The Nernst Equation

Standard potentials assume 1 M solutions. Real cells drift away from that as they run, and their voltage changes. Combining ΔG = −nFE with ΔG = ΔG° + RT ln Q gives the correction.

Key idea

$$E = E^\circ - \frac{RT}{nF}\ln Q$$ or, at $25\ ^\circ\text{C}$, $$E = E^\circ - \frac{0.0592\ \text{V}}{n}\log Q$$ $Q$ is products over reactants, with solids left out. As the cell runs, $Q$ grows toward $K$ and $E$ falls toward zero: a dead battery is a cell at equilibrium.

Why it works

Substitute $\Delta G = -nFE$ and $\Delta G^\circ = -nFE^\circ$ into $\Delta G = \Delta G^\circ + RT\ln Q$ and divide by $-nF$. When products are scarce ($Q < 1$), the log is negative and $E$ is larger than $E^\circ$; when products have built up, $E$ is smaller. At equilibrium $Q = K$ and $E = 0$.

Slide the ion concentrations. Lowering $[\mathrm{Cu^{2+}}]$ (a reactant) lowers the voltage; lowering $[\mathrm{Zn^{2+}}]$ (a product) raises it.

Worked examples

Worked example: A cell running down

A zinc–copper cell ($E^\circ = 1.10\ \text{V}$) has $[\mathrm{Zn^{2+}}] = 1.0\ \text{M}$ and $[\mathrm{Cu^{2+}}] = 0.010\ \text{M}$. Find $E$ at 25 °C.

$$\mathrm{Zn(s)} + \mathrm{Cu^{2+}(aq)} \to \mathrm{Zn^{2+}(aq)} + \mathrm{Cu(s)}$$: $Q = \dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} = \dfrac{1.0}{0.010} = 100$ (the metals are left out).
$$E = 1.10 - \dfrac{0.0592}{2}\log 100 = 1.10 - (0.0296)(2) = 1.04\ \text{V}$$

Answer: $1.04\ \text{V}$.

Worked example: A concentration cell

Both half-cells are $\mathrm{Cu^{2+}/Cu}$, one with $0.0010\ \text{M}$ and one with $1.0\ \text{M}$ $\mathrm{Cu^{2+}}$. Find $E$.

$E^\circ = 0$: the two half-reactions are the same. The dilute side is the anode (copper dissolves there, raising its concentration); the concentrated side is the cathode.
Overall: $$\mathrm{Cu^{2+}(1.0\ M)} \to \mathrm{Cu^{2+}(0.0010\ M)}$$ so $Q = \dfrac{0.0010}{1.0} = 1.0 \times 10^{-3}$.
$$E = 0 - \dfrac{0.0592}{2}\log(1.0 \times 10^{-3}) = (0.0296)(3) = 0.089\ \text{V}$$

Answer: $0.089\ \text{V}$.

The correction is small but not negligible: every factor of 10 in $Q$ changes $E$ by $0.0592/n$ volts. For a two-electron cell, that is 30 mV per factor of 10.

Common mistake

Putting the reactant concentrations on top of $Q$, or including solids. $Q$ has the same form as $K$: products over reactants, aqueous ions and gases only. Check: making more reactant should raise the voltage; if your $Q$ says otherwise, it is upside down.

A pH meter uses the same law. Its glass electrode responds to $[\mathrm{H_3O^+}]$, and at 25 °C its voltage changes by about $0.0592\ \text{V}$ (59 mV) for each unit of pH, the Nernst slope for one electron per proton. That is why meters are calibrated with buffers of known pH, and why they correct for temperature.

Your nerve cells use concentration differences the same way: a membrane potential of roughly $-70\ \text{mV}$ comes mainly from a difference in potassium ion concentration across the cell membrane.

The Nernst Equation

Worked examples

Practice

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Sources and further reading