Electrochemistry 1 Question 51
50. A current of $3.7 \mathrm{~A}$ is passed for $6 \mathrm{~h}$ between nickel electrodes in $0.5 \mathrm{~L}$ of a $2.0 \mathrm{M}$ solution of $\mathrm{Ni}\left(\mathrm{NO}{3}\right){2}$. What will be the molarity of solution at the end of electrolysis?
$(1978,2 \mathrm{M})$
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Solution:
- During electrolysis, $\mathrm{Ni}^{2+}$ will be reduced at cathode and $\mathrm{H}_{2} \mathrm{O}$ will be oxidised at anode.
Number of Faraday’s passed $=\frac{3.7 \times 6 \times 60 \times 60}{96500}=0.828$
$\Rightarrow \quad 0.828$ g equivalent of $\mathrm{Ni}^{2+}$ will be deposited at cathode.
Initial moles of $\mathrm{Ni}^{2+}$ ion $=2 \times 0.5=1.0$
Moles of $\mathrm{Ni}^{2+}$ ion remaining after electrolysis $=1.0-\frac{0.828}{2}$
$$ \begin{gathered} =0.586 \ \Rightarrow \quad \text { Molarity of } \mathrm{Ni}^{2+} \text { in final solution }=\frac{0.586}{0.50}=1.172 \mathrm{M} \end{gathered} $$
