One of the properties of the standard cell potential E, or E-naught, is that when you multiply it by a factor of 2, 3, and so on, it's going to remain unchanged. So this is kind of what I'm talking about. So here's an example.
We have lithium turns into lithium plus an electron. It is an exothermic reaction with a delta G of 289.5 kJ per mole and a standard cell potential of 3 volts. If we were to double the chemical reaction, we simply multiply this all by a factor of two, right? So we get two, instead of one lithium, we have two Li.
Instead of one Li plus, we have two Li two plus, excuse me, Li plus, two electrons and so on. So the question is, how much energy do we get if we were to double the quantities that we started out with? Well, we're going to double the amount of energy, which is our delta G.
However, the standard cell potential is going to remain unchanged. So that's going to be... a thing to remember. Now why doesn't it change?
Well, partly because this is a potential energy, right? It's called cell potential because it is a potential energy. Why is it a potential energy?
Well, this is just joules per coulomb. That's the unit of volt, right? So it is the amount of energy per electrons reacting inside.
whatever chemical reaction. So in this case, if we have one electron, the amount of energy produced is this amount. If you have two electrons, the amount of energy is double, but the energy per electron is exactly the same. Now, of course, this is also expressed kind of mathematically, and yes, there's some derivation here that I'm going to show in just a tiny bit. You can do it both kind of understanding what it is, as well as do this mathematically.
That is, the Nernst equation is delta G naught. is equal to negative NF E naught, right? So that's kind of what we have.
And if we want to convert delta G into cell potential, we simply have to rearrange the equation. This becomes negative delta G naught divided by NF. Okay, so let's try to convert these into, how do we get three volt in this case? And I'll try to show how these number remains the same.
So in order to do this, we have to apply this equation. So It's going to be a negative of the delta G. The delta G itself is negative 289,500 joules, because it needs to be in joules, per mole. So that's our delta G, divided by N. How many electrons are involved in this particular process?
Well, we have one electron here, right? So that's one. So that's our N.
And what is F, or the Faraday constant? Well, this is just going to be 9,6485 coulomb per mole. And you can see that the unit cancels out nicely. We have joules per coulomb, which is the unit of the fault.
Okay, so how about the second one? Well, if we were to double the chemical reaction, what I want you to see is that we double the delta G, right? So this is technically, this 579 is 2 times negative 289,500.
I've missed a 5 there. Joules per mole, divided by... Well, not only that we double the amount of energy when we calculate the delta G, we also double the amount of electrons involved. See, there's a 2, factor of 2 from the 2 electrons, times 96485 coulombs per mole.
The unit cancels out nicely as usual, but here's kind of what I want to highlight. Notice in this case, we have 1 electron, and this is kind of multiplied by a factor of 1. But once we double it, so in this case we multiply it by a factor of two, we double the amount of energy, or the delta G doubles, but also we double the amount of electrons. So in this case it cancels out.
So that cancels out. So that's why the cell potential remains the same. If it kind of makes sense to you that this is just the standard energy per electron, that's why it doesn't change, because now you double the electrons.
Or if you're more kind of mathematically driven. then that's kind of the basis of it is that the delta g doubles and then the number of electrons also doubles in the same factor and that's why it cancels out precisely the only time that you have to concern and change the e is that when you flip the chemical reaction right so in this case when you flip if you flip e becomes also flipped So for example, if we were to have our lithium reaction as it was before, so we have lithium turns into Li plus electrons. And then suppose that we flip, right? So let's say flip and multiply it by a factor of two, right?
So that's negative two. So what happens is that in this case, we have two electrons plus 2 Li plus turns into 2 Li, right? So we multiply by a factor of 2 and then we also flip the chemical reaction, whatever the product turns into reactants and vice versa.
Well, what happens to the delta G in this case? Well, it was negative 289.5 kilojoules per mole. If we multiply, we flip the chemical reaction and multiply it by a factor of 2, we're also going to multiply by a factor of 2 and then flip the sign.
So this is going to be 579 kilojoules per mole. So what happens to the E in this case? Well, in this case, the E is going to be, originally, is 3 volts, so that's the same. But if you were to flip, notice that whatever on the reactant side becomes a product, then product becomes reactants, so we also have to flip the sign. So this is going to be negative 3 volts.
But we ignore the factor of 2 because, you know, the energy per electron is the same. It's just the sign, or the reaction progress is flipped, and that's why we add this negative sign in. Okay.
You can do that mathematically as we did from before. I don't feel like doing it just because it's just more math. So the point that I'm trying to make here is that this is the derivation of why E remains unchanged when you multiply it by a factor.
But when you flip the chemical reaction, don't forget to put in that negative sign in there. Okay, thanks everyone.