The half cell reactions for the corrosion are, 
$2H^{+}\frac{1}{2}{O}_2+2e^{-}\to H_{2}O$$E^0=1.23V$    
 $F{e}^{+2}+2e^{-}\to F{e}_{\left(s\right)}{E}^0=-0.44V$
Find the  $\Delta G$ (in KJ) for the over all reaction. 
 

$F{e}_{\left(s\right)}\to F{e}^{+2}+2e^{-}\Delta G$

$2H^{+}+2e^{-}+\frac{1}{2}{O}_2\to H_2{O}_{\left(l\right)}\Delta G_2$

$Fe\left(s\right)+2H^{+}+\frac{1}{2}{O}_2\to F{e}^{+2}+H_{2}O\Delta G_3$

Applying  $\Delta G_1+\Delta G_2=\Delta G_3$

$\Delta G_3=\left(-2F\times 0.44\right)+\left(-2F\times 1.23\right)$

$=\left(-2\times 96500\times 0.44\right)+\left(-2\times 96500\times 1.23\right)$

$=-322310J$

$=-322kJ$