resistance, water is kind of different because water is everywhere in the cell. But let's say there was a different reaction happening, where you needed ions from the solution to come to the surface, right? And just like before we have we had diffusion limits, you know, that can add additional resistance, and you might see this voltage drop even lower in order to get that required current that you need because of the slow diffusion of ions to the surface. That's one way. There's other other other things that can increase this resistance, such as the surface area of your electrode.  But if you use a large sheet, a large film of platinum, for example, has a larger surface area, you have more active redox sites on the surface. So you don't have to have as much over potential to create the same amount of current. Okay, I like to think of it, and I'm not sure if this is correct, although I'm pretty sure you can be thought of this way as Ohm's law. So if we require a certain amount of current, but there's a certain amount of resistance in our cell, for example, the diffusion resistance or the electrical resistance, then that creates a certain amount of over potential, which is the potential drop in the cell, beyond of like this equilibrium position. Okay. So as I goes up, as I goes up, like this red line has a peak, then the voltage drop also goes up. Okay. Okay, so since we were talking about Prussian blue, this will be the last thing we talked about, actually. All right. I'll let me introduce a little bit more history about this material that I've been doing research on. So Prussian blue is the trade name. The chemical name is iron hexacyano ferrate. And it's a quite, quite interesting history. It was discovered in the early 1700s by a paint maker, okay, who's who is mixing chemicals together to make a paint. And it was first the synthesis was first published in 1731. And by this guy, George Stahl, and I could be wrong, I thought George Stahl might have been from, from England, working in Berlin. And what was interesting is I found the original publication that includes the synthesis of what was called at the time Berlin blue is discovered named after the city was made in, which at the time Berlin was part of the Prussian Empire, which is why it's called Prussian blue, and not German blue. And you guys know, have any idea what language this was published in? So again, it was I think a English guy working in Germany. And what language did he publish this, this article and have the chat open? Hold on. If you guys have been asking questions, I haven't seen any of them because the chat wasn't open. Okay, no, no question. Okay. Well, if you can't read it, the yes, that's correct, Aaron. The language is Latin. And if you're if you know that the language of science during this time period was Latin, even though this guy was English, and he was in Germany. Everything was published in Latin. And so now now these days, the language of science is English. So convenient for us. That was kind of interesting. So I found this paper on like the Google books. And you know how you can search through Google books like word by word. And so I had translated Berlin blue into Latin to find exactly what page it was on, which was kind of kind of cool. I don't know what the rest of it says. But anyway, so it's significant, this material was discovered because it became the first synthetic paint pigment before for at least for blue, I mean, for blue paint pigment. Before the blues were made out of crushed up minerals, which could be quite expensive. And this was provided a cheaper alternative that can also be mass produced. And so it's used in a lot of different famous paintings. A lot of paintings is very common paint pigment actually, such as the Starry Night by Vincent van Gogh. And there's a bunch of these different paints are named after the materials that are used to make them like titanium oxide white or lead oxide white. If you guys remember a couple weeks ago, I was talking about Gero site as a material, Gero site has a yellow color. And it also has a paint pigment called Gero site yellow or something like that. So it's kind of interesting. So this is let me go back here. This is the crystal structure of the Prussian blue. And it has a cage like structure where we have iron FCC lattice. So face centered cubic lattice of iron in the high spin state. So these are the corners in the face center. And then also it's another FCC lattice of low spin iron and the low spin iron is coordinated octahedrally by six cyanide ligands. So cyanide as you probably know is a toxic material. And the reason why it's toxic is similar to carbon monoxide. So if we breathe in carbon monoxide, it binds to the iron in our in our blood, the heme group, and that prevents it from delivering oxygen to different parts of our body. And the same idea for cyanide is it binds the iron in our blood if we breathe it in and then prevents the oxygen transport. However, the iron Prussian blue is relatively non hazardous. Okay, even though it contains this cyanide ion, and that's exactly the same reason because the bonding between the iron, the cyanide is very strong. So it's very safe to handle. Unless you unless you eat it, and then maybe perhaps the it can break down in your stomach acid, but I'm not sure. Actually, actually, I could be wrong. I might be wrong. Actually, I think I'm wrong, because I know for a fact that Prussian blue, or at least this kind of the material is used as a medicine for radioactive poisoning. So for example, if you ingest radioactive element, particularly, it's just for I think, cesium, radioactive cesium. And if it's in your body, you would take this medicine Prussian blue and eat it. I guess you eat it. This is not medical advice. I'm not a doctor.  