The Nernst equation

A cell that is not at standard conditions

The standard emf of the cadmium and silver cell is $1.20\ \text{V}$. That holds only when both ions are at $1\ \text{mol/L}$. What happens if the silver solution is very dilute?

Think of the reaction $\text{Cd} + 2\text{Ag}^+ \rightarrow \text{Cd}^{2+} + 2\text{Ag}$. Less silver ion means less push towards the products, so the cell should give a smaller emf. And as the cell works, silver ions are used up and cadmium ions are made. So the emf of a working cell must fall. We want an equation that says by how much.

The Nernst equation

In Lesson 2 we used the Gibbs energy of the reaction at standard conditions. For any conditions, thermodynamics gives:

Here $Q$ is the reaction quotient, which has the form of $K$ but uses the concentrations of the moment. Replace $\Delta G$ by $-nFE$ and $\Delta G^\circ$ by $-nFE^\circ$:

We want $E$, so divide by $-nF$:

This is the Nernst equation. At $298\ \text{K}$, change to base 10 and put in the constants. The factor $\dfrac{2.303RT}{F}$ is $0.0591\ \text{V}$, as we found in Lesson 2:

Two checks tell us the equation is sensible:

• When every concentration is $1\ \text{mol/L}$, $Q = 1$ and $\log Q = 0$. Then $E = E^\circ$.
• At equilibrium, $E = 0$ and $Q = K$. Then $E^\circ = \dfrac{0.0591\ \text{V}}{n}\log K$, which is the equation of Lesson 2.

Writing the reaction quotient

The quotient has products on top and reactants below. Each concentration is raised to the power of its coefficient in the balanced equation. Four rules keep it correct:

• A pure solid or a pure liquid is left out. Its concentration does not change.
• A dissolved ion goes in as its concentration in $\text{mol/L}$.
• A gas goes in as its pressure in bar.
• Water as the solvent is left out.

Each concentration is compared with the standard value, $1\ \text{mol/L}$, and each pressure with $1\ \text{bar}$. So $Q$ is a pure number with no unit.

Here are three examples.

• $\text{Cd}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cd}^{2+}(aq) + 2\text{Ag}(s)$ gives $Q = \dfrac{[\text{Cd}^{2+}]}{[\text{Ag}^+]^2}$.
• $\text{Fe}(s) + 2\text{H}^+(aq) \rightarrow \text{Fe}^{2+}(aq) + \text{H}_2(g)$ gives $Q = \dfrac{[\text{Fe}^{2+}]\,p_{\text{H}_2}}{[\text{H}^+]^2}$.
• $2\text{Fe}^{2+}(aq) + \text{Cl}_2(g) \rightarrow 2\text{Fe}^{3+}(aq) + 2\text{Cl}^-(aq)$ gives $Q = \dfrac{[\text{Fe}^{3+}]^2[\text{Cl}^-]^2}{[\text{Fe}^{2+}]^2\,p_{\text{Cl}_2}}$.

A worked example

The cell $\text{Cd}(s) \mid \text{Cd}^{2+}(aq, 0.20\ \text{mol/L}) \parallel \text{Ag}^+(aq, 0.020\ \text{mol/L}) \mid \text{Ag}(s)$ is at $298\ \text{K}$. Find its emf.

1. The reaction is $\text{Cd} + 2\text{Ag}^+ \rightarrow \text{Cd}^{2+} + 2\text{Ag}$, with $n = 2$ and $E^\circ = 1.20\ \text{V}$.
2. The reaction quotient is $Q = \dfrac{0.20}{(0.020)^2} = 500$.
3. Then $\log Q = 2.70$.
4. $E = 1.20\ \text{V} - \dfrac{0.0591\ \text{V}}{2} \times 2.70 = 1.20\ \text{V} - 0.0798\ \text{V} = 1.12\ \text{V}$.

The emf is a little lower than the standard value. It must be: there is more of the product ion and less of the reactant ion than at standard conditions.

The Nernst equation is a straight line of $E$ against $\log Q$. The slope is $-0.0591\ \text{V}/n$. The line meets $E^\circ$ at $Q = 1$, and it reaches zero at $\log Q = \log K$. This cell would need $\log Q = 40.6$ before it went flat.

Try this. Set both ions to $1.00\ \text{mol/L}$ and see that the emf equals $1.20\ \text{V}$. Now make the silver ion ten times more dilute. By how many volts does the emf fall? Compare with the slope in the picture.

Which way does the emf move?

You do not need the numbers to know the direction. Use this rule.

• More product, or less reactant: $Q$ gets larger, and the emf falls.
• Less product, or more reactant: $Q$ gets smaller, and the emf rises.

