Question

Molten $AlC{l}_3$ is electrolysed with a current of 0.5 ampere to produce 27 g $Al.$ The number of electrons are used are $x{N}_A$. The value of $x$ is :

Answer 1

$A{l}^{3+}+3e^{-}\to Al$

From Faraday's law of electrolysis, we know that

$w=\frac{EQ}{96500}$.

$E=27/3=9$.

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

One Farad is the charge on 1 mole of electrons. Hence $3N_A$ electrons are used.

So, the value of $x$ is $3$