But the idea is, just like in a lithium or sodium ion battery, it has these large cage like structures, I have another slide here, large cage structures that can absorb large monovalent and even divalent ions. Okay, so cesium is monovalent cesium plus. And so it's a very large ion. So this, this material can absorb the radioactive cesium from it. And then I guess you pee or poop it out layer and gets rid of it from your body. But that's why how it's used as a medicine for radio radiation poisoning. But the same idea, you know, the advantage of this structure is that has these very open cages. So there's a lot of volume interstitial volume for large ions to be intercalated in and out of it at very fast rates. Right. So the diffusion coefficient of these ions in the material are fairly high, which is good for a battery material. And not only that, but such a large space, you can you can incorporate even larger ions other than like lithium. For example, sodium plus has good diffusion, potassium ion has been investigated for this material potassium ion battery that is. I forget about also, there's a, there's a research group in Portland, I forget what university or in Oregon, and they're they're investigating ammonium ion batteries. So instead of a single atom, it's a molecule of NH four plus that gets intercalated in and out of this material. And the advantage of using different, well, one of the factors changing the intercalation ion, well, the size of it changes the redox potential of this material. And the larger the ion this has been studied that the larger the ion that's intercalated, the higher the redox potential, which is good for a cathode material to want high higher potential. And the ammonium ion is the largest of them all. And it has the highest redox potential for Prussian blue if you intercalate it in. And one, one thing I should point out is that, you know, in this diagram, I draw the the metal plus ion the monovalent ion right in the center of these cages, but just is not necessarily true. Because these cages are really open. And it's just, if you guys remember from like msd 170, when you calculate the coordination of different materials, you know, if the ion size is a certain ratio to the cat ion size, and I on the cat ion size, it'll either be tetrahedral octahedral, right, basically, you don't want to have a small ion surrounded in a big cage, because then it's it can rattle around, it's not very energetically favorable. And so the same cases here is that this cage is so big, it's not necessarily the most energetic energetically favorable site is in the middle. Actually, there's a lot of studies that show simulations that the most energetic sites might actually be, you know, inside one of these squares or against one of these squares or against one of these corners. So it's kind of just a simplification in this diagram. And also, well, I'll get into that later. There's there's a lot going on here. Okay, so like I said before, the energy state of the two different irons for excuse me, the, yeah, the electron energies of two different irons are different because of what's coordinated to them. So the iron on the outside of this equation here that's bonded to the nitrogen of the cyanide complex, that is in what's called the high spin state. Okay, so these are the d orbitals that the electron orbitals. And if you take a single ion, ion that's not surrounded by anything, all the energy is symmetric all around, and there's nothing around it, if you take a single iron ion, all the energies are equivalent, they're all equal, they're all lined up with each other. But as soon as you start putting things around it, like ligands, for example, the cyanide ligand, then that's going to change the energy states of these different orbitals. And for octahedral symmetry, it makes these this kind of deviation of energy states of the d orbital electrons. Okay. And so depending on what is bonded to it, and also like the bond length of the ligands, that changes the energy splitting of these two energy levels of the d orbitals. And if that energy splitting is large enough, then you start to start to pair up these electrons, because it's more energetically favorable to make pairs of electrons before distributing them to all these single unoccupied states. And this is called the low spin state. I think I go into that a bit more later. So this is kind of schematic of how the sodium ion battery works for Prussian blue. So actually, when I synthesize the material, most often, it's synthesized as the partially reduced state. Alright, so it already has a sodium inside of it. And it's in the mixed valence between three plus and two plus to the different irons. And so usually the first step when I characterize my batteries, I have to either discharge it or charge it first. But so the charging process is when we're taking sodium ions out of the material, and we're also taking electrons out. Remember, I said we have to maintain charged neutrality. That's the golden rule about everything. You have to if you take an electron out of the system, you also have to take a positive ion out of the system. So we take a sodium out, we also take an electron out that were that oxidizes this iron two plus iron three plus so it's empty. This is the fully charged state or fully oxidized state. I guess I should be using the terms oxidized versus reduced or instead of charged versus discharged, because you could be saying talking about the same thing for an anode material and it'd be the opposite. But this is a cathode material. So when we take the ions out, it's charged. And then we go through the test of discharging the battery. Okay, and so the there's different potentials. Like I said, the first, the low spin has the higher potential. So we introduce an electron and a sodium and triculates, and that reduces the low spin iron. And then we have another step where we introduce a second sodium ion, and that reduces the high spin state. So those are the two redox steps in this material. This is a bit more explanation about the crystal field splitting, which is the you know, the energy splitting of these D orbitals. I think I was just showing. I don't have to go in too much detail, but showing that the chemical species, the ligand that surrounds this iron ion affects that energy splitting and eventually get to a high a large explosion of splitting that these electrons will go into the low spin state. You guys remember from chemistry class, what this principle is called when you're adding electrons to orbitals, it's the poly exclusion principle, I think I think that's it, poly exclusion, poly exclusion, poly exclusion principle, where, you know, you're when you add electrons to these orbitals, it's more energetically favorable to have them unoccupied. So you're just going to add one to each orbital. But in this case for low spin, right, it's more energetically favorable to pair them before adding them to the this empty higher energy state. Okay, so that when when you add electron, when you pair electron together, there's a pairing energy involved. So, you know, when I add this electron in here, the energy state of these orbitals actually increases, and which is not indicated by these diagrams. But in general, you know, if that that added energy is larger than what these energies are at, then it'll go into the high spin because it's more energetically favorable to put in that position. Anyways, this is kind of a more advanced in organic chemistry that we don't need to know too much about just kind of skim over. Oh, there's more, there's more. I made these slides. So I like to show them because I spent a lot of work, you know, putting the fine detail. It's important when you make presentations that the extra detail, like for example, I put a lot of work into these. These are all handmade these diagrams, these reference electrodes.  