Question

Aluminium oxide may be electrolysed at $1000^{\circ }C$ to furnish aluminium metal
(Atomic mass $=27u;1F=96500C$ ).
The cathode reaction is
$A{l}^{3+}+3e^{-}⟶A{l}^0$
To prepare $5.12kg$ 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)

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