Nernst Equation Reaction Quotients

Introduction

Standard electrode potentials apply only when every dissolved species is at 1 mol dm⁻³ and every gas at 1 bar. Real batteries, sensors and biological membranes almost never meet these conditions. The Nernst equation corrects a standard potential for the actual composition. It is the electrochemical form of the relation between ΔG and the reaction quotient, and most errors in using it come from building Q wrongly or from choosing the wrong electron number.

Core explanation

Origin. Since ΔG = ΔG° + RT ln Q and ΔG = −nFE, dividing through by −nF gives the Nernst equation:

E = E° − (RT/nF) ln Q

At 298 K, 2.303RT/F = 0.0592 V, so the working form is

E = E° − (0.0592 V/n) log₁₀ Q

Building Q. Write Q from the overall balanced equation exactly as for an equilibrium expression: products over reactants, each concentration raised to its stoichiometric coefficient. Solids, pure liquids and water as solvent are omitted. Gases appear as partial pressures in bar. For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), Q = [Zn²⁺]/[Cu²⁺].

Choosing n. n is the number of electrons cancelled in the same balanced equation from which Q was written. If you double the equation, Q is squared and n doubles: (0.0592/2n) × log Q² = (0.0592/n) × log Q. The potential is unchanged, as it must be for an intensive quantity. This is a powerful self-check: n and Q must always come from the same equation.

Direction of change. When Q < 1 (reactants relatively concentrated), log Q is negative and E exceeds E°. When Q > 1, E falls below E°. At equilibrium E = 0 and Q = K, recovering log K = nE°/0.0592.

Half-cell form. For a single reduction half-equation, Ox + ne⁻ → Red, the Nernst equation reads E = E° − (0.0592/n) log([Red]/[Ox]). Electrons do not appear in Q. For Ag⁺ + e⁻ → Ag, E = 0.80 − 0.0592 log(1/[Ag⁺]).

pH-dependent couples. When H⁺ appears in a half-equation, it enters Q with its coefficient, often a large one. For MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, Q = [Mn²⁺]/([MnO₄⁻][H⁺]⁸). The eighth power makes the oxidising strength of permanganate strongly dependent on pH.

Concentration cells. With identical electrodes, E° = 0 and the whole potential comes from the concentration ratio. Electrons flow so as to equalise the concentrations: the dilute side is the anode.

Formulae

E = E° − (RT/nF) ln Q; at 298 K, E = E° − (0.0592/n) log Q; at equilibrium, log K = nE°/0.0592.

Step-by-step reasoning

1. Write the balanced overall (or half) equation and read off n. 2. Write Q from that same equation, omitting solids and pure liquids. 3. Substitute concentrations in mol dm⁻³ and pressures in bar. 4. Evaluate log Q, then E = E° − (0.0592/n) log Q. 5. Check the direction: does E rise when reactants are more concentrated?

Visual explanation

Plot E against log Q. The graph is a straight line with intercept E° at log Q = 0 and gradient −0.0592/n. A two-electron couple gives a line half as steep as a one-electron couple, so its potential responds more gently to concentration changes.

Real-world analogy

A standard potential is like the advertised speed of a train on an empty track. The Nernst term adjusts for the actual traffic: a crowded platform of reactants pushes the train harder, a jam of products slows it down.

Real-world example

A glass pH electrode relies on Nernst-type behaviour: its potential changes by close to 59 mV for every unit change in pH at 25 °C. Meters are calibrated with two buffers precisely to measure this slope and intercept.

Why?

Why is the correction logarithmic rather than linear? The Gibbs energy of a dissolved species depends on RT ln(concentration), because it reflects the number of ways particles can be arranged. Potential is Gibbs energy per unit charge, so it inherits the logarithmic dependence.

Common misconception

"Use n = 1 because the Nernst constant is 0.0592 V." The constant 0.0592 V is divided by n. For a Daniell cell with Q = 100, using n = 1 gives 0.98 V instead of the correct 1.04 V.

Worked example

Question: Calculate the potential at 298 K of the half-cell MnO₄⁻/Mn²⁺ (E° = +1.51 V) at pH 3.00 when [MnO₄⁻] = [Mn²⁺].

Reasoning: n = 5. Q = [Mn²⁺]/([MnO₄⁻][H⁺]⁸) = 1/(1.0 × 10⁻³)⁸ = 1.0 × 10²⁴, so log Q = 24.0. E = 1.51 − (0.0592/5) × 24.0 = 1.51 − 0.284 = 1.23 V.

Answer: E ≈ +1.23 V; permanganate is a noticeably weaker oxidant at pH 3 than in strongly acidic solution.

Quick check

1. A Daniell cell has [Zn²⁺] = 1.0 mol dm⁻³ and [Cu²⁺] = 0.010 mol dm⁻³. What is its potential at 298 K if E° = 1.10 V? Answer: Q = 1.0/0.010 = 100, so E = 1.10 − (0.0592/2) × 2 = 1.04 V.

Exam focus

Write Q explicitly before substituting, state n with a reason, and check that Q and n come from the same equation. Remember that solids are excluded and that H⁺ carries its full coefficient as a power. Quote E to two decimal places unless the data justify more.

Advanced insight

Strictly, Q should contain activities rather than concentrations. In moderately concentrated ionic solutions, activity coefficients fall well below 1, so Nernst predictions using concentrations can be off by tens of millivolts. Precise work uses formal potentials measured in a fixed ionic medium, which absorb the activity corrections.

Summary

The Nernst equation, E = E° − (0.0592/n) log Q at 298 K, adjusts a standard potential for actual concentrations. Q is written from the balanced equation, excluding solids and pure liquids, and n must come from the same equation. Scaling the equation leaves E unchanged. At equilibrium E = 0 and Q = K.

Practice questions

1. Calculate the potential of an Ag⁺/Ag half-cell (E° = +0.80 V) when [Ag⁺] = 0.010 mol dm⁻³. Answer: E = 0.80 − 0.0592 × log(1/0.010) = 0.80 − 0.118 = +0.68 V. 2. Find the potential of the concentration cell Cu Cu²⁺(0.0010 mol dm⁻³) Cu²⁺(0.10 mol dm⁻³) Cu. Answer: E° = 0 and Q = 0.0010/0.10 = 0.010, so E = −(0.0592/2) × (−2) = +0.059 V. 3. Calculate the hydrogen-electrode potential at pH 4.00 with p(H₂) = 1 bar. Answer: E = −0.0592 × pH = −0.0592 × 4.00 = −0.237 V. 4. A student doubles the Daniell equation and uses Q = [Zn²⁺]²/[Cu²⁺]² with n = 4. Will the answer differ from using the single equation? Explain. Answer: No; log Q doubles and n doubles, so (0.0592/n) log Q is unchanged and E is the same.