9.9 Cell Potential & ΔG°

Cell Potential and Free Energy

Pick two half-reactions to set cathode and anode. The cell is drawn from them, electrodes and all, then E°cell, ΔG° = −nFE° and K follow step by step.

E°cell = E°cat − E°anΔG° = −nFE°Half-Reaction TableFavored vs Not
Topic 9.9

Cell Potential and Free Energy

Explain whether an electrochemical cell is thermodynamically favored, based on its standard cell potential and the constituent half-reactions within the cell.

Electrochemistry studies redox reactions in electrochemical cells. The reactions are either thermodynamically favored (positive voltage) or thermodynamically unfavored (negative voltage, requiring an externally applied potential).

The standard cell potential is calculated by identifying the oxidation and reduction half-reactions and their standard reduction potentials:

E°cell = E°(cathode) − E°(anode)

where both values are taken from a table of standard reduction potentials. Equivalently, reverse the anode half-reaction, flip the sign of its E°, and add.

How to assign electrodes: the half-reaction with the more positive standard reduction potential has the greater tendency to be reduced, so it runs as the reduction — at the cathode. The other is forced to run in reverse, as the oxidation, at the anode. Doing this always yields a positive E°cell for the galvanic arrangement.

EK 9.9.A.3 gives the thermodynamic link, on the equation sheet:

ΔG° = −nFE°

ΔG° is proportional to the negative of the cell potential. A cell with positive E° involves a thermodynamically favored reaction; a cell with negative E° involves an unfavored one. F = 96 485 coulombs per mole of electrons.

The rule students break most often: E° is an intensive property. When you multiply a half-reaction to balance electrons, do not multiply E°. Voltage is energy per unit charge, and scaling the reaction scales both. The stoichiometry enters through n in ΔG° = −nFE°.

Combining all three relationships gives the master picture:

E° > 0 ⇔ ΔG° < 0 ⇔ K > 1 ⇔ favored

Key points

  • E°_cell = E°(cathode) − E°(anode), both from a reduction-potential table.
  • Never multiply E° when scaling a half-reaction — it is intensive.
  • ΔG° = −nFE°, where n is moles of electrons transferred.
  • E° > 0 ⇔ ΔG° < 0 ⇔ K > 1 ⇔ thermodynamically favored.

Equations

  • on the exam sheet
    • moles of electrons transferred
    • 96 485 C mol⁻¹
    • standard cell potential (V)
  • not on the sheetBoth taken as reduction potentials.

Common mistakes

  • Multiplying E° when scaling a half-reaction. The most common electrochemistry error.
  • Forgetting to flip the sign of the anode’s reduction potential when writing it as an oxidation.
  • Using the wrong n. n is the number of electrons actually transferred in the balanced overall reaction.
  • Unit slips. ΔG° = −nFE° gives joules; divide by 1000 for kJ.

Worked example

For the cell Al(s) | Al³⁺(1 M) ‖ Ag⁺(1 M) | Ag(s), with E°(Al³⁺/Al) = −1.66 V and E°(Ag⁺/Ag) = +0.80 V: (a) write the balanced overall reaction, (b) calculate E°_cell, and (c) calculate ΔG°.

(a) Balanced reaction. Ag⁺/Ag has the more positive reduction potential, so silver is reduced at the cathode and aluminum is oxidized at the anode.

Cathode (reduction): Ag⁺ + e⁻ → Ag ×3
Anode (oxidation): Al → Al³⁺ + 3 e⁻

Multiply the silver half-reaction by 3 so the electrons cancel:

Al(s) + 3 Ag⁺(aq) → Al³⁺(aq) + 3 Ag(s), with n = 3

(b) E°cell.
E°cell = E°(cathode) − E°(anode) = 0.80 − (−1.66) = +2.46 V

Note: even though the silver half-reaction was multiplied by 3, its E° stays at +0.80 V. Voltage is intensive.

(c) ΔG°.
ΔG° = −nFE° = −(3)(96 485 C/mol)(2.46 V)
ΔG° = −712 060 J/mol = −712 kJ/mol

Strongly negative, consistent with a large positive E°: the reaction is thermodynamically favored, and this is a galvanic cell.

Full notes for topic 9.9 →