So it's nice to have nice, nice diagrams. But in this case, I talk about how the octahedral splitting energy, which is this, this, this energy gap depends on different things like the metal ion that you're looking at. So in general, I think you might be able to relate this to the periodic table. But valence state is definitely one of them. So like manganese two plus versus manganese four plus, that energy gap will change depending on the valence state. And then I think more importantly is the ligand. Okay, so what is being surround, what is surrounding these ions either in liquid or in solid will change that energy. And so, you know, typically, you're used to seeing all the d orbitals lined up like this, right? In this case, there's it's symmetric electric field, there's nothing around it coordinated to it. And then you start coordinating things to it. And then that changes the energy. Okay. Yeah, so here's carbon monoxide and cyanide as strong ligands. Oh, here's some more, some more diagrams. This just talks about the different orbitals between the metal ion, metal center, and the ligand and what pairs they make. And whether it's, you know, if you have these pi orbitals that make a pi bond, right, so if you have an empty pi orbital for the ligand, and then it's accepting electrons from the from the metal pi t two g orbitals, then it's a pi acceptor. And that correlates with a larger energy gap. Anyways, don't need to I was one more. This is molecular orbital diagram of iron cyanide. So you have your iron ion your six cyanides and how how the orbitals of the cyanides pair up with the orbitals of the the iron to make these bonds, right? So you have sigma bonds and pi bonds. And then the energy gap, so on. Oh, one thing I want want to bring up is not it's loosely related. When I was when I was making these orbital diagrams learning more about this, these energies, you know, in chemistry, like high school chemistry and freshman chemistry, we we always learn, you know, when we're adding the the energies, like counting up the electrons and these different energies in orbitals, we always learn to fill the four s orbitals before the 3d orbitals. I was wondering, I was trying to figure out why that was. And because, you know, actually, I found some papers related to chemical chemistry education, so saying that why we should not teach it that way.  And that's the controversy, right? I'm saying this is four s has a higher energy than 3d. And the proof is that if you had, let's say, a neutral iron atom, so you have, you have two electrons in the four s. And you have, you know, these were the six electrons in the 3d. So, you know, of course, yeah, that makes sense.  But which electron has the highest energy is the question. And I'm saying the four s has higher energy than 3d. And the proof is that if you were to ionize this iron, let's say we're ionizing it to iron two plus or iron one plus, which doesn't exist. Which electron do you take away? Are you taking away the 3d electron? Because according to what we learned in chemistry class, we would take away the 3d electron because it has a higher energy. But I'm saying, well, actually, four s has higher energy. Anyways, that's the proof that yeah, you take away from the four s at orbital, when you ionize something, you always take away the highest energy electron, and it's coming from the four s. That's the brief aside. But something you might want to investigate for yourself. Here's another related side note, but it's related to like these energy levels and materials as well. So we have three materials here, corundum, ruby and sapphire. Do you guys know what, what chemical equation this is for these materials? What's the chemical formula of these materials? Or at least the corundum, for example. And it's okay if you don't know. But I hope after today, you'll know. But ruby or sapphire, you know, if I say, Hey, what's what's the chemical equation for sapphire? Anyway, the chemical equation is aluminum oxide for all three of these materials. Okay, so the base materials all aluminum oxide is evident, you know, they all have some similar crystal, crystal graphic orientations and the crystal shapes. So they're all the same material. The difference between ruby and sapphire is that we've added a dopant. So a dopant is a small concentration of ions that's substituted with the aluminum in this material. And the dopant for ruby is chromium. So adding a little bit chromium, like less than 1% will make corundum red ruby. Okay, and then for sapphire to make it blue, it's tight, you have to add both titanium and iron to plus. And there's some kind of exchange and electrons when it absorbs light. So that's why you need to. So I guess the question is, well, let me move on. So here's another example. Where we compare corundum and another mineral called barrel. Okay, another the brilliant aluminum silicate. In both of these materials, if you substitute, I believe it's aluminum from barrel, that could be wrong, substitute about 1% chromium into both of these materials. It turns the corundum red, but it turns barrel green. And that's called emerald. So the question is why how can the same ion that is being substituted into these two materials turn one red and one green? Okay. And it has to do with the the d orbital splitting of the chromium, which is responsible for the absorbing the light in the material. Okay, just like I talked before, you know, it's all about these energy levels of these d electrons and what is