Gibbs energy and the equilibrium constant

Why a battery goes flat

A torch cell lights the bulb for hours. Then it stops. Nothing has been used up in the sense of disappearing. The reaction has simply run until it can go no further. The cell has reached equilibrium, and at equilibrium the emf is zero.

This tells us that the emf is a measure of how far the reaction is from equilibrium. A large emf means a reaction that is far from equilibrium and has a lot of energy to give. This lesson turns that idea into two equations.

Electrical work and Gibbs energy

When one mole of the reaction runs in a cell, $n$ moles of electrons pass through the wire. One mole of electrons carries a charge of $F = 96500\ \text{C/mol}$, the Faraday constant. So the charge that moves is $nF$.

Each coulomb of charge falls through a potential difference of $E$. A volt is a joule per coulomb, so the electrical work is:

A cell gives the most work when it runs so slowly that it stays at equilibrium with its surroundings at each step. From thermodynamics, the most useful work (not counting expansion) that a reaction can do at constant temperature and pressure is $-\Delta G$. Put the two together:

For standard conditions we write $\Delta G^\circ = -nFE^\circ$. The units are joules per mole of reaction, or kilojoules per mole of reaction when $F$ is written as $96.5\ \text{kJ/(V mol)}$.

Look at the cadmium and silver cell from Lesson 1. Here $n = 2$ and $E^\circ = 1.20\ \text{V}$:

• $\Delta G^\circ = -2 \times 96500\ \text{C/mol} \times 1.20\ \text{V} = -231\,600\ \text{J/mol} = -231.6\ \text{kJ/mol}$

The value is negative, so the reaction can run. Now write the same reaction with every coefficient halved: $\tfrac{1}{2}\text{Cd} + \text{Ag}^+ \rightarrow \tfrac{1}{2}\text{Cd}^{2+} + \text{Ag}$. Then $n = 1$ and $\Delta G^\circ = -115.8\ \text{kJ/mol}$. The emf is still $1.20\ \text{V}$. The Gibbs energy depends on how much reaction you write. The emf does not.

From the emf to the equilibrium constant

For any reaction, $\Delta G^\circ = -RT\ln K$. Now we have two expressions for the same $\Delta G^\circ$:

Cancel the minus signs and change to base-10 logarithms, using $\ln K = 2.303\log K$:

At $298\ \text{K}$ the factor is a fixed number. Using $R = 8.314\ \text{J/(K mol)}$ and $F = 96500\ \text{C/mol}$:

• $2.303 \times 8.314\ \text{J/(K mol)} \times 298\ \text{K} = 5706\ \text{J/mol}$
• $\dfrac{5706\ \text{J/mol}}{96500\ \text{C/mol}} = 0.0591\ \text{V}$

The units work out: joule per coulomb is volt. So at $298\ \text{K}$:

Solve for $\log K$:

For the cadmium and silver cell, $\log K = \dfrac{2 \times 1.20\ \text{V}}{0.0591\ \text{V}} = 40.6$. So $K \approx 4 \times 10^{40}$. The reaction goes essentially to completion. The Daniell cell, with $E^\circ = 1.10\ \text{V}$ and $n = 2$, has $\log K = 37.2$, which is nearly as large.

The graph shows why a cell with a good emf makes such a huge $K$. Every extra $0.0591\ \text{V}$ for one electron multiplies $K$ by ten.

Move the emf down to $0.05\ \text{V}$ and use one electron. The value of $\log K$ is below 1, so $K$ is only about 7. The reaction still goes forward, but not completely.

Reading the three quantities together

The three quantities say the same thing about a reaction.

• $E^\circ > 0$, then $\Delta G^\circ < 0$ and $K > 1$. The reaction goes forward under standard conditions.
• $E^\circ = 0$, then $\Delta G^\circ = 0$ and $K = 1$. Reactants and products are equal at the standard state.
• $E^\circ < 0$, then $\Delta G^\circ > 0$ and $K < 1$. The reverse reaction is favoured.

