Standard potential, free energy, and the equilibrium constant | AP Chemistry | Khan Academy

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:

ΔG = -nFE.

ΔG is the instantaneous difference in free energy between the reactants and the products. n 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 "°" to ΔG and the superscript "°" to E. So we can use this equation to calculate ΔG° from E° or vice versa. We could calculate E° from ΔG°. Remember that ΔG° is connected to the equilibrium constant K by this equation:

ΔG° = -RT ln(K),

where R is the ideal gas constant and T is the temperature. Because ΔG° is a part of both of these two equations, these two equations connect the standard potential E° to the standard change in free energy ΔG°, 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°. First, we're going to calculate ΔG° and then from ΔG°, we're going to calculate the equilibrium constant K. Here's our redox reaction, and we're going to calculate ΔG° at 25 degrees Celsius.

For this redox reaction, E° is equal to +1.54 volts. Our goal is to calculate ΔG° at 25 degrees Celsius. In our redox reaction, Ag⁺ is being reduced to solid silver, and solid chromium is being oxidized to Cr³⁺. Now that we've identified what's being oxidized and what's being reduced, let's write the oxidation and reduction half-reactions.

For our reduction half-reaction, we have Ag⁺ gaining an electron to form solid silver. For the oxidation half-reaction, solid chromium is turning into Cr³⁺ 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 Ag⁺ cation is gaining an electron, but in the oxidation half-reaction, solid chromium is losing three electrons. We need to make the number of electrons equal; therefore, we need to multiply the first half-reaction through by a factor of three.

Now, when we add together our reactants — 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 are being transferred in our redox reaction, n is equal to three. 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 ΔG°.

I went ahead and rewrote our redox reaction. Our goal is to calculate ΔG°. We already know what n is; we already know what E° 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. 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, which gives us joules per mole of reaction. After we do the math, we find that ΔG° at 25 degrees Celsius for this redox reaction is equal to -4.46 × 10⁵ joules per mole of reaction. We could also write this in terms of kilojoules per mole of reaction, so ΔG° 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. When ΔG° is negative, the reaction is thermodynamically favored, which means at 25 degrees Celsius, Ag⁺ cations will react with solid chromium to form solid silver and Cr³⁺ ions. What this example shows us is that whenever the voltage is positive, that means a thermodynamically favored reaction because ΔG° will be negative.

Now that we've calculated ΔG° for our reaction, let's calculate the equilibrium constant K at 25 degrees Celsius. We need to plug in our value for ΔG°, the ideal gas constant, the temperature in Kelvin, and we will solve for the equilibrium constant. I've gone ahead and plugged everything into our equation.

So here's the value for ΔG° 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. Solving for K, the equilibrium constant is approximately equal to 10⁷⁸. Since K is such an extremely large number, that tells us the reaction essentially goes to completion and all the reactants turn into products.

For a reaction like this, it's not necessary to have equilibrium arrows; we can just draw an arrow going to the right. Remember, we started with a positive standard potential for this redox reaction. To summarize what we've seen with these calculations:

A positive value for the standard potential leads to a negative value for ΔG°, which means a thermodynamically favored redox reaction. A negative value for ΔG° leads to an equilibrium constant K that is greater than one. We've just seen what happens to ΔG° 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° is less than zero. Let's think about the sign for ΔG°. If we plug in a negative value for the standard potential, we will see that if we have a negative voltage, the two negative signs in the equation will give us a positive value for ΔG°.

A positive value for ΔG° means that the forward reaction of reactants turning into products is a thermodynamically unfavorable reaction. If we were to plug in a positive value for ΔG°, we would find that the equilibrium constant K would be less than one. 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° and the standard change in free energy ΔG° are related by the equation on the left, and ΔG° and the equilibrium constant K are related by the equation on the right.

When E° is greater than zero, ΔG° is less than zero, and the reaction is thermodynamically favored, which means at equilibrium there will be more products than reactants, and the equilibrium constant K will be greater than one. However, when E° is less than zero, ΔG° is greater than zero, and the reaction is thermodynamically unfavored, which means at equilibrium there will be more reactants than products, and the equilibrium constant K will be less than one.