around it affects that energy level. This is a Tanabe-Sugano diagram that's used often with these kind of kind of light absorbent materials. And actually, a TA from the morning group helped explain this because I wasn't too familiar. I made these slides a long time ago, I knew it at the time, but I can explain a little bit. So but I'm kind of paraphrasing what he said. So the x-axis is the ligand field energy, okay, or crystal field energy, you could also say is the is the energy splitting of these d electrons in the this is the octahedral configuration of the d electrons. So if you're a crystal field energy of zero, what that essentially is saying is that nothing is surrounding your metal ion, you know, it's just saying it's like this, this has a splitting energy of zero, all of these energy orbitals are at the same level, there's nothing surrounding the ion. But as soon as you start surrounding the ion, then you start splitting that those energy levels. And depending on the ligand strength and the bond length, the bond strength of the the ions and ligands, that affects the ligand field energy. Okay, and then this is the energy of light that's absorbed by the material. Okay, so for chromium, I found I found these this information for chromium that the ligand field energy of chromium in aluminum oxide is you know, 2.2 volts and then in the barrel, it's a little bit less. And that affects the the energy level of the light absorbed. So in aluminum oxide, you're absorbing these two like spectrums of light and then brilliant, you're absorbing different colors of light, which means different colors of light is transmitted. Alright, so this helps explain why chromium is responsible for both the red light and green light in both of these materials, as different because the ligand field energy and ligand field energy is different because the local environment surrounding the chromium atom ion is different in these two different materials. Alright, so now you kind of see how everything is related. I talked about, you know, how it's related in the battery materials, right? That changes the redox potential of the battery materials, it changes the optical properties of these materials. There's one more I thought about I forget. Anyways, here's another chart just relates the dopant concentration, the color. So I guess chromium oxides green aluminum oxides clear, but yeah, chromium makes it red. So just quickly go over, we're about about a bit over time, that's fine. I was kind of talk about some of the data from my, my publication. So again, this was looking at Prussian blue is looking at the synthesis of Prussian blue. Now and typically the synthesis, if you add a ferro cyanide salt, such as sodium ferro cyanide and dissolve it in water, and then add a ferric or ferrous salt, the press undergoes a precipitation reaction. Okay, the solubility product of Prussian blue in water is very low. It's like 10 to the negative 200. It's very, very low. So that means as soon as these ions are in contact, they're going to precipitate into a solid. Okay. And because of such high precipitation or low solubility product, what ends up happening happening is, you get very small nanoparticles, you get very high nucleation. And if you have high nucleation, you usually have low growth. Alright, because you're taking the ions away from solution, ions allow you to grow the crystal. And so you what ends up having you have these really rough nanoparticles and they're more or less grown not grown under thermodynamic conditions. So they don't have nice facets or cubic structure. And there's a lot of vacancies in water because of the fast reaction as well. The idea was to add a collating agent to this reaction that would collate to one of the iron ions. So essentially, we're reducing the activity of the iron species in solution. And that would help slow down the precipitation reaction, and then also adjust the pH because the collation strength of this chelate is affected by the pH. And so what ended up happening, right, you get a different pH, you get different sized particles, you get different vacancy content, different water content. And then more important, most importantly, you had a different redox characterization. So here's another view of it. These are the particles grown at lower pH at higher pH, the particles were much larger. But what's interesting, if you look at these particles, they're not necessarily cubic. Right? I mean, they're definitely faceted.  Okay. And this in this crystal structure, it's going to be a cube. But like just looking at it, you know, you have like a rectangular prism here, you have something with like a step, you have like an L shape here is kind of weird. And you can see the same thing in the smaller one, but they're just much smaller. And so the what I had done some literature research, and it seems like it's fairly common with the Prussian blue material grown under thermodynamic conditions to make a what's called a mesocrystal mesocrystals when you take smaller crystals, and then they they aggregate together in an oriented matter, and then they can fuse together to make another bigger single crystal Audi small nano crystals. Okay. And so that's sure that makes sense. And it's supported by literature. But the other question was, you know, why, why do I have small mesocrystals versus large mesocrystals? And also, saying that they're mesocrystals of speculation, I haven't, I don't have any hard evidence that they are mesocrystals, but other than leisure. But the question, why is the small and why are these big? And so my, my proposal, my proposed mechanism was, well, because the the solubility product of this material is so low, and they precipitate so faster, such high nucleation, that the particles growing at low pH have much higher nucleation. And then that results in very small crystals. And in those small crystals, together can aggregate to form smaller mesocrystals. On the other hand, the higher pH would have low nucleation. Okay, and that allow for more growth, larger particles, and then that would aggregate into a larger mesocrystals. And I wanted to I wanted to prove that the pH 3.8, you know, I had some faster growth is what I want to show. And then at 4.4, I had slower growth. So what I did is I took a UV vis spectrum of the material. So these are actually four different pH is my only show two on here. And it