When $ceMnO{4}^{-}$ oxidises $ceF{e}^{2+}$ in acidic solution, the balanced overall reaction is $ceMnO{4}^{-}+5F{e}^{2+}->M{n}^{2+}+5F{e}^{3+}$. Given the standard potentials $E_{ceMnO{4}^{-}/M{n}^{2+}}^{circ}=+1.51$ V and $E_{ceF{e}^{3+}/F{e}^{2+}}^{circ}=+0.77$ V, calculate the equilibrium constant $K$ for this reaction.

Step-by-step solution

1. Step 1: The cell potential for the reaction is $E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$. The cathode is $ceMnO{4}^{-}+8H++5e-->M{n}^{2+}+4H2O$ with $E^{\circ }=+1.51$ V.
2. Step 2: The anode is the reverse of the Fe³⁺/Fe²⁺ couple, so $E_{\text{anode}}^{\circ }=-0.77$ V.
3. Step 3: Thus $E_{\text{cell}}^{\circ }=1.51-(-0.77)=0.74$ V.
4. Step 4: The overall reaction involves 5 electrons, so $n=5$. Use the relation $lo{g}_{10}K=\frac{n{E}_{\text{cell}}^{\circ }}{0.0592}$ at 25 °C.
5. Step 5: Compute $lo{g}_{10}K=\frac{5\times 0.74}{0.0592}=62.5$.
6. Step 6: Convert to $K$: $K=10^{62.5}\approx 3.8\times 10^{62}\approx 4\times 10^{62}$.
7. Final Answer: $Kapprox4\times 10^{62}$

Concept tested: Equilibrium constant from cell potential

Published 10 October 2026 · How this solution was prepared and checked