Chemistry - Electrochemistry Question with Solution | TestHub

The given half-cell reactions are:

${\text{Fe}}^{3+}+3e^{-}\to \text{Fe}$, $E_1^{\circ }=-0.04$ V
$\Delta G_1^{\circ }=-n_{1}F{E}_1^{\circ }=-3\times F\times (-0.04)=0.12F$

${\text{Fe}}^{2+}+2e^{-}\to \text{Fe}$, $E_2^{\circ }=-0.44$ V
$\Delta G_2^{\circ }=-n_{2}F{E}_2^{\circ }=-2\times F\times (-0.44)=0.88F$

The required half-cell reaction is:
${\text{Fe}}^{3+}+e^{-}\to {\text{Fe}}^{2+}$

This reaction can be obtained by subtracting the second reaction from the first reaction:
$\Delta G_3^{\circ }=\Delta G_1^{\circ }-\Delta G_2^{\circ }$

$-1\times F\times E_3^{\circ }=0.12F-0.88F$

$-F{E}_3^{\circ }=-0.76F$

$E_3^{\circ }=+0.76$ V

Answer: $+0.76$ V