Electrochemistry - Result Question 47
####49. On the basis of the information available from the reaction
$\frac{4}{3} Al+O_2 \to \frac{2}{3} Al_2 O_3, \Delta G=-827 kJ mol^{-1}$ of $O_2$. The minimum e.m.f required to carry out an electrolysis of $Al_2 O_3$ is $(F=96500 C mol^{-1})$
(a) $8.56 V$
(b) $2.14 V$
(c) $4.28 V$
(d) $6.42 V$
[2003]
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Solution:
- (b) $\Delta G=-n E F, \quad A l \to A l^{3+}+3 e^{-}$
For $1 mol$ of $Al, n=3$
$\therefore$ for $\frac{4}{3} mol$ of $A l, n=3 \times \frac{4}{3}=4$
According to question,
$827 \times 1000=4 \times E \times 96500$
$E=\frac{827 \times 1000}{4 \times 96500}=2.14 V$
