Latimer diagram: skipping an oxidation state

Worked example

Part of the Latimer diagram for manganese in basic solution reads MnO₄⁻ —(0.568 V)→ MnO₄²⁻ —(2.284 V)→ MnO₂, with one and two electrons transferred in the two steps. Calculate E° for MnO₄⁻ → MnO₂.

1. 1. Gibbs energy and cell potential: solve for ΔG°₁ = −1·F·E°₁

   $\Delta G^\circ = -nFE^\circ$

   = −54.80 kJ mol⁻¹

2. 2. Gibbs energy and cell potential: solve for ΔG°₂ = −2·F·E°₂

   $\Delta G^\circ = -nFE^\circ$

   = −440.7 kJ mol⁻¹

3. 3. Add ΔG°₁ = −1·F·E°₁ + ΔG°₂ = −2·F·E°₂ → ΔG° overall

   = −495.5 kJ mol⁻¹

4. 4. Gibbs energy and cell potential: solve for E° overall (3e⁻)

   $E^\circ = -\dfrac{\Delta G^\circ}{nF}$

   = 1.712 V

Answer: E°(MnO₄⁻/MnO₂) = 1.71 V

Method

Convert each step to ΔG° = −nFE°, add the Gibbs energies, and convert back using the total number of electrons: E° = (n₁E°₁ + n₂E°₂)/(n₁ + n₂). Because the right-hand potential is higher, MnO₄²⁻ is unstable to disproportionation here.

Common mistakes

• Potentials are not averaged equally: weight each by its electrons, E = (1·E°₁ + 2·E°₂)/3.
• Potentials do not add; Gibbs energies do. Convert with ΔG° = −nFE°, add, then convert back with n = 3.

Formulas used

• Gibbs energy and cell potential

  $\Delta G^\circ = -nFE^\circ$

More electrochemistry questions

• ΔG° from E°
• Nernst equation
• Mass deposited by electrolysis
• E°cell from reduction potentials
• Equilibrium constant from E°cell
• Electrode potential away from standard conditions
• Disproportionation constant from a Latimer diagram
• Daniell cell away from standard conditions