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Q. On the basis of the information available from the reaction.
$\frac{4}{3} Al + O _{2} \rightarrow \frac{2}{3} Al _{2} O _{3}, \Delta G =-827\, KJ\, mol ^{-1}$
of $O_2, $ the minimum EMF required to carry out the electrolysis of $Al_2O_3$ is
($F = 96500\, C\, mol^{-1})$

AIPMTAIPMT 2003Electrochemistry

Solution:

$\frac {4}{3}Al+O_2 \rightarrow \frac {2}{3}Al_2O_3,$
$ \Delta G=-827 \,kJ \, mol^{-1} $
$ \Delta G=-nEF^0 \, \, \, \, (n=4) $
$-827 \times 10^3 \, J=-4 \times E^0 \times 96500 $
$ E= \frac {827 \times 10^3}{4 \times 96500 } $
$ E^0=2.14 \, V $