A solution of
Question:

A solution of $\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}$ is electrolysed between platinum electrodes using $0.1$ Faraday electricity. How many mode of Ni will be deposited at the cathode?

1. $0.05$

2.  $0.20$

3.  $0.15$

4. $0.10$

Correct Option: 1

Solution:

According to the Faraday’s law of electrolysis, $\mathrm{nF}$ of current is required for the deposition of $1 \mathrm{~mol}$ According to the reaction,

$\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \longrightarrow \mathrm{Ni}^{2+}+2 \mathrm{NO}_{3}^{-}$

$2 \mathrm{~F}$ of current deposits $=1 \mathrm{~mol}$

$\therefore 0.1 \mathrm{~F}$ of current deposits $=\frac{0.1}{2}=0.05 \mathrm{~mol}$