Question

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

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

Answer 1

The correct option is D + 0.772 V

$\Delta G^{\circ }=-n{E}^{\circ }F$
$F{e}^{2+}+2e^{-}\to Fe$ $.......\left(i\right)$
$\Delta G^{\circ }=-2\times F\times (-0.440V)=0.880F$
$F{e}^{3+}+3e^{-}\to Fe$ $......\left(ii\right)$
$\Delta G^{\circ }=-3\times F\times (-0.036)=0.108F$
On substracting equation $\left(i\right)$ from $\left(ii\right)$
$F{e}^{3+}+e^{-}\to F{e}^{2+}$
$\Delta G^{\circ }=0.108F-0.880F=-0.772F$
$E^{\circ }$ for the reaction $=-\frac{\Delta G^{\circ }}{nF}=-\frac{(-0.772F)}{1\times F}=+0.772V$