$\mathrm{FeO}_{4}^{2-} \xrightarrow{+2.0 \mathrm{~V}} \mathrm{Fe}^{3+} \xrightarrow{0.8 \mathrm{~V}} \mathrm{Fe}^{2+} \xrightarrow{-0.5 \mathrm{~V}} \mathrm{Fe}^{0}$ In the above diagram, the standard electrode potentials are given in volts (over the arrow). The value of $\mathrm{E}_{\mathrm{FeO}_{4}^{2-} / \mathrm{Fe}^{2+}}^{\Theta}$ is

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Redox Reactions and Electrochemistry

FeO_{4}^{2 -} + 2.0 V Fe^{3 +} 0.8 V Fe^{2 +} - 0.5 V Fe^{0}

In the above diagram, the standard electrode potentials are given in volts (over the arrow).
The value of

E_{FeO_{4}^{2 -} / Fe^{2 +}}^{Θ}

Answer: (A)

Standard electrode potentials cannot be added directly, since each corresponds to a different number of electrons transferred; only

Δ G^{\circ}

values are additive.

Fe O_{4}^{2 -} \to F e^{3 +}

n_{1} = 3

E_{1}^{\circ} = 2.0 V

F e^{3 +} \to F e^{2 +}

n_{2} = 1

E_{2}^{\circ} = 0.8 V

For the combined 4-electron reduction

Fe O_{4}^{2 -} \to F e^{2 +}

n_{4} E_{4}^{\circ} = n_{1} E_{1}^{\circ} + n_{2} E_{2}^{\circ} = 3 (2.0) + 1 (0.8) = 6.8

E_{4}^{\circ} = \frac{6.8}{4} = 1.7 V

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option