Cell EMF and the Nernst Equation: Quick Revision

A brief note on Cell EMF and the Nernst Equation. Read it before a test. Try each question at the end before you open its answer.

Cell notation and emf

• Anode on the left, cathode on the right. A bar $\mid$ is a phase boundary and a double bar $\parallel$ is the salt bridge.
• A comma separates species in the same phase. When no solid metal takes part, write an inert platinum electrode.
• Example: $\text{Pt} \mid \text{Fe}^{2+}, \text{Fe}^{3+} \parallel \text{Cl}^- \mid \text{Cl}_2(g) \mid \text{Pt}$.
• $E^\circ_{cell} = E^\circ_{right} - E^\circ_{left}$. A positive emf means the cell runs as written.
• The emf is the potential difference at zero current. Under load, $V = E - Ir$.
• The emf does not change when the equation is multiplied. The number of electrons $n$ does.

Gibbs energy and equilibrium

• $\Delta G^\circ = -nFE^\circ$, with $F = 96500\ \text{C/mol}$.
• $E^\circ = \dfrac{2.303RT}{nF}\log K$, which is $\dfrac{0.0591\ \text{V}}{n}\log K$ at $298\ \text{K}$.
• $E^\circ > 0$: $\Delta G^\circ < 0$ and $K > 1$.
• When half-reactions with different numbers of electrons are added, add the Gibbs energies, not the potentials.

The Nernst equation

• $E = E^\circ - \dfrac{RT}{nF}\ln Q = E^\circ - \dfrac{0.0591\ \text{V}}{n}\log Q$ at $298\ \text{K}$.
• $Q$ is products over reactants, to the power of the coefficients. Solids and pure liquids are left out. Gases enter as pressures in bar.
• At $Q = 1$, $E = E^\circ$. At equilibrium, $E = 0$ and $Q = K$.
• One electrode: $E = E^\circ + \dfrac{0.0591\ \text{V}}{n}\log[\text{M}^{n+}]$.

A worked example

The cell $\text{Cd} \mid \text{Cd}^{2+}(0.20\ \text{mol/L}) \parallel \text{Ag}^+(0.020\ \text{mol/L}) \mid \text{Ag}$ has $E^\circ = 0.80\ \text{V} - (-0.40\ \text{V}) = 1.20\ \text{V}$ and $n = 2$.

1. Gibbs energy. $\Delta G^\circ = -2 \times 96.5\ \text{kJ/(V mol)} \times 1.20\ \text{V} = -231.6\ \text{kJ/mol}$.
2. Equilibrium constant. $\log K = \dfrac{2 \times 1.20\ \text{V}}{0.0591\ \text{V}} = 40.6$.
3. Emf now. $Q = \dfrac{0.20}{(0.020)^2} = 500$, so $E = 1.20\ \text{V} - \dfrac{0.0591\ \text{V}}{2} \times 2.70 = 1.12\ \text{V}$.

Practice with new numbers

A program to try

The program works out the Gibbs energy, $\log K$ and the Nernst emf of the cell above.

Mistakes to avoid

• Left minus right. The emf is the right-hand potential minus the left-hand one.
• Doubling $E^\circ$ with the equation. The emf stays the same, and $n$ doubles.
• Dropping the minus sign in $\Delta G^\circ = -nFE^\circ$.
• A solid in $Q$. Solids and pure liquids are left out.
• An upside-down $Q$. Products go on top.
• Using $0.0591$ with a natural logarithm. It goes with $\log_{10}$.

Check yourself

1. A cell has $E = 1.50\ \text{V}$, $r = 0.50\ \Omega$ and delivers $1.0\ \text{A}$. What is the voltage across the terminals?

2. A reaction has $n = 2$ and $E^\circ = 0.30\ \text{V}$. Find $\Delta G^\circ$.

3. A one-electron reaction has $E^\circ = 0.35\ \text{V}$. Is $K$ larger or smaller than 1?

4. What does the emf do as a cell delivers current, and why?