The half‑reaction $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$ has $E°=+1.51$ V at 25 °C. The Fe²⁺/Fe³⁺ couple has $E°=+0.77$ V. Write the balanced overall redox reaction when $Mn{O}_4^{-}$ is the oxidising agent and Fe²⁺ the reducing agent. Then calculate the equilibrium constant $K$ for this reaction using ${\log }_{10}K=nE{°}_{cell}/0.0592$.

Step-by-step solution

1. Step 1: Write the reduction of permanganate: $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$.
2. Step 2: Write the oxidation of Fe²⁺ (reverse of the given reduction): $F{e}^{2+}\to F{e}^{3+}+e^{-}$.
3. Step 3: Balance electrons – multiply the Fe²⁺ oxidation by 5 to supply 5 electrons: $5F{e}^{2+}\to 5F{e}^{3+}+5e^{-}$.
4. Step 4: Add the two half‑reactions; electrons cancel giving the overall reaction: $Mn{O}_4^{-}+8H^{+}+5F{e}^{2+}\to M{n}^{2+}+4H_{2}O+5F{e}^{3+}$.
5. Step 5: Identify the cathode and anode potentials. Cathode (reduction) $E{°}_{\text{cathode}}=+1.51$ V, anode (oxidation) $E{°}_{\text{anode}}=-0.77$ V.
6. Step 6: Compute $E{°}_{cell}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}=1.51-0.77=0.74$ V.
7. Step 7: Total electrons transferred $n=5$ (from the permanganate half‑reaction).
8. Step 8: Calculate $lo{g}_{10}K=nE{°}_{cell}/0.0592=5\times 0.74/0.0592=62.5$.
9. Step 9: Hence $K=10^{62.5}$.

Concept tested: Balancing redox equations and calculating equilibrium constants

Published 13 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked