Q. The rusting of iron takes place as follows ${2H}^{+}\ + \ {2e}^{-}\ + \frac{1}{2}\ O_{2}\ \ \longrightarrow \ H_{2}\ O\ (l);\ \$ $E^{{^\circ}}\ = \ + 1.23\ V$ ${Fe}^{2 +}\ + \ {2e}^{-}\ \longrightarrow \ Fe\ (s);\ \ \ \ \ E^{{^\circ}}\ = \ - 0.44\ V$ Calculate $\Delta G{^\circ}$ for the net process.

A) -322 kJ ${mol}^{- 1}$
B) -161 kJ ${mol}^{- 1}$
C) -152 kJ ${mol}^{- 1}$
D) -76 kJ ${mol}^{- 1}$
Explanation

(a)
Reactions
(i) $Fe(s) \rightarrow \ {Fe}^{2 +}\ + \ {2e}^{-},\ \ \ \ \ \ E^{{^\circ}}\ = \ + \ 0.44\ V$
and ${\mathrm{\Delta}G}_{1}^{{^\circ}} = - \ {nE}^{{^\circ}}F = \ - \ 2\ \times \ 0.44\ \times \ F$
(ii)$\ {2H}^{+} + {2e}^{-} + \frac{1}{2}O_{2} \longrightarrow H_{2}O(l);\ \ \ E^{{^\circ}} = + 1.23\ V$
and ${\mathrm{\Delta}G}_{2}^{{^\circ}} = - \ 2\ \times ( + \ 1.23) \times F$
Net reaction,
$Fe\ (s) + {2H}^{+} + \frac{1}{2}O_{2}\ \rightarrow {Fe}^{2 +} + H_{2}O(l)$
${\mathrm{\Delta}G}_{3}^{{^\circ}} = \ {\mathrm{\Delta}G}_{1}^{{^\circ}}\ + \ {\mathrm{\Delta}G}_{2}^{{^\circ}}$
$= - 2 \times ( + \ 0.44)\ F + ( - \ 2 \times 1.23 \times F)$
$= - \ 0.88\ F\ \times - \ 2.46\ F = - \ 3.34\ F$
$= - \ 3.34 \times 96500\ J$
$= - 322.31\ kJ = - \ 322\ kJ\$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

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