Cell Voltage and Gibbs Energy

Introduction

The equation ΔG=−nFE is often memorized, but its meaning is richer than a sign rule. It connects thermodynamic work per reaction extent to charge moved across a potential difference. Deriving it clarifies why voltage is intensive, why electron count matters and why an operating battery delivers less than its reversible limit.

Core explanation

For a balanced cell reaction, n moles of electrons pass through the external circuit per mole of reaction as written. Their charge magnitude is q=nF, with F≈96,485 C mol⁻¹ of electrons. If the charge moves through a reversible potential difference E, electrical work delivered by the cell has magnitude qE. At constant temperature and pressure, maximum non-expansion work delivered by a spontaneous reaction equals −ΔᵣG, so ΔᵣG=−nFE. This is a relation for the reversible cell potential under the stated composition.

The signs follow physical direction. If E is positive for the forward galvanic reaction, ΔᵣG is negative and the cell can do useful electrical work. Reversing the balanced reaction changes the signs of ΔᵣG and E, while n remains a positive electron count by convention. A driven electrolytic process can proceed in the reverse direction when an external source supplies sufficient energy; the reaction's spontaneous direction at a given state still follows the Gibbs-energy sign.

Scaling the balanced equation by a factor k makes k times as many electrons pass per mole of the newly written reaction event and makes its ΔᵣG k times as large. E=−ΔᵣG/(nF) stays unchanged. This intensive/extensive distinction also explains why combining half-reactions requires adding Gibbs energies, not adding arbitrary potential numbers without electron accounting.

Standard quantities obey ΔᵣG°=−nFE°. At nonstandard composition, ΔᵣG=ΔᵣG°+RT ln Q. Equating the two forms produces the Nernst relation. This derivation shows that a voltage changes with reactant and product activities because their chemical potentials change. E° remains tied to a standard state, even when the actual E differs.

The open-circuit EMF approaches the reversible potential when the system is near equilibrium at each electrode and the meter draws negligible current. Under a real load, ohmic loss, charge-transfer overpotential and concentration polarization reduce delivered galvanic terminal voltage. Multiplying loaded voltage by charge gives actual electrical energy delivered, while −ΔG represents the ideal maximum under a specified reversible path. Heat generation and entropy production account for the difference.

Reaction enthalpy ΔH is not generally equal to maximum electrical work. At constant temperature and pressure, ΔG=ΔH−TΔS. Some reaction energy may appear as heat exchange even in a reversible cell, depending on entropy change. Equating “total chemical energy” directly with voltage can obscure this distinction.

Step-by-step reasoning

1. Balance reaction and count electrons n. 2. Convert electron amount to charge nF. 3. Multiply by reversible E for work magnitude. 4. Apply ΔᵣG=−nFE with direction sign. 5. Scale Gibbs energy and n together if the reaction equation is rescaled, and keep operating losses separate.

Visual explanation

Draw an energy box labeled reaction Gibbs-energy decrease feeding a wire with charge q through potential E. Label ideal electrical work qE. Add a second branch from the box to heat or dissipative losses for real current flow, making clear why terminal voltage is lower than reversible E.

Real-world analogy

Lifting one kilogram through a height requires energy proportional to mass and height; height itself does not double if two kilograms are lifted. Voltage resembles energy per charge, while total charge resembles mass. The analogy clarifies scaling but not the chemical-potential origin of E.

Real-world example

A battery may have a measured open-circuit voltage near its reversible value, then a lower terminal voltage when powering a motor. The chemical reaction can still be favorable, but internal resistance and kinetic losses convert some available free energy to heat instead of delivered electrical work.

Why?

Why is ΔG tied to reversible rather than arbitrary loaded voltage? Gibbs energy gives the maximum non-expansion work available from the state change. A loaded cell produces entropy through finite-rate transport and resistance, so its delivered electrical work is less than that maximum.

Common misconception

“A cell with twice the electrode area has twice the equilibrium voltage.” More area may allow greater current at lower overpotential, but the reversible voltage per unit charge is set by reaction activities and temperature, not by geometric size.

Worked example

A reaction transfers three electrons and has reversible E=0.50 V. Then ΔᵣG=−3×96,485×0.50≈−144,728 J mol⁻¹, or −145 kJ mol⁻¹. If a real device delivers only 0.40 V during discharge over the same idealized charge amount, delivered electrical energy magnitude is about 116 kJ mol⁻¹; the difference is not a new ΔᵣG value for the original reversible state.

Quick check

1. What is the charge magnitude for one mole of a reaction transferring n=2 electrons? Answer: 2F≈193,000 C per mole of reaction. 2. Does doubling reaction coefficients double E? Answer: No; it doubles ΔG and n together.

Exam focus

Include “per mole of reaction as written” with ΔᵣG and n, convert volts to joules per coulomb and verify signs. Distinguish reversible E from a loaded terminal voltage and Gibbs energy from enthalpy.

Advanced insight

The formal differential relation dG=−S dT+V dP+Σμᵢdnᵢ links composition to chemical potentials. Electrochemical cell voltage is the Gibbs-energy derivative with respect to transferred charge along a balanced reaction path. This derivative view explains why E changes as the cell discharges and activities evolve.

Summary

Reversible cell voltage is maximum Gibbs work per transferred charge, giving ΔᵣG=−nFE. Potential is intensive while reaction Gibbs energy and charge scale with extent. Actual devices deliver less electrical work under current because irreversible losses produce entropy.

Practice questions

1. A two-electron cell has E=1.20 V. Estimate ΔᵣG. Answer: −2×96,485×1.20≈−232 kJ mol⁻¹ of reaction as written. 2. Why does a loaded cell not generally deliver all of −ΔG as electrical work? Answer: Resistance, reaction overpotential and transport losses dissipate part of the available free energy. 3. What changes in ΔᵣG and E when the forward reaction is reversed? Answer: Both change sign for the reverse direction; the positive count n remains the same.