Question

The half-cell reactions for the corrossion are
$2H^{+} + 1/2 O_{2} + 2e^{-} \to H^{2}O ; E^{\circ} = - 1.23 V$
$Fe^{2+} + 2e^{-} \to Fe; E^{\circ} = - 0.44 V$
Find the $\Delta G^{\circ}$ (in $kJ$) for the overall reaction.

• A. $ - 76$
• B. $-322$
• C. $ -161$
• D. $-152$

Answer 1

Answer: (B)

The reduction potential of two half reactions suggest that reduction of $H^{+}$ and oxidation of $Fe$ take place, so
$E^{\circ}_{\text{cell}} =E^{\circ}_{\text{cathode}} -E^{\circ}_{\text{anode}}$
$ =1.23-\left(-0.44\right) =1.67V$
$\Delta G^{\circ} = - nFE^{\circ} n =2$
$\Delta G^{\circ} = -2 \times 96500 \times 1.67 = -322310 \,J$
$= -322.31 \,kJ$