The half‑reaction $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$ has E° = +1.51 V at 25 °C. The $F{e}^{3+}/F{e}^{2+}$ couple has E° = +0.77 V. Write the balanced overall redox reaction when $Mn{O}_4^{-}$ acts as oxidising agent and $F{e}^{2+}$ as reducing agent. Then calculate the equilibrium constant K for this overall reaction. Use the relation $lo{g}_{10}K=n{E}_{cell}^{\circ }$/0.0592.

Step-by-step solution

1. Write the two half‑reactions:
2. (1) $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$ (reduction, E° = +1.51 V)
3. (2) $F{e}^{2+}\to F{e}^{3+}+e^{-}$ (oxidation, reverse of the given reduction, E° = -0.77 V)
4. Multiply the oxidation half‑reaction by 5 to match the electrons:
5. 5 $F{e}^{2+}$ → 5 $F{e}^{3+}+5e^{-}$
6. Add the two half‑reactions to obtain the overall balanced equation:
7. 5 $F{e}^{2+}+Mn{O}_4^{-}+8H^{+}$ → 5 $F{e}^{3+}+M{n}^{2+}+4H_{2}O$.
8. Identify the cell potential: $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }$ = 1.51 V - 0.77 V = 2.28 V.
9. The total number of electrons transferred in the balanced equation is n = 5.
10. Use the formula $lo{g}_{10}K=n{E}_{cell}^{\circ }$/0.0592.
11. Calculate the exponent: $n{E}_{cell}^{\circ }$/0.0592 = (5 × 2.28)/0.0592 ≈ 192.6.
12. Hence $lo{g}_{10}K$ ≈ 192.6, so K ≈ $10^{192.6}\approx 10^{193}$.

Concept tested: Standard cell potential and equilibrium constant

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked