If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:9:08

Standard potential, free energy, and the equilibrium constant

AP.Chem:
ENE‑6 (EU)
,
ENE‑6.B (LO)
,
ENE‑6.B.1 (EK)
,
ENE‑6.B.3 (EK)

Video transcript

- [Instructor] For a generic redox reaction where the reactants turn into the products, the free energy is related to the potential for the redox reaction. The equation that relates free energy and potential is given by delta G is equal to negative nFE. Delta G is the instantaneous difference in free energy between the reactants and the products and refers to the number of electrons that are transferred in the redox reaction. F is Faraday's constant, which tells us the charge carried by one mole of electrons. So one mole of electrons carries 96,485 coulombs. E refers to the potential of the redox reaction. You could also call this the voltage of the redox reaction. If the reactants and products are both in their standard states, we can add the superscript naught to delta G and the superscript naught to E. So we can use this equation to calculate delta G naught from E naught or vice versa. We could calculate E naught from delta G naught. And remember that delta G naught is connected to the equilibrium constant K by this equation. Delta G naught is equal to negative RT natural log of K, where R is the ideal gas constant and T is the temperature. And because delta G naught is a part of both of these two equations, these two equations connect the standard potential E naught to the standard change in free energy delta G naught, which is then connected to the equilibrium constant K. So let's look at an example of a redox reaction where we know the value for the standard potential E naught, and first we're gonna calculate delta G naught. And then from delta G naught, we're gonna calculate the equilibrium constant K. So here's our redox reaction, and we're gonna calculate delta G naught at 25 degrees Celsius from E naught. So for this redox reaction, E naught is equal to positive 1.54 volts, and our goal is to calculate delta G naught at 25 degrees Celsius. In our redox reaction, silver one plus is being reduced to solid silver, and solid chromium is being oxidized to chromium three plus. Now that we've identified what's being oxidized and what's being reduced, let's write the oxidation and reduction half reactions. So for our reduction half reaction, we have silver one plus gaining an electron to form solid silver. And for the oxidation half reaction, solid chromium is turning into chromium three plus and losing three electrons. Next, we need to figure out how many electrons are being transferred in our redox reaction. In the reduction half reaction, the silver one plus cation is gaining an electron, but in the oxidation half reaction, solid chromium is losing three electrons. And we need to make the number of electrons equal. Therefore, we need to multiply the first half reaction through by a factor of three. So now when we add together our reactants, so all the reactants on one side and all the products on the other side, the three electrons cancel out and give us our overall redox reaction. Because three electrons, three moles of electrons were being transferred in our redox reaction, n is equal to three. And since we know that n is equal to three and the standard potential is equal to +1.54 volts, we're ready to plug into our equation to calculate delta G naught. So I went ahead and rewrote our redox reaction, and our goal is to calculate delta G naught. We already know what n is, we already know what E naught is, and F is Faraday's constant. Here's our equation with everything plugged in. N is equal to three. There were three moles of electrons transferred in our redox reaction per mole of reaction. F is Faraday's constant, which is 96,485 coulombs per mole of electrons. And the standard voltage was equal to 1.54 volts. One volt is equal to one joule per coulomb, so I went ahead and changed the units from volts to joules per coulomb. Next, we can cancel out some units. So moles of electrons will cancel out, and coulombs will cancel out, and that gives us joules per mole of reaction. And after we do the math, we find that delta G naught at 25 degrees Celsius for this redox reaction is equal to -4.46 times 10 to the 5th joules per mole of reaction. We could also write this in terms of kilojoules per mole of reaction. So delta G naught is equal to -446 kilojoules per mole of reaction. Notice that a positive voltage and the negative in the equation means that we get a negative sign in the final answer. And when delta G naught is negative, the reaction is thermodynamically favored, which means at 25 degrees Celsius, silver cations will react with solid chromium to form solid silver and chromium three plus ions. And what this example shows us is whenever the voltage is positive, that means a thermodynamically favored reaction because delta G naught will be negative. Now that we've calculated delta G naught for our reaction, let's calculate the equilibrium constant K at 25 degrees Celsius. So we need to plug in our value for delta G naught, the ideal gas constant, the temperature in Kelvin, and we'll solve for the equilibrium constant. I've gone ahead and plugged everything into our equation. So here's the value for delta G naught in joules per mole of reaction. The ideal gas constant is equal to 8.314 joules per Kelvin mole of reaction. And 25 degrees Celsius, if we add 273 to that, we get 298 Kelvin. Looking at the units, Kelvin cancels out and so does joules per mole of reaction. And solving for K, the equilibrium constant is approximately equal to 10 to the 78th power. And since K is such an extremely large number, that tells us the reaction essentially goes to completion, and all the reactants turn into products. So for a reaction like this, it's not necessary to have equilibrium arrows in here. We can just draw an arrow going to the right. And remember, we started with a positive standard potential for this redox reaction. So to summarize what we've seen with these calculations, a positive value for the standard potential leads to a negative value for delta G naught, so a thermodynamically favored redox reaction. And a negative value for delta G naught leads to an equilibrium constant K that is greater than one. We've just seen what happens to delta G naught and the equilibrium constant K when the standard potential is positive or greater than zero. Now let's think about a generic redox reaction where the standard potential is negative, so E naught is less than zero. So let's think about the sign for delta G naught if we plug in a negative value for the standard potential. So if we have a negative voltage, the two negative signs in the equation will give us a positive value for delta G naught. A positive value for delta G naught means that the forward reaction of reactants turning into products is a thermodynamically unfavorable reaction. And if we were to plug in a positive value for delta G naught, we would find that the equilibrium constant K would be less than one. And when K is less than one, that tells us whenever the reaction reaches equilibrium, there will be a lot more reactants than products. Let's look at a quick summary of what we've learned from these examples. The standard potential E naught and the standard change in free energy delta G naught are related by the equation on the left. And delta G naught and the equilibrium constant K are related by the equation on the right. When E naught is greater than zero, delta G naught is less than zero, and the reaction is thermodynamically favored, which means at equilibrium, there'll be more products than reactants, and the equilibrium constant K will be greater than one. However, when E naught is less than zero, delta G naught is greater than zero, and the reaction is thermodynamically unfavored, which means at equilibrium, there'll be more reactants than products, and the equilibrium constant K will be less than one.
AP® is a registered trademark of the College Board, which has not reviewed this resource.