Gibbs Energy and Cell Potential

Introduction

A cell potential tells you how much energy each coulomb of charge carries, while a Gibbs-energy change tells you how much useful energy a reaction releases per mole of reaction. Linking the two lets you turn a voltmeter reading into thermodynamic data, predict equilibrium constants and calculate the electrical work that a battery can deliver. The link is a single equation, but applying it correctly depends on getting the electron number and the amount of reaction right.

Core explanation

The key relation. For a reaction written as a balanced equation,

ΔG = −nFE

where n is the number of moles of electrons transferred per mole of reaction as written, F = 96 485 C mol⁻¹ and E is the cell potential in volts. Under standard conditions, ΔG° = −nFE°. Because 1 V × 1 C = 1 J, the product nFE comes out in joules per mole of reaction; divide by 1000 for kJ mol⁻¹.

Sign logic. A positive E corresponds to a negative ΔG: the reaction is spontaneous and the cell can do electrical work. A negative E corresponds to a positive ΔG, so the reaction as written is not spontaneous.

Where n comes from. Find n from the balanced half-equations after electrons have been cancelled. For 2Ag⁺ + Cu → 2Ag + Cu²⁺, two electrons move for each mole of reaction, so n = 2. If the same chemistry is written as Ag⁺ + ½Cu → Ag + ½Cu²⁺, then n = 1 and ΔG° is halved, although E° is unchanged. This is the numerical signature of the difference between intensive potential and extensive Gibbs energy.

Reaction extent. ΔG in kJ mol⁻¹ means "per mole of reaction". If a cell runs for an extent ξ (in mol of the reaction as written), the Gibbs-energy change is ξΔG and the maximum electrical work available is −ξΔG. Equivalently, the charge passed is Q = nξF, and the work is QE. Both routes must agree, which provides a useful check.

Equilibrium constant. Since ΔG° = −RT ln K, combining the relations gives

ln K = nFE°/(RT)

At 298 K a cell potential of only about 0.1 V with n = 2 already corresponds to K of roughly 10³, which is why cells with potentials of a volt or more represent reactions that go essentially to completion.

Combining half-reactions. Potentials of two half-reactions cannot be added to give the potential of a third half-reaction, because electrons do not cancel. Instead convert each to ΔG° = −nFE°, add the Gibbs energies (which are extensive and additive, like Hess's law enthalpies), then divide by −nF for the new half-reaction.

Formulae

ΔG° = −nFE°; ΔG = −nFE; ln K = nFE°/(RT); Q(charge) = nξF; maximum work = −ξΔG = nξFE.

Step-by-step reasoning

1. Write the balanced overall equation and identify n from the half-equations. 2. Calculate E°cell = E°(cathode) − E°(anode). 3. Evaluate ΔG° = −nFE° in J mol⁻¹, then convert to kJ mol⁻¹. 4. Multiply by the extent of reaction if an actual amount of reactant is consumed. 5. If needed, find K from ln K = −ΔG°/(RT).

Visual explanation

Picture a staircase. Each mole of electrons is a ball carried down a step of height E. The energy released per ball is FE; the number of balls per mole of reaction is n; and the number of complete "reactions" is ξ. The total energy is the product of all three.

Real-world analogy

Voltage is like the price per kilogram of fruit and nF is like the number of kilograms in one standard bag. The cost of one bag is price times mass; the cost of a whole order is the cost per bag times the number of bags, which corresponds to the extent of reaction.

Real-world example

A lead–acid car battery has a cell potential of about 2.0 V, with n = 2 for its discharge reaction. Each cell therefore releases roughly 2 × 96 485 × 2.0 ≈ 390 kJ of Gibbs energy per mole of reaction, which is why a relatively compact battery can crank a large engine.

Why?

Why does ΔG depend on how the equation is written while E does not? ΔG is an energy per mole of reaction, so halving the equation halves the amount of chemical change being described. E is energy per unit charge; halving the equation halves both energy and charge, leaving their ratio unchanged.

Common misconception

"ΔG° = −FE°, with n = 1 for every cell." Omitting n is the most common error. For the Daniell cell, n = 2, and using n = 1 would underestimate ΔG° by half and give ln K half its true value.

Worked example

Question: For the Daniell cell, Zn + Cu²⁺ → Zn²⁺ + Cu, E° = +1.10 V. Calculate ΔG°, the maximum electrical work when 0.0500 mol of zinc is consumed, and K at 298 K.

Reasoning: n = 2. ΔG° = −2 × 96 485 × 1.10 = −2.12 × 10⁵ J mol⁻¹ = −212 kJ mol⁻¹. Consuming 0.0500 mol Zn is an extent ξ = 0.0500 mol, so the work is 0.0500 × 212 = 10.6 kJ. Check: Q = 2 × 0.0500 × 96 485 = 9650 C, and 9650 × 1.10 = 1.06 × 10⁴ J. For K: ln K = 2.12 × 10⁵ ÷ (8.314 × 298) ≈ 85.6, so K ≈ 1.5 × 10³⁷.

Answer: ΔG° = −212 kJ mol⁻¹; maximum work 10.6 kJ; K ≈ 10³⁷.

Quick check

1. For 2Ag⁺ + Cu → 2Ag + Cu²⁺ with E° = +0.46 V, what is ΔG° in kJ per mole of reaction? Answer: n = 2, so ΔG° = −2 × 96 485 × 0.46 J mol⁻¹ ≈ −88.8 kJ mol⁻¹.

Exam focus

Show n explicitly and justify it from the half-equations. State units at every step, converting J to kJ at the end. When a question mentions a mass or amount consumed, multiply by the extent of reaction, not by the coefficient again. Signs matter: a spontaneous cell must give a negative ΔG.

Advanced insight

Because ΔG = ΔH − TΔS, measuring E at several temperatures yields ΔS° = nF(dE°/dT) and hence ΔH°. Electrochemical measurements are therefore one of the most precise routes to thermodynamic data. The same Gibbs-energy bookkeeping explains why E° for Fe³⁺ + 3e⁻ → Fe is about −0.04 V: it is the charge-weighted combination of +0.77 V (one electron) and −0.44 V (two electrons), not their sum.

Summary

Cell potential and Gibbs energy are linked by ΔG = −nFE, where n is the electron number per mole of reaction as written. ΔG scales with how the equation is written and with the extent of reaction, whereas E does not. The same relation gives ln K = nFE°/(RT) and allows half-reaction potentials to be combined through additive Gibbs energies.

Practice questions

1. A cell reaction with n = 3 has E° = +0.62 V. Calculate ΔG°. Answer: ΔG° = −3 × 96 485 × 0.62 ≈ −1.79 × 10⁵ J mol⁻¹ = −179 kJ mol⁻¹. 2. A reaction has ΔG° = +38.6 kJ mol⁻¹ with n = 2. Find E° and comment on spontaneity. Answer: E° = −ΔG°/(nF) = −38 600 ÷ (2 × 96 485) = −0.20 V; negative, so it is not spontaneous under standard conditions. 3. How much charge passes, and what is the maximum work, when a 1.10 V Daniell cell consumes 0.0200 mol of Cu²⁺? Answer: Q = 2 × 0.0200 × 96 485 ≈ 3860 C; work = 3860 × 1.10 ≈ 4.25 kJ. 4. Use Gibbs energies to find E° for Fe³⁺ + 3e⁻ → Fe, given E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(Fe²⁺/Fe) = −0.44 V. Answer: ΔG° = −F(0.77) − 2F(−0.44) = +0.11F; E° = −ΔG°/(3F) = −0.11/3 ≈ −0.04 V.