JEE Main 2025 Jan 24 Shift 1, Chemistry Q1: Electrochemical Cell And Nernst Equation

The $\mathrm{F}{\mathrm{e}}^{3+}/F{\mathrm{e}}^{2+}$ couple is not given directly. Combine the two iron half-cells using Gibbs energies (which are additive) rather than potentials (which are not).

For the path $\mathrm{Fe}\to \mathrm{F}{\mathrm{e}}^{2+}\to \mathrm{F}{\mathrm{e}}^{3+}$ versus the direct $\mathrm{Fe}\to \mathrm{F}{\mathrm{e}}^{3+}$:

$\Delta G^0(\mathrm{Fe}\to \mathrm{F}{\mathrm{e}}^{3+})=\Delta G^0(\mathrm{Fe}\to \mathrm{F}{\mathrm{e}}^{2+})+\Delta G^0(\mathrm{F}{\mathrm{e}}^{2+}\to \mathrm{F}{\mathrm{e}}^{3+})$

Using $\Delta G^0=-nF{E}^0$ for each step (oxidation potentials are negatives of the reduction values given):

$-3F(-z)=-2F(-y)+(-1F)E_{F{e}^{2+}/F{e}^{3+}}^0$

Solving: $E_{F{e}^{2+}/F{e}^{3+}}^0=2y-3z$, so the reduction potential $E_{F{e}^{3+}/F{e}^{2+}}^0=3z-2y$.

For the full cell, Ag${}^{+}$/Ag is the cathode and Fe${}^{3+}$/Fe${}^{2+}$ is the anode:

$E_{cell}^0=E_{cathode}^0-E_{anode}^0=x-(3z-2y)=x+2y-3z$

$\therefore$ Option (2) is correct.