Question

The half cell reaction for rusting of iron are
$2\,H^{+}+\frac{1}{2}O_{2}+2e^{-} \rightarrow H_{2}O;E^\circ =1.23\,V$
$Fe^{2 +}+2e^{-} \rightarrow Fe;E^\circ =-0.44\,V$
$ΔG^\circ $ for the reaction
$4H^{+}+O_{2}+2Fe \rightarrow 2Fe^{2 +}+2H_{2}O$ is :-

• A. $-76\,kJ$
• B. $-644\,kJ$
• C. $-122\,kJ$
• D. $-176\,kJ$

Answer 1

Answer: (B)

$4H^{+}+O_{2}+4e^{-} \rightarrow 2H_{2}O$ $1.23$
$\frac{2 Fe \rightarrow 2 Fe^{2 +} + 4 e^{-}}{4 H^{+} + O_{2} + 2 Fe \rightarrow 2 Fe^{2 +} + 2 H_{2} O}\frac{0.44}{E^\circ =1.67}$
$ΔG^\circ =-nFE^\circ =-4\times 96500\times 1.67$
$=-644\,kJ$