Electrochemistry Result Question 19
19. The half cell reactions for rusting of iron are :
$2 \mathrm{H}^{+}+2 e^{-}+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O}(l) ; E^{\circ}=+1.23 \mathrm{~V} $
$\mathrm{Fe}^{2+}+2 e^{-} \longrightarrow \mathrm{Fe}(s) ; E^{\circ}=-0.44 \mathrm{~V}$
$\Delta G^{\circ}$ (in kJ) for the reaction is
(2005, 1M)
(a) -76
(b) -322
(c) -122
(d) -176
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Answer:
Correct Answer: 19. ( b )
Solution:
- The net reaction is
$ \begin{gathered} 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} \mathrm{F}=-\frac{2 \times 1.67 \times 96500}{1000} \mathrm{~kJ}=-322.31 \mathrm{~kJ} \end{gathered} $
BETA
