Topic 4 of 7
Gibbs energy and combined potentials
Use electron amounts to combine energy changes correctly.
A-Level 9476 (2026-2027)
Add Gibbs energies when combining half-reactions
The electron count connects an intensive potential to an extensive energy change.
ΔG° = -nFE°cell. Here n is the number of moles of electrons transferred per mole of the cell reaction as written, and F is the charge per mole of electrons. With F in C mol-1 and E in V, ΔG° is in J mol-1 of reaction because 1 V = 1 J C-1.
Worked example
Convert a cell voltage into free energy
Find ΔG° for Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), with E°cell = 1.10 V and F = 9.65 × 104 C mol-1.
- The balanced reaction transfers two electrons, so n = 2.
- ΔG° = -2(9.65 × 104)(1.10) = -212300 J mol-1.
- Divide by 1000 to express the result in kJ mol-1.
ΔG° = -212 kJ mol-1. Doubling the reaction would double n and ΔG°, but E°cell would remain 1.10 V.
When two reduction steps combine to make a new reduction half-equation, their E° values cannot simply be added. Convert each to its corresponding -nFE° energy contribution relative to the same reference, add those contributions with the required signs, then divide by -nF for the combined half-equation.
Worked example
Find the potential for a combined iron reduction
Given Fe3+ + e- → Fe2+, E°1 = +0.77 V, and Fe2+ + 2e- → Fe, E°2 = -0.44 V, find E° for Fe3+ + 3e- → Fe.
- The first step contributes -F(0.77); the second contributes -2F(-0.44).
- Add: ΔG°combined = -F[0.77 + 2(-0.44)] = +0.11F.
- The combined step transfers three electrons, so -3FE°combined = +0.11F.
- E°combined = [0.77 + 2(-0.44)]/3 = -0.0367 V from the rounded supplied data.
Approximately -0.04 V, consistent with the Data Booklet. The result is electron-weighted, not the sum +0.33 V or an unweighted average.