shows the peak at 700 nanometers 700 nanometers is the main absorption peak of Prussian blue. And so 700 nanometers is like a red, right? So it's kind of almost infrared. Yeah. And so if you're absorbing red, that means you transmit all the other colors and it ends up being blue. That's why Prussian blue is blue. But what I did is I took my my material, and I put it into a cuvette. And I added the two components together and started the test right away, as it was reacting. And then every five seconds or so, I think it was five or 10, 10 seconds, every 10 seconds, I would take a scan or a measurement just at 700 nanometers of the intensity of the solution. Because as this material precipitates, it's going to absorb more light, or it's going to absorb blue light or excuse me, it's going to absorb the red light at 700 nanometers. So here we see a pH 3.8. As soon as I add the two solutions together, you know, it becomes saturated very quickly. So that means that the the nucleation growth is very quick. But at lower higher pH, it's much more gradual, which means that the growth of the particles and nucleation is much more slow. And that makes sense because at higher pH, the collision strength of this chelate is higher. So it's more it's inhibiting the nucleation of this reaction. So it slows it down. And that's why you get larger growth. Okay, so moving on to electrochemical properties. So here is the capacity, the discharge and charge capacity over many cycles. And then also I increase the current rate after every 10 cycles. So remember, this is galvanostatic cycling. So this is like the profile of each one of these dots is like this. This is actually the last dot of each of each series is this profile. So if you just look at one curve, this is the discharge curve, you start at high voltage after charging, you're discharging at a constant rate. So the first section is only 100 milliamps per gram, right, and you measure the voltage as a discharges. And like I said, there's two different iron species in this material, and that represent two different redox reactions. So here's one redox reaction is a higher voltage corresponds to the low spin iron, and then a lower redox potential corresponding the high spin iron. And then the other the other chart the other plot line is just the charge profile of the same material. Okay, and I said before, you know, as we increase the current of the battery, as we're drawing more current, things become there's at there's the capacity decreases and the voltage decreases. Remember, I said, you know, I was kind of relating it to V equals IR, that there's these resistances inside the battery, such as diffusion resistance, or electrical resistance. And as we increase the current, you know, resistance more or less stays the same. But as we increase current, that means the voltage, in this case, voltage drop increases. So that's why you see this voltage as we increase current rate, the voltage drops. Okay. So if we just one one idea, obviously, you'll see that the capacity of the smaller particles is larger than the capacity of the larger particles. Okay, and now, I've said this before, in a previous lecture, you know, it's always better to make nano materials. Okay, if you make something nano, it means the diffusion distance of lithium ions or sodium ions and solution into the materials shorter. Alright, it doesn't have to travel as far. And that results in less resistance for diffusion during discharge. Okay, and that would result in higher capacity as well. So, you know, right away, you look at this, and you look at the different materials, and you say, Well, you know, this has higher capacity, because these are smaller, smaller smaller particles, it makes sense, right? And so if that was the case, if if capacity was kinetically limited by the diffusion of ions, and due to the particle size, you would expect to see as we increase the current rate, that the current the capacity retention of larger particles would decrease even more. But if you compare the initial capacities at slow rate to the capacities at high rate between the small particles and the large particles to the ratio of this number to this number, basically, they're they're pretty much the same. Alright, which would show which would indicate that kinetics did not play a major role in this material as far as the capacities go. However, if you investigate it a little bit more in detail, but in these charts here, I separate an estimated capacity contribution, you know, for each one of these lines, just the discharge ignoring the charge of the low spin iron, which is this higher voltage plateau here, versus the high spin iron, which is the lower voltage plateau here. And so the red is the low spin iron capacity contribution, you see it's this is a bit smaller than this line here. And the black is the high spin. And if you compare the two particle sizes together, you'll see that as you increase the current, the capacity contribution of the high spin iron is more or less constant. Okay, and the same for the larger particles, it's more or less constant, which would indicate that the redox mechanism related to the high spin iron is not dependent, or largely dependent, everything's dependent, but it's not largely dominated by kinetics of the reaction, right? It's a we're able to supply it electrons and we're able to supply it ions through fast diffusion effectively. Okay, there's no big decrease in that that level. But if you look at the low spin capacity contribution, that's where you see a decrease. All right, which is kind of unusual, because you have one material. And there's a difference in redox mechanism between the you know, this plateau and the second plateau, which is a bit interesting in my opinion. And I don't have I don't have an explanation for why this redox mechanism of low spin is different. Other than that, perhaps it's not what we think it is. So we always think, you know, we add sodium ions in and we're reducing iron, three plus iron two plus, and it was the same for the both. But perhaps that's not the case for this higher voltage case. So I'm looking into this. When I get back to lab, this is the next thing I'm going to look at is investigate this redox potential at different rates of different materials. I speculate that actually the water content might have something to do with it, actually, because if you if you