Explanation

(c)
Given,
${Fe}^{3 +} + \ {3e}^{-}\ \longrightarrow Fe;\$ $E_{1}^{{^\circ}}$ = - 0.036 V ….(i)
${Fe}^{2 +} + \ {2e}^{-}\ \longrightarrow Fe;\ \$ $E_{2}^{{^\circ}}$ = - 0.439 V ….(ii)
We need to calculate
${Fe}^{3 +} + \ e^{-}\ \longrightarrow \ {Fe}^{2 +}\ E_{3}^{{^\circ}}\ = \ ?\ \ \ \$ …..(iii)
We can obtain the (III) by subtracting II from I but $E_{3}^{{^\circ}}$ , we can not obtain that way because electrode potential is intensive property. That’s when we determine $E_{3}^{{^\circ}}$ calculating
${\mathrm{\Delta}G}_{3} = \ {\mathrm{\Delta}G}_{1}\ - \ {\mathrm{\Delta}G}_{2}$
(△G is an extenwise property)
$\ \ \ \ {\mathrm{\Delta}G}_{3}\ = \ 3\ \times \ 0.036\ F\ –\ 2\ \times \ 0.439\ F$
$\ {\mathrm{\Delta}G}_{3}\ = \ 0.108\ F\ –\ 0.878\ F$
$\ \ - 1\ \times \ F\ \times \ E_{3}^{{^\circ}}\ = \ - 0.770\ F$
$\ \ \ \ \ \ \ \ \ \ \ \ E_{3}^{{^\circ}}\ = \ 0.770\ F$