Question

A current of $96.5\, A$ is passed for $18\, \min$ between nickel electrodes in $500\, mL$ solution of $2\, M\, Ni \left( NO _{3}\right)_{2}$. The molarity of solution after electrolysis would be

• A. 0.46 M
• B. 0.92 M
• C. 0.625 M
• D. 1.25 M

Answer 1

Answer: (B)

Moles of $Ni \left( NO _{3}\right)_{2}$ in $500\, mL$ of $2M\, Ni \left( NO _{3}\right)_{2}$
is
$=\frac{2 \times 500}{1000}=1\, mol$
Charge $=96.5 \times 18 \times 60=104220\, C$
$Ni ^{2+}+2 e^{-} \longrightarrow Ni$
$2 \times 96500\, C$ deposits $1$ mole of $Ni \left( NO _{3}\right)_{2}$
$\therefore 104220\, C$ will deposite
$=\frac{104220}{2 \times 96500}=0.54\, mol$
So, moles of $Ni \left( NO _{3}\right)_{2}$ left
$=1.0-0.54=0.46\, mol$
Thus, molarity of $Ni \left( NO _{3}\right)_{2}$ solution
$=2 \times 0.46=0.92\, mol / L$