Remember the limits. A positive $E^\circ$ says a reaction can happen. It does not say how fast. A big $K$ does not mean a quick reaction either. A mixture may sit unchanged for years because the reaction has no easy path.

Practise with new numbers

Why we add $\Delta G^\circ$ and not $E^\circ$

Two half-reactions can be combined to give a third, and then a mistake is easy. Take iron:

• $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$, with $E^\circ = 0.77\ \text{V}$ and $n = 1$.
• $\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}$, with $E^\circ = -0.44\ \text{V}$ and $n = 2$.

Add them and you get $\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}$. What is its $E^\circ$? It is not $0.77 + (-0.44)$, and it is not the average either. The potential is a push per electron, and the two half-reactions have different numbers of electrons.

The Gibbs energy is different. Energy is added when reactions are added. So:

• $\Delta G^\circ_1 = -1 \times 96.5\ \text{kJ/(V mol)} \times 0.77\ \text{V} = -74.3\ \text{kJ/mol}$
• $\Delta G^\circ_2 = -2 \times 96.5\ \text{kJ/(V mol)} \times (-0.44\ \text{V}) = +84.9\ \text{kJ/mol}$
• Total: $\Delta G^\circ = +10.6\ \text{kJ/mol}$, for $n = 3$.

Now turn the total back into a potential:

The potential of $\text{Fe}^{3+}/\text{Fe}$ is about $-0.04\ \text{V}$. When the numbers of electrons differ, add the Gibbs energies and not the potentials. When two half-cells are joined in a cell, the electrons cancel and the potentials can be subtracted.

A cell at equilibrium

Put the cell to work and the reaction moves forward. Products build up and reactants fall. The emf falls with them. In the end, the reaction quotient reaches $K$ and the emf reaches zero. Then $\Delta G = 0$, the current stops, and the cell is flat. The next lesson gives the equation that follows the emf at every step along the way.

Try it in Python

The program turns the standard emf of the cadmium and silver cell into a Gibbs energy and an equilibrium constant. It also works out the factor $2.303RT/F$ from the constants.

Common mistakes

• Dropping the minus sign in $\Delta G^\circ = -nFE^\circ$. A positive emf gives a negative Gibbs energy.
• Using the wrong $n$. Count the electrons in the balanced equation as it is written.
• Mixing joules and kilojoules. With $F = 96500\ \text{C/mol}$ the result is in joules per mole. Divide by 1000 for kilojoules.
• Writing $E^\circ = 0.0591 \log K$ without the $n$. The factor is $0.0591\ \text{V}/n$.
• Adding $E^\circ$ values of two half-reactions with different numbers of electrons. Add the Gibbs energies instead.

Remember these

• $\Delta G^\circ = -nFE^\circ$, with $F = 96500\ \text{C/mol}$. A positive $E^\circ$ gives a negative $\Delta G^\circ$.
• $E^\circ = \dfrac{0.0591\ \text{V}}{n}\log K$ at $298\ \text{K}$.
• $E^\circ$ does not change when the equation is multiplied. $\Delta G^\circ$ and $n$ do, and $K$ is raised to a power.
• At equilibrium the emf is zero, $\Delta G = 0$ and $Q = K$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A cell reaction moves 2 mol of electrons and has $E^\circ = 0.30\ \text{V}$. Find $\Delta G^\circ$ in kilojoules per mole.

2. A reaction has $E^\circ = -0.20\ \text{V}$. Is its equilibrium constant larger or smaller than 1?

3. The equation of a cell reaction is multiplied by 2. What happens to $E^\circ$, to $\Delta G^\circ$ and to $K$?

4. A reaction has a large positive $E^\circ$, yet a mixture of the reactants stays unchanged for years. How can that be?

5. Why can a flat torch cell show a normal emf on a voltmeter yet not light a bulb?