Question

Given, standard electrode potentials of following reactions, $Fe ^{2+}+2 e^{-} \longrightarrow Fe ; E^{\circ}=-0.440\, V $ $Fe ^{3+}+3 e^{-} \longrightarrow Fe ; E^{\circ}=-0.036\, V$ The standard electrode potential $\left(E^{\circ}\right)$ for $Fe ^{3+}+e^{-} \longrightarrow Fe ^{2+}$, is

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

Answer 1

The Correct Option is D

$\Delta G^{\circ}=-n F E^{\circ}$
$ Fe ^{2+}+2 e^{-} \longrightarrow Fe$ ... (i)
$ \Delta G^{\circ} =-2 \times F \times(-0.440 \,V ) $
$=0.880\, F $
$ Fe ^{3+}+3 e^{-} \longrightarrow Fe $ ... (ii)
$ \Delta G^{\circ}=-3 \times F \times(-0.036)=0.108\, F $
On subtracting E (i) from E (ii)
$ Fe ^{3+}+e^{-} \longrightarrow Fe ^{2+}$
$\Delta G^{\circ}=0.108\, F -0.880 \,F =-0.772 \,F$
$E_{\text {cell }}^{\circ}=\frac{-\Delta G^{\circ}}{n F}$
$=-\frac{(-0.772 F)}{1 \times F}=+0.772 \,V$