Question

Aluminium oxide may be electrolysed at $1000^{\circ} C$ to furnish aluminium metal
(Atomic mass $=27 u ; 1 F =96500 C$ ).
The cathode reaction is
$Al ^{3+}+3 e^{-} \longrightarrow Al ^{0}$
To prepare $5.12 \,kg$ of aluminium metal by this method would require

• A. $5.49 \times 10^{1}\, C$ of electricity
• B. $5.49 \times 10^{4}\, C$ of electricity
• C. $1.83 \times 10^{7}\, C$ of electricity
• D. $5.49 \times 10^{7}\, C$ of electricity

Answer 1

Answer: (D)

$Al ^{3+}+3 e^{-} \longrightarrow Al$
$w=z Q$
where $w=$ amount of metal
$=5.12\, kg$
$=5.12 \times 10^{3} g$
$z=$ electrochemical equivalent
$=\frac{\text { equivalent weight }}{96500}$
$=\frac{\text { atomic mass }}{\text { electrons } \times 96500}$
$=\frac{27}{3 \times 96500} $
$5.12 \times 10^{3} =\frac{27}{3 \times 96500} \times Q$
$Q =\frac{5.12 \times 10^{3} \times 3 \times 96500}{27} C$
$=5.49 \times 10^{7}\, C$