Question

A direct current deposits $54g$ of silver during electrolysis. How much amount of aluminium would be deposited from aluminium chloride solution by the same amount of electricity?
( $\text{Atomic mass of Ag}=108g/mol$)
($\text{Atomic mass of Al}=27g/mol$)

• A. $4.5g$
• B. $5.4g$
• C. $54g$
• D. $2.7g$

Answer 1

The correct option is A $4.5g$

Faradays second law of electrolysis
If equal amount of charge (Q) is passed through two different solutions, the amount of substance deposited/liberated (W) is proportional to their chemical equivalent weights (E).

Thus,
$\frac{W_1}{W_2}=\frac{E_1}{E_2}=\frac{Z_1}{Z_2}$

$\text{Equivalent mass of silver}=108gequi{v}^{-1}$
$\text{Equivalent mass of aluminium in}AlC{l}_3=\frac{27}{3}=9gequi{v}^{-1}$

$\frac{W_{Ag}}{W_{Al}}=\frac{E_{Ag}}{E_{Al}}$

$\frac{54}{W_{Al}}=\frac{108}{9}$

$W_{Al}=4.5g$