Question

Aluminium oxide may be electrolysed at $\text{1000}{}^{\text{o}}\text{C}$ to furnish aluminium metal (atomic mass $=27u;1F=96,500C$ ).
The cathode reaction is $A{l}^{3+}+3e^{-}\to A{l}^0$
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)

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