Question
On the basis of the information available from the reaction $$\frac{4}{3}Al + {O_2} \to \frac{2}{3}A{l_2}{O_3},\Delta G = - 827\,kJ\,mo{l^{ - 1}}$$ of $${O_2}$$ the minimum $$e.m.f$$ required to carry out an electrolysis of $$A{l_2}{O_3}$$ is $$\left( {F = 96500\,C\,mo{l^{ - 1}}} \right)$$
A.
8.56$$\,V$$
B.
2.14$$\,V$$
C.
4.28$$\,V$$
D.
6.42$$\,V$$
Answer :
2.14$$\,V$$
Solution :
$$\eqalign{
& \Delta G = - nEF \cr
& {\text{For }}1{\text{ }}mole{\text{ of }}Al,{\text{ }}n = 3 \cr
& \therefore \,\,{\text{for}}\,\frac{4}{3}mole\,{\text{of}}\,Al,n = 3 \times \frac{4}{3} = 4 \cr
& {\text{According to question,}} \cr
& - 827 \times 1000 = - 4 \times E \times 96500 \cr
& E = \frac{{827 \times 1000}}{{4 \times 96500}} = 2.14\,V \cr
& \therefore \,\,{\text{ minimum }}e.m.f.{\text{ required}} = 2.14\,V \cr} $$