JEE Advanced 2022 Paper 1, Chemistry Section 1 Q2: Electrochemical Cell And Nernst Equation

Combine three half-reactions using $\Delta G^{\circ }=-nF{E}^{\circ }$ (Gibbs energies add; potentials do not add directly when electron counts differ).

(1) ${\text{MnO}}_4^{-}+4{\text{H}}^{+}+3e^{-}\to {\text{MnO}}_2+2{\text{H}}_2\text{O}$,   $\Delta G_1^{\circ }=-3F(1.68)=-5.04F$.

(2) ${\text{MnO}}_2+4{\text{H}}^{+}+2e^{-}\to {\text{Mn}}^{2+}+2{\text{H}}_2\text{O}$,   $\Delta G_2^{\circ }=-2F(1.21)=-2.42F$.

(3) ${\text{Mn}}^{2+}+2e^{-}\to \text{Mn(s)}$,   $\Delta G_3^{\circ }=-2F(-1.03)=+2.06F$.

Adding gives ${\text{MnO}}_4^{-}+8{\text{H}}^{+}+7e^{-}\to \text{Mn(s)}+4{\text{H}}_2\text{O}$ with $n=7$.

$\Delta G^{\circ }=(-5.04-2.42+2.06)F=-5.40F=-7F{E}^{\circ }$.

$E^{\circ }=\frac{5.40}{7}=0.77$ V.