Question

The same amount of electricity was passed through two cells containing molten $ Al_{2}O_{3} $ and molten $ NaCl $ . If $1.8 \,g$ of $Al$ were liberated in one cell, the amount of $Na$ liberated in other cell is

• A. 2.8 g
• B. 3.2 g
• C. 4.6 g
• D. 6.8 g

Answer 1

Answer: (C)

Key Idea According to Faraday second law of electrolysis
$W \propto E $
$\frac{W_{1}}{W_{2}}=\frac{E_{1}}{E_{2}}$
Equivalent weight $=\frac{\text { atomic weight }}{\text { valency }}$
$W_{ Al }=1.8 g$
$E_{ Na }=23$
$E_{ N }=\frac{27}{3}=9$
$\because \frac{W_{ Na }}{W_{ N }}=\frac{E_{ Na }}{E_{ Al }}$
$ \therefore \frac{W_{ Na }}{1.8}=\frac{23}{9} $
$ W_{ Na } =4.6 \,g $