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Q. Aluminium oxide may be electrolysed at 1000 $^{\circ}$C to furnish aluminium metal (Atomic mass = 27 amu: 1 faraday = 96,500 Coulombs). The cathode reaction is
$Al^{3+} + 3e^- \to Al^{\circ}$
To prepare 5.12 kg of aluminium metal by this method would require :

AIEEEAIEEE 2005

Solution:

$Al^{3+} + 3e^- \to Al$
$w = zQ$
where w = amount of metal
$= 5.12\, kg = 5.12 \times 10^3\,g$
z = electrochemical equivalent
$=\frac{Equivalent \,weight}{96500} = \frac{Atomic \,mass}{Electrons \times 96500}$
$= \frac{27}{3\times96500}\times Q$
$5.12\times10^{3} = \frac{27}{3\times 96500}\times Q$
$Q = \frac{5.12\times10^{3}\times3\times96500}{27}C$
$= 5.49 \times 10^{7} \,C$