$\mathrm{FeO}_{4}^{2-} \xrightarrow{+2.2 \mathrm{~V}} \mathrm{Fe}^{3+} \xrightarrow{+0.70 \mathrm{~V}} \mathrm{Fe}^{2+} \xrightarrow{-0.45 \mathrm{~V}} \mathrm{Fe}^{0}$ $\mathrm{E}_{\mathrm{FeO}_{4}^{2-} / \mathrm{Fe}^{2+}}^{\theta}$ is $\mathrm{x} \times 10^{-3} \mathrm{~V}$. The value of x is ____

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Redox Reactions and Electrochemistry

FeO_{4}^{2 -} + 2.2 V Fe^{3 +} + 0.70 V Fe^{2 +} - 0.45 V Fe^{0}

E_{FeO_{4}^{2 -} / Fe^{2 +}}^{θ}

x \times 1 0^{- 3} V

. The value of x is ____

Answer: 1825

Standard potentials cannot be added directly, but the quantity

n E^{\circ}

, which is proportional to free energy, is additive. Going from ferrate(VI) to iron(III) the oxidation number of iron falls from

+ 6

+ 3

, a three-electron change with

E^{\circ} = 2.2

V, while going from iron(III) to iron(II) is a one-electron change with

E^{\circ} = 0.70

V. For the combined four-electron change from ferrate(VI) to iron(II),

4 E^{\circ} = 3 (2.2) + 1 (0.70) = 7.3

, so

E^{\circ} = 7.3/4 = 1.825

V. Written as

x \times 1 0^{- 3}

V this gives

x = 1825

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Integer answer