Question

The half cell reactions for rusting of iron are :
$2 H ^{+}+2 e^{-}+\frac{1}{2} O _{2} \longrightarrow H _{2} O (l) ; E^{\circ}=+1.23\, V$
$Fe ^{2+}+2 e^{-} \longrightarrow Fe (s) ; E^{\circ}=-0.44\, V$
$\Delta G^{\circ}($ in $kJ )$ for the reaction is

• A. -76
• B. -322
• C. -122
• D. -176

Answer 1

Answer: (B)

The net reaction is
$2 H ^{+}+\frac{1}{2} O _{2} + Fe \longrightarrow H _{2} O + Fe ^{2+} ; E^{\circ}=1.67\, V$
$\Delta G^{\circ} =-n E^{\circ} F =-\frac{2 \times 1.67 \times 96500}{1000} kJ$
$=-322.31 \,kJ$