Given : \(\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}}^{\circ}=-0.036 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.439 \mathrm{~V}\). The value of standard electrode potential for the change, \(\mathrm{Fe}_{(\mathrm{aq})}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})\) will be :

JEE Main · 2009 · 12 Apr 2009 · Offline · Redox Reactions and Electrochemistry

Given :

E_{Fe^{3 +} / Fe}^{\circ} = - 0.036 V, E_{Fe^{2 +} / Fe}^{\circ} = - 0.439 V

. The value of standard electrode potential for the change,

Fe_{(aq)}^{3 +} + e^{-} \to Fe^{2 +} (aq)

will be :

Answer: (C)

Potentials do not add directly, but free energies do, so convert each through

Δ G^{\circ} = - n F E^{\circ}

. For

F e^{3 +} + 3 e^{-} \to Fe

Δ G_{1}^{\circ} = - 3 F (- 0.036) = + 0.108 F

, and for

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

Δ G_{2}^{\circ} = - 2 F (- 0.439) = + 0.878 F

. Subtracting gives the free energy of

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

0.108 F - 0.878 F = - 0.770 F

, and since this equals

- 1 \cdot F \cdot E^{\circ}

, the electrode potential is

0.770

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option