Question

A current of 4.0 A is passed for 5 hours through 1L of 2M solution of nitrate using two nickel electrodes. The molarity of the solution at the end of the electrolysis will be:

• A. 1.5 M
• B. 1.2 M
• C. 2.5 M
• D. 2.0 M

Answer 1

Answer: (A)

1.5 M

$N{i}^{2+}+2e^{-}⟶Ni$

Faraday $=\frac{4\times 5\times 60\times 60}{96500}=0.746$

$\therefore$ Moles of $Ni$ reduced $=\frac{0.746}{2}=0.373$ $mol$

Moles of $N{i}^{2+}$ before reaction $=2\times 1=2$ $mol$

Left out $=2-0.373=1.626$

$\therefore Molarity=\frac{Moles}{Volume}=\frac{1.626}{1}=1.626$ $\frac{mol}{Lit.}$