Electrochemistry 2 Question 20
19. The half cell reactions for rusting of iron are :
$$ \begin{aligned} 2 \mathrm{H}^{+}+2 e^{-}+\frac{1}{2} \mathrm{O}{2} \longrightarrow \mathrm{H}{2} \mathrm{O}(l) ; & E^{\circ} & =+1.23 \mathrm{~V} \ \mathrm{Fe}^{2+}+2 e^{-} \longrightarrow \mathrm{Fe}(s) ; & E^{\circ} & =-0.44 \mathrm{~V} \end{aligned} $$
$\Delta G^{\circ}$ (in $\mathrm{kJ}$ ) for the reaction is
(a) -76
(b) -322
(c) -122
(d) -176
(2005, 1M)
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Solution:
- The net reaction is
$$ \begin{array}{r} 2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}{2}+\mathrm{Fe} \longrightarrow \mathrm{H}{2} \mathrm{O}+\mathrm{Fe}^{2+} ; E^{\circ}=1.67 \mathrm{~V} \ \Delta G^{\circ}=-n E^{\circ} F=-\frac{2 \times 1.67 \times 96500}{1000} \mathrm{~kJ}=-322.31 \mathrm{~kJ} \end{array} $$