compare a non aqueous Prussian blue to aqueous Prussian blue battery, you always see that this plateau is much higher in the aqueous state than the non aqueous state. So it makes me think that water plays a large role in the redox reaction of this material, even in this non aqueous battery, because I still have water inside my material. So anyways, so the point is, is that, well, there's still a lot to learn. And I think once I figure this out, I can publish that next paper and I can get out of this university and then move on to the next big thing. I'm gonna end it here. It's been about an hour and a half. Any questions before we we we leave? Oh, thanks for sticking with me. I believe on Friday, you have a journal review. Do excuse me, literature, literature summary is due on Friday. So remember, for your your review paper, you need to have at least 10 literature sources since 2019. But there's only five literature summaries, summary assignments. So if you want to be proactive, you can actually make two summaries, each assignment, but you only you only need to submit one to get full credit and it is for credit. So please write out a summary. I'm going to try to write a summary example tonight and send it to you guys to show you what I'm kind of expecting. But basically, the more the more you do, the easier it's going to be for when you want to compile your your review paper. Okay. Yeah. And then Tuesday will be the first day of lab. So we'll meet here again at the same time. In that day, the TA will lead most of the lecture, and I'll be here in support. So we'll go through some discussion questions. There's no pre lab. So you don't need to prepare anything. But, you know, just just be ready to learn, I guess, be ready to learn. And be sure you can take notes during the lecture. So that at by the end of the lecture, you can submit a summary of of our lab lab lecture. So the TA will discuss some some background and discussion questions of the topic. Then we will have a video of the experiment that Tatiana has has been busy putting together these videos. And now there's an issue with streaming the video, I tried streaming videos on zoom. And it doesn't really work because it there's like a lag. So I think what we're going to have to do is once we get to that point, you know, we're going to say, Okay, take 20 minutes to watch the video and the regroup, and you watch it on your own, and then we'll regroup and then we'll have discussion questions. And then the TA has data, you'll be given the data. And the TA will go over a data set on how to like plot or analyze the data. And you guys do that together. So it'll be it'll be interesting. Now it goes. I'm quite curious myself, it'll be the first time. Okay, any questions? And if not, we'll, you guys can get out of here. Okay, well, I'll see you guys next week. Have a good weekend, everyone. Try to get some sunlight. But you know, stay away from other people, but try to get outdoors, do some exercise.


What I will do first is, let me just point out some of the important things that we do at surgery when you want to mark a flap. The scalp flap in general, we try and make it cosmetically acceptable. So, you do not want to get it on to the forehead and stay behind the hairline and you keep the base of the flap broader than what the length is going to be for the vascularity. And if you know all the vessels that come in from the base, you base your flap onto one of those vessels, all right. Now, in terms of marking the flap based on the MRI showing you the lesion, there are some bony landmarks which sometimes we find confusing. So, let us just go through them one by one. The sagittal, the midline, quite often you find that once you position the head and somebody tilts the head a little bit, the other person comes and marks it completely off the midline. So, please take a pen, feel the lesion and mark it along the midline.  So, midline is one thing that is extremely important. I assume that all of you practice the timeout. Do you have a timeout? You know what a timeout in basketball is. You have everybody sitting and playing and then suddenly the coach notices that there is something going wrong. He calls it a timeout.  So, at surgery when you are doing before you start, the anesthetists are busy starting the lines. Nurses are getting the equipment together. The surgeons are putting up the scans. And so finally, nobody knows whether you are operating on the right patient. Are you doing the right side and what procedure, all right. And so, I have a notice in my theatre saying please practice the three P's. Patient position and in Tamil, side is pakam, so, patient position pakam. Make sure that you do these three things. So, side is important, all right. It seems silly, but please do it. I have seen one situation in our theatre where a senior professor showed the, on the left of patients, the head is fixed. He showed the junior consultant, he marked the coronal and because he marked the coronal on the left side, the consultant thought that that was the side that he wanted to operate and they opened that side, okay. So, the person who is going to be marking the flap, please make sure that you do these the timeout. Now, and that is why the midline is going to be important. And then you have to say, this is a left-sided tumor, this is the left side. Sometimes when you turn the patient prone, you are going to get your side mixed up. So, it is not silly. Next is the coronal suture. Now, where would you look for the coronal suture, feeling for it? Because you cannot see it on the bone. Feel it next to the midline, all right. The further lateral you go, the less prominent it is and quite often it is not felt at all. So, near the midline you start from anterior, start feeling posteriorly and the first bump that you feel that is the coronal suture. Now, I just wanted to talk to you about the tereon. The tereon is the junction between the greater wing of sphenoid. It is an H-shaped suture. The greater wing of the sphenoid down below, the frontal in front, the parietal superiorly and the temporal posteriorly, greater wing of sphenoid, frontal, parietal and temporal. Now, what I want you to do is, I want you to put your finger on the tereon and then turn the skull inwards and see where your finger points. Tell me where you see it. If you make your burr hole just in front of the tereon, front