Q. Given standard electrode potentials ${Fe}^{2 +} + \ {2e}^{-}\ \longrightarrow Fe\$ $E^{{^\circ}}$ = -0.440 V ${Fe}^{3 +} + \ {3e}^{-}\ \longrightarrow Fe$ $E^{{^\circ}}$ = -0.036 V The standard electrode potential ($E^{{^\circ}}$) for $\ {Fe}^{3 +} + \ e^{-}\ \ \longrightarrow {Fe}^{2 +}\$ is

A) +0.772 V
B) -0.772 V
C) +0.417 V
D) -0.414 V
Explanation

(a)
${\mathrm{\Delta}G}^{{^\circ}}\ = \ - \ {nFE}^{{^\circ}}$
${Fe}^{2 +} + \ {2e}^{-}\ \longrightarrow \ \ Fe\ \ \ \ \ \$
${\mathrm{\Delta}G}^{{^\circ}}\ = \ - \ 2\ \times \ F\ \times \ ( - \ 0.440\ V)$
$\ \ \ \ \ \ \ \ \ \ = \ 0.880\ F\$ …(i)
${Fe}^{3 +} + \ {3e}^{-}\ \ \longrightarrow \ \ Fe$
${\mathrm{\Delta}G}^{{^\circ}}\ = \ - \ 3\ \times \ F\ \times \ ( - \ 0.036\ V)$
$\ \ \ \ \ \ \ \ \ \ = \ 0.108\ F\ \ \ \$ …(ii)
On subtracting Eqs. (i) from (ii)
${Fe}^{3 +} + \ e^{-}\ \ \rightarrow \ {Fe}^{2 +}$
${\mathrm{\Delta}G}^{{^\circ}}\ = \ 0.108\ F\ –\ 0.880\ F\ = \ - \ 0.772\ F$
$E^{{^\circ}} = \ - \ \frac{{\mathrm{\Delta}G}^{{^\circ}}}{nF}\ = \ - \ \frac{- \ 0.772\ F}{1\ \times F}\ = \ + \ 0.772\ V$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.