Question
The rusting of iron takes place as follows
$$\eqalign{
& 2{H^ + } + 2{e^ - }{ + ^{\frac{1}{2}}}{O_2} \to {H_2}O\left( l \right);\,{E^o} = + 1.23\,V \cr
& F{e^{2 + }} + 2{e^ - } \to Fe\left( s \right);\,{E^o} = - 0.44\,V \cr} $$
Calculate $$\Delta {G^o}$$ for the net process
A.
$$ - 322\,kJ\,mo{l^{ - 1}}$$
B.
$$ - 161\,kJ\,mo{l^{ - 1}}$$
C.
$$ - 152\,kJ\,mo{l^{ - 1}}$$
D.
$$ - 76\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ - 322\,kJ\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& Fe\left( s \right) \to F{e^{2 + }} + 2{e^ - };\,\,\,\,\,\,\,\,{E^o} = 0.44\,V \cr
& \frac{{2{H^ + } + 2{e^ - } + \frac{1}{2}{O_2} \to {H_2}O\left( l \right);\,\,\,\,{E^o} = + 1.23\,V}}{{Fe\left( s \right) + 2{H^ + } + \frac{1}{2}{O_2} \to F{e^{2 + }} + {H_2}O;}} \cr
& E_{cell}^o = 0.44 + 1.23 = 1.67V \cr
& \therefore \,\,\Delta {G^o} = - nFE_{cell}^o = - 2 \times 96500 \times 1.67 = - 322kJ \cr} $$