or inferior to it, you will expose three things, the frontal dural, the temporal and you will also get the periorbiter, all right. So, that is a very useful point. Some people call it the psychopathic point because if you do not get that point right, the surgeon becomes a psychopath if you are assisting him. Now, quite often I have noticed that people put the burr hole out here, you know, junction between just behind the psychomatic process of the frontal. Now, that is not right because that will only give you the frontal dural. Put your finger on it and have a look at it yourself and secondly, that will give you quite a cosmetically bad result unless you plug it, all right. Now, how many burr holes would you place for a craniotomy, whether you put one or multiple, that it kind of varies for older people, generally put in more because the dura is stuck and so you might tear the dura. But otherwise, if you have one burr hole and you have got a craniotomy, you can manage quite well, particularly if it is frontal or parietal. If you want to go across the sphenoid ridge, you may want to put one on the tearyon, in front of the tearyon and then one frontal so that you can go across the lesser wing easily. Whether you put it on the midline, suppose you need an inter hemispheric approach, you want to get as close to the fox as possible. Would you put it on the midline, just off the midline, what is your feeling about that? Yeah, well, you know, what happens is that if you are careful with these perforators, getting onto the sinus, you will never damage it. In fact, when you are doing a posterior fossa, if you want to do a posterior fossa craniotomy, if you put a burr hole, bang on the tocular, you will practically never damage it, perfectly placed on it. But if you are a little worried about it, you could put it just off the midline. You may not get too close to the fox, but try getting it onto the midline, you will find that it manages quite well. As far as burr holes and craniotomy is concerned, I will just show you some cases now that we have done and then I will get some feedback from you about what you think you would do. This is a boy who came to us, his left hand was totally clenched like this, but he was able to move his right hand well. He had dystonic movements, we made a walk, you can see his left leg circumducting a bit and his left hand totally clenched like this. So, a neurologist saw him and felt that this was more of a dystonic problem, felt that he had a lesion, this basal ganglia, got an MRI done and this is the MRI. You can see that it is located deep in the capsule ganglionic area, it is partly solid, partly cystic. It is not really extending into any of the opecula, but the superior part of it is cystic, there is a central area that is enhancing, the solid portion. The cyst wall does not seem to be enhancing, that is a sagittal view which gives you the impression that it is deep, it is not really into the either frontal or temporal. That is the coronal view which shows that it is really kind of displaced all the deep structures. The question is how would you approach this, what sort of craniotomy would you do to get at this and what are your concerns? We have got this DTI and the yellow portion there is the, that is solid portion of the tumor and here you can see all the, the archaea fasciculis, the superior and inferior longitudinal fasciculi and the internal capsule there is medial to the tumor. Let us go a little inside, you can see the internal capsule medial to it, that is the unsinnet fasciculis, but here essentially you see that all the fibers have been displaced out of the side and it appears that going through the sylvanes seems to be a good option rather than going through frontal or temporal. So that is what we did, made a large flap there, did a frontal temporal craniotomy, got the sylvanes fissure in the middle of the field. In general I like to keep the head kind of horizontal so that your orientation is okay. So here we are opening up the sylvanes fissure, you could either use a needle and then very gently open up the whole sylvanes fissure, but finally you do have to make a cotysectomy onto the insula. So that is what we are doing right now, all slowly, now we have got a tubular retractor which we have inserted, this is about 1.9 centimeters in diameter and I found that actually this branch of the middle several artery was coming into the retractors a little too big which is why I finally removed it and just had one retractor placed in and here we have and then of course the whole principle is to stay within the tumor rather than do any dissection outside for fear of damaging the white fiber tracks. Now you could use the QSA but I find that sometimes particularly at the dips the QSA goes and gets away tumor too fast and then you might end up in white matter earlier but I do use it from time to time. So this is one situation at the end of it here we are just putting in an ultrasound to see how much of tumor we left behind, we did leave behind tumor medially onto the thing but ultimately when you look at it the frontal and the temporal opecular quite nicely preserved that is the postoperative scan we see medially on the area where the internal capsule fibers were displaced we have left tumor behind it, this was a pylositic astrocytoma. So we have left him alone and he looks happy, he started smiling after surgery and more importantly his hand opened up that was good and this is another guy who somebody asked him to show how he shaves his beard and he is doing all sorts of stuff maybe that is how he really shaves his beard that is his image. So we had a huge big debate before surgery wondering whether this was an extra axial lesion is it inside the ventricle because there is you can see this tumor in the quadigeminal system it is extending into the atrium of the lateral ventricle it was going into the body of the lateral ventricle I mean finally our conclusion was that this was an intraventricular tumor then somebody pointed out that it was attached to the tent over there is it a meningioma what do you guys think that is a T2 flare sagittal coronal with gado any other differentials you think of I mean this is the ones that we thought of probably in a pandemoma or some other glauoma or a meningioma. So how would you get at this the left dominant hemisphere one thing you must understand there