The Nernst Equation and Non-Standard Conditions

The Nernst equation, E = E° − (RT/nF) ln Q, gives the cell potential under non-standard conditions; at 25 °C it can be written as E = E° − (0.0592/n) log Q. In brief: for the Zn–Cu cell with [Zn²⁺] = 1.0 M and [Cu²⁺] = 0.010 M, E = 1.10 − (0.0592/2) log(100) = 1.04 V, and the cell voltage falls to zero when Q reaches the equilibrium constant.

Real cells are rarely at standard conditions, and their concentrations change as they operate. The Nernst equation shows how the potential depends on the concentrations of the ions and the pressures of the gases. It is the basis for the pH electrode, for the potentials of nerve cells and for the explanation of why batteries gradually lose voltage. This article presents the equation, gives a table of values and shows examples from chemistry and biology.

What is the Nernst equation?

From thermodynamics, ΔG = ΔG° + RT ln Q. Since ΔG = −nFE and ΔG° = −nFE°, dividing by −nF gives:

E = E° − (RT / nF) ln Q

At 25 °C (298.15 K), the constant RT/F = 0.02569 V, and converting to common logarithms (ln x = 2.303 log x):

E = E° − (0.0592 V / n) log Q

Here Q is the reaction quotient for the cell reaction, written with products over reactants and without pure solids and liquids, and n is the number of moles of electrons transferred. For the reaction Zn + Cu²⁺ → Zn²⁺ + Cu, Q = [Zn²⁺] / [Cu²⁺]. When Q < 1, E is greater than E°. When Q > 1, E is smaller. A tenfold change in Q changes the potential by 0.0592/n volts, which is only a small effect for most cells.

How does E depend on Q?

Worked examples

Example 1: the zinc–copper cell

Find E for Zn(s) | Zn²⁺(1.0 M) || Cu²⁺(0.010 M) | Cu(s) at 25 °C. n = 2, Q = 1.0 / 0.010 = 100, so E = 1.10 − (0.0592/2) log 100 = 1.10 − 0.0296 × 2 = 1.04 V.

Example 2: an iron(III)/iron(II) half-cell

The half-reaction Fe³⁺ + e⁻ → Fe²⁺ has E° = +0.77 V. With [Fe³⁺] = 0.10 M and [Fe²⁺] = 1.0 M, Q = [Fe²⁺] / [Fe³⁺] = 10, so E = 0.77 − 0.0592 × log 10 = 0.77 − 0.059 = 0.71 V. Less Fe³⁺ makes it a weaker oxidizing agent.

Example 3: a hydrogen electrode and pH

For 2H⁺ + 2e⁻ → H₂ at P(H₂) = 1 bar, E = 0 − (0.0592/2) log(1/[H⁺]²) = −0.0592 × pH. In a solution of pH 4.00, E = −0.0592 × 4.00 = −0.237 V. This shows how a potential measurement can find the pH.

Example 4: a cell with a gas

Consider the cell Zn(s) | Zn²⁺(1.0 M) || H⁺(0.10 M) | H₂(1 bar) | Pt. E° = 0 − (−0.76) = +0.76 V. The reaction is Zn + 2H⁺ → Zn²⁺ + H₂, and Q = [Zn²⁺] P(H₂) / [H⁺]² = 1.0 × 1 / 0.010 = 100. E = 0.76 − 0.0296 × 2 = 0.70 V.

Example 5: concentration needed for a given potential

A Zn–Cu cell has [Zn²⁺] = 1.0 M and a measured potential of 1.07 V. Then 1.07 = 1.10 − 0.0296 log Q, so log Q = 1.01 and Q = 10.3, giving [Cu²⁺] = 1.0 / 10.3 = 0.097 M.

What happens at equilibrium?

As the cell reaction proceeds, Q increases (products accumulate and reactants are used up), and E decreases. The cell stops producing current when E = 0, and the reaction is at equilibrium: Q = K. Setting E = 0 in the Nernst equation gives E° = (0.0592/n) log K, which is the formula used earlier to find K from E°. A flat (dead) battery is a cell at equilibrium.

How does a pH electrode work?

A glass electrode contains a thin glass membrane that separates the sample from a reference solution of fixed acidity. The potential across the membrane depends on the hydrogen ion activities on the two sides according to the Nernst equation: it changes by 59.2 mV per pH unit at 25 °C. A reference electrode (such as Ag/AgCl) completes the cell, and a high-impedance voltmeter converts the potential to a pH reading. Because the slope depends on temperature (it is 61.5 mV per pH unit at 37 °C), pH meters are calibrated with buffer solutions of known pH and have a temperature compensation.

How does the equation apply to cells?

The resting potential of a nerve cell arises from the unequal distribution of ions across the membrane. For an ion of charge z, the equilibrium (Nernst) potential is E = (RT/zF) ln([ion]_out / [ion]_in). For K⁺ with a typical concentration of 5 mM outside and 140 mM inside the cell at 37 °C (310 K), RT/F = 0.0267 V, so E = 0.0267 × ln(5/140) = 0.0267 × (−3.33) = −0.089 V, or about −89 mV. This is close to the resting potential of a neuron, which is about −70 mV, because the membrane is mainly permeable to potassium at rest.

Who derived the equation?

Walther Nernst derived the equation in 1889, when he was a young researcher in Ostwald’s laboratory in Leipzig, and applied it to the potentials of electrodes in solutions of different concentration. It was a key part of the foundation of physical chemistry, and Nernst received the Nobel Prize in Chemistry in 1920 for his work in thermochemistry. The use of the glass electrode for pH measurement was developed by Fritz Haber and Zygmunt Klemensiewicz in 1909, and became a laboratory standard in the 1930s.

Which mistakes are common?

• Inverting Q: products go in the numerator.
• Using 0.0592 at other temperatures: the value applies at 25 °C only.
• Forgetting n: the coefficient is 0.0592/n.
• Including pure solids: metals and solids are not in Q.
• Confusing ln and log: use 0.0592 with log₁₀ and RT/F = 0.0257 with ln.