Question

Molten $AlC{l}_3$ is electrolysed using platinum electrode with the current of $0.5$ ampere to produced $27g\text{of}Al$. Find the number of faradays used for deposition of $Al$.

• A. 3
• B. 5
• C. 6
• D. 4

Answer 1

The correct option is A 3

Electrolysis of molten $AlC{l}_3$ using platinum electrode.
At cathode,
$A{l}^{3+}\left(l\right)+3e^{-}\to Al\left(s\right)$
Thus, 3 moles of electron is required to produce 1 mole of $Al$

Given :
$\text{Current, I}=0.5A$
$\text{Mass deposited}=27g$

From Faraday's law of electrolysis, we know that
$W=\frac{EQ}{96500}$

$\text{Equivalent weight, E}=\frac{\text{Molar Mass}}{n-factor}$
$\text{Equivalent weight, E}=\frac{\text{27}}{3}$
$\text{Equivalent weight, E}=9$

$\frac{Q}{96500}=\frac{W}{E}$

$\frac{Q}{96500}=\frac{27}{9}Q=3F$

Hence, option A is the right answer.