In acidic solution, potassium permanganate oxidises ferrous ions. Given $E_{Mn{O}_4^{-}/M{n}^{2+}}^{\circ }$ = +1.51 V and $E_{F{e}^{3+}/F{e}^{2+}}^{\circ }$ = +0.77 V, which of the following statements is correct?

Step-by-step solution

1. Write the two half‑reactions:${\text{nMnO}}_4^{-}+8H^{+}+5e^{-}\to$ $M{n}^{2+}+4H_{2}O$ \quad $E^{\circ }$=+1.51\,V${\text{nFe}}^{3+}+e^{-}\to$ $F{e}^{2+}$ \quad $E^{\circ }$=+0.77\,V.
2. Identify the cathode (higher reduction potential): the $Mn{O}_4^{-}$ reduction is the cathode; $F{e}^{2+}\to$ $F{e}^{3+}$ is the oxidation at the anode (reverse of the given reduction).
3. Calculate the standard cell potential: $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }$=1.51-0.77=0.74\,V.
4. Balance electrons: multiply the Fe half‑reaction by 5 so that $5e^{-}$ are exchanged.\n5($F{e}^{2+}\to$ $F{e}^{3+}+e^{-}$) gives $5F{e}^{2+}\to 5F{e}^{3+}+5e^{-}$.
5. Add the half‑reactions to obtain the overall equation:${\text{nMnO}}_4^{-}+8H^{+}+5F{e}^{2+}\to$ $M{n}^{2+}+4H_{2}O+5F{e}^{3+}$.
6. Number of electrons transferred n =5. Use the relation ${\log }_{10}K=n{E}_{cell}^{\circ }$/0.0592.$\text{n}{\log }_{10}K$ = 5$\times 0$.74/0.0592 = 62.5.
7. Therefore K = $10^{62.5}$ $\approx$ $10^{62}$.
8. Select the option that matches both $E_{cell}^{\circ }$=0.74\,V and K $\approx$ $10^{62}$.

Concept tested: Overall redox and equilibrium constant

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