Question

Given standard electrode potentials

$Fe^{2 +} + 2e^{-} \rightarrow Fe - \overset{E^{0}}{0.440V}$; $Fe^{3 +} + 3e^{-} \rightarrow Fe - 0.036V$

The standard electrode potential $(E^{0})$ for

$Fe^{3 +} + e \rightarrow Fe^{2 +}$is

• A. – 0.476 V
• B. –0.404 V
• C.
  • 0.404V
• D. +0.772V

Answer 1

Answer: (D)

+0.772V

Answer 2

$\Delta G^{0} = - nFE^{0}$

$Fe^{2 +} + 2e^{-} \rightarrow Fe$ .....(i)

$\Delta G^{0} = - 2 \times F \times ( - 0.440V) = 0.880F$;

$Fe^{3 +} + 3e^{-} \rightarrow Fe$ .....(ii)

$\Delta G^{0} = - 3 \times F \times ( - 0.036) = 0.108F$,

On subtracting equation (i) from equation (ii) we get

$\mathbf{\Delta}\mathbf{G}^{\mathbf{0}}\mathbf{= 0.108F}\mathbf{-}\mathbf{0.880F}$ =– 0.772F

$E^{0}$for the reaction $= \frac{\Delta G^{0}}{nF} = - \frac{( - 0.772F)}{1 \times F} = + 0.772V$