is no one correct method how would you open the head now in this this approach actually is quite good except that when you are dealing with something down here the distance is huge and so finally when you have got a vascular it is going to be vascular you are going to be operating at a great depth you might get away with it so it is an option right wait hold it you have already done your craniotomy and now you have started the craniotomy what would you do you may need to cut the tent but you may not but I agree that that is an option. So what we did was we had a look at the DTI again and so here you worried about the optic radiation and here you see the tumors in yellow and wherever you look there are white fibers wherever you go isn't it and so what we decided was we would do this inter hemispheric posterior inter hemispheric approach and what we did was we kept the left side down so that it would aid in gravity falling away and marked it right on to the midline that is the sagel sinus we opened the bone we found that the dura was really really tense so what would you do tap the ventricle so that is what we did over here and you know you be surprised at how difficult it is to to tap a ventricle so we had put in ultrasound to see where it is especially when the patient is positioned and turn the head but a good rule to follow would be to try and go perpendicular wherever you are when you perpendicular to dura you should hit the ventricle but when you have to take into account that the ventricle would have shifted because of the mass so we usually make sure that we have the ultrasound into to guide us when you want to tap but how much help removing CSF does just look at that absolutely nice and like so now when you open the dura you find that actually the without even putting in a retractor the occipital lobe is nicely away from the forks and then you get at this turned out to be a very soft extremely vascular tumor completely surrounding the vein of Galen etc it was a pendomoma so we left behind tumor on the vein of Galen and the basal vein of Rosenthal that area but we managed to take off all the portion laterally now here's another boy almost the same situation 17 year old one generalized seizure no focal deficits that's his MRI and it's draining superficially and so we decided to do a similar posterior intramasphoric approach in exactly the same position as the previous one now in terms of opening the dura if you are going to get to the sagittal sinus you can quite safely go right up to the edge and make sure you do that posteriorly you'll find that there are several veins that that can actually get into it and preserve all veins try and preserve all veins when you get to the sagittal sinus don't hesitate to even extend the dura opening either superior inferior if you find there are veins there you might be lucky all you need is maybe 1 to 1.5 centimeters of access because with the angulation of the microscope you can take out fair amounts of the AVM so we've basically got some veins draining into the fox over there and we're not really sure whether that is the main area so we've just putting a occlusion on it to kind of see whether there's swelling hold it on there for about a minute or so and there was no swelling so we cut it and then we found that actually the main draining vein was was there inferior and got this there you go down there so we find this approach and positioning quite quite useful what I do is I just keep them lateral and keep the head up 45 degrees in this turn the head down in our situation we need to keep things simple I think a three quarter prone would require a lot of people to make sure that there's the patient's position well good support so he doesn't fall forward so the lateral is one thing that even the technicians know how to position in and so we prefer to keep it simple this is a girl who had raised intracranial pressure obviously with left cerebellar dysfunction that's her MRI looks like it's a fourth ventricular mass or probably from the worm is there you can see a little bit of fourth ventricle seems to be going out into the foramen lusca on the right side so a regular midline posterior fossa mass and girl could be a medulloblastoma or ependomoma or something what I mean we've been doing is once you do your craniotomy typically how do you open the dura how do you open the dura and the posterior fossa and what are the things you need to watch out for so basically when you're opening the dura and the posterior fossa you've got to watch out for in children particularly where the occipital sinus might be quite big and even the circular sinus can be quite big and so you cut the dura on both sides put clamps and then cut and then tie and this sometimes I find quite cumbersome and ultimately when you open up the whole both cerebellar hemispheres and you're getting into the ventricle through one side one Tilo Velo tonsil approach it almost seems wasteful to open both so what I've been trying recently over the last year it seems to work so far is open the thing on one side and here we've got rise and cut the circular sinus we're far away from the midline and so we haven't gotten to the occipital sinus at all and we open the dura on one side alone and curve it towards the midline like a flap and so when you get it to the midline here reflected with silk sutures medially and then you have the entire midline even some of the opposite side you can see and then the axis for a Tilo Velo tonsila approach is between the vermus and the tonsil all right and we're going into the right side because you remember the tumor was getting into the lusca on the right side on the scan so that's the vermus there we're cutting open the arachnoid and getting into the Tilo Velo tonsila groove and it's amazing how if you open this groove from the lower end to the upper end you can actually see the entire flow of the fourth ventricle to the aqueduct you don't need to split the vermus at all and then you have a very good axis and the rest of the you wouldn't believe that I've not opened the dura on the left side because if it's I'm effectively in the midline so I find it useful doing it on one side and the great thing about it is that you can close the dura completely I'm a great believer in Dural closure and the vermus is safe so all this mutism and all that is this was a medulloblastoma all right thanks a lot