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Current time:0:00Total duration:5:22

Video transcript

- [Voiceover] We've already seen the equation on the left, which relates the standard change of free energy, delta G zero, to the standard cell potential, E zero. The equation on the right is from thermodynamics and it relates the standard change in free energy, delta G zero, to the equilibrium constant, K. So, we can set these equal to each other to relate the standard cell potential to the equilibrium constant. Since both of these are equal to delta G zero, we can say that this is equal to this. So, now we have negative NFE zero is equal to negative RT natural log of the equilibrium constant, K. Now let's solve for E zero. Let's solve for the standard cell potential. So, to do that, we need to divide both sides by negative NF. So, we get E zero is equal to positive RT over NF natural log of our equilibrium constant. Next, we're gonna solve for what this is equal to. So, if we're at 25 degrees C, alright, so our temperature under standard conditions for our cells that we've been talking about, T is in Kelvin, so we need to convert degrees Celsius into Kelvin, and to do that, you need to add 273.15. So, that gives us 298.15 Kelvin. So, that's what this T is, it's our absolute temperature in Kelvin. R is the gas constant. So, R is equal to 8.314 joules over mole times Kelvin, here. So, we're gonna multiply that by our absolute temperature. And so, our absolute temperature was 298.15, so, this is 298.15 Kelvin. This is all over Faraday's constant. Alright, so remember, F is Faraday's constant. So, this F right here, Faraday's constant, which is 96,500 coulombs per mole, so, the charge of one mole of electrons. So, this give us RT over F. And so, let's get out the calculator and find what this is equal to, here. So, we have 8.314 times 298.1, lemme go back here, .15, and then we're going to divide that by Faraday's constant, 96,500. And so, we get .0257. So, let's round that. .0257, so this give us .0257. What will the units be? Well, Kelvin would cancel out, here, and, let's see, what else cancels out, the moles would cancel out, and that gives us joules over coulombs. Which, of course, is equal to volts. Alright, so we can rewrite our equation, so, the one we had up here. So, we're gonna plug in for RT over F, now. So, now we would have the standard cell potential, E zero, is equal to, well this would be .0257 and that was volts over N. Remember, N is the number of moles that are transferred in your redox reaction. And this is times the natural log of K, our equilibrium constant, here. So, this is one form of the equation that relates the standard cell potential, alright, the standard cell potential E zero, to the equilibrium constant, K. We can write this in a different way. Alright, so what we could do is we could take that .0257, .0257, and we can multiply that by the natural log of ten. So, let's do that. So, we have .0257 times the natural log of ten. And that gives us .0592. So, we get .0592. And the reason why we could do this is to write our equation in log form. Alright, up here, we have natural log, so up here, we have a natural log, but now, we can write it in log form. So, now we have the standard cell potential, E zero, is equal to, well now, we'd have .0592 volts. So, we have .0592 volts, once again, over N, the number of moles of electrons transferred in our redox reaction. And this time, it would be times the log of K. So, not the natural log, the log of K. So, we've taken care of that in our calculation. So, this is just another form of the same equation. Alright, relating the standard cell potential, E zero, to the equilibrium constant, K.