This is Le Chatelier's principle in voltage form. A cell with a larger concentration of the reacting ion is a stronger cell.

A working cell runs down slowly

When the cell delivers current, silver ions are removed and cadmium ions are made. Take the cell above with the cadmium ion at $1\ \text{mol/L}$ at the start and equal volumes on both sides. Each two silver ions used make one cadmium ion.

The result may surprise you. With the silver ion starting at $1\ \text{mol/L}$, the emf has fallen by only about $0.12\ \text{V}$ after $99$ per cent of it is used. Because the equation uses a logarithm, even a very large change in $Q$ moves the emf only a little. A real torch cell fades mainly because its internal resistance grows, as you saw in Lesson 1, and because the reaction slows.

One electrode at a time

The same equation works for a single electrode. For $\text{M}^{n+}(aq) + ne^- \rightarrow \text{M}(s)$ the quotient is $Q = \dfrac{1}{[\text{M}^{n+}]}$, because the solid is left out. Then:

Take a silver electrode, $\text{Ag}^+ + e^- \rightarrow \text{Ag}$, with $E^\circ = 0.80\ \text{V}$, in a solution with $[\text{Ag}^+] = 0.0050\ \text{mol/L}$.

• $E = 0.80\ \text{V} + 0.0591\ \text{V} \times \log 0.0050 = 0.80\ \text{V} + 0.0591\ \text{V} \times (-2.30) = 0.664\ \text{V}$

A more dilute ion lowers the electrode potential. The silver ion is then a weaker oxidising agent. For each tenfold dilution, the potential falls by $0.0591\ \text{V}/n$.

When the half-reaction uses acid

Some half-reactions use $\text{H}^+$ ions. Take the permanganate ion in acid, which has $E^\circ = 1.51\ \text{V}$:

Here $Q = \dfrac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][\text{H}^+]^8}$ and $n = 5$. The hydrogen ion is raised to the power 8. So a tenfold drop in $[\text{H}^+]$ lowers the potential by $\dfrac{8 \times 0.0591\ \text{V}}{5} = 0.095\ \text{V}$. Permanganate is a much weaker oxidising agent in weak acid than in strong acid.

The effect of temperature

The factor $\dfrac{2.303RT}{nF}$ depends on $T$. At $298\ \text{K}$ it is $0.0591\ \text{V}/n$. At $313\ \text{K}$ it is $0.0621\ \text{V}/n$. We assume $E^\circ$ itself changes very little over such a range, so the main effect is on the slope. Raise the temperature in the sim and watch the line tilt more steeply.

Practise with new numbers

Try it in Python

The program finds the emf of the cadmium and silver cell for a falling silver-ion concentration.

Common mistakes

• Putting $Q$ upside down. Products go on top and reactants below.
• Leaving a solid in $Q$. Metals such as cadmium and silver are not in the quotient.
• Forgetting the exponent. In $Q = \dfrac{[\text{Cd}^{2+}]}{[\text{Ag}^+]^2}$ the silver ion is squared, because its coefficient is 2.
• Using $\ln$ with $0.0591$. The number $0.0591\ \text{V}$ goes with $\log_{10}$. With $\ln$, use $RT/F$ instead.
• Using the Nernst equation with $E^\circ$ of the wrong cell. The value $E^\circ$ must be the standard emf of the reaction as written.

Remember these

• $E = E^\circ - \dfrac{RT}{nF}\ln Q = E^\circ - \dfrac{0.0591\ \text{V}}{n}\log Q$ at $298\ \text{K}$.
• $Q$: products over reactants, to the power of their coefficients. Solids, pure liquids and the solvent are left out. Gases go in as pressures in bar.
• At $Q = 1$, $E = E^\circ$. At equilibrium, $E = 0$ and $Q = K$.
• Larger $Q$ means a smaller emf. For one electrode, $E = E^\circ + \dfrac{0.0591\ \text{V}}{n}\log[\text{ion}]$.

Check yourself

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

1. Write the reaction quotient for $\text{Zn}(s) + 2\text{H}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{H}_2(g)$.

2. In the cadmium and silver cell, the silver ion is made ten times more dilute and the cadmium ion is unchanged. Does the emf go up or down, and by how much?

3. Find the potential of a copper electrode in $0.010\ \text{mol/L}$ copper sulphate at $298\ \text{K}$. The standard potential is $0.34\ \text{V}$.

4. Why does the emf of a cell fall as it delivers current?

5. In the permanganate half-reaction, $n = 5$ and eight hydrogen ions are on the left. What happens to the potential when $[\text{H}^+]$ rises tenfold?