Electrochemistry - Result Question 102
50. A current of $3.7 A$ is passed for $6 h$ between nickel electrodes in $0.5 L$ of a $2.0 M$ solution of $Ni\left(NO _3\right) _2$. What will be the molarity of solution at the end of electrolysis?
$(1978,2 M)$
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Answer:
Correct Answer: 50. ($1.172 $ $M$)
Solution:
- During electrolysis, $Ni^{2+}$ will be reduced at cathode and $H _2 O$ will be oxidised at anode.
Number of Faraday’s passed $=\frac{3.7 \times 6 \times 60 \times 60}{96500}=0.828$
$\Rightarrow 0.828$ g equivalent of $Ni^{2+}$ will be deposited at cathode.
Initial moles of $Ni^{2+}$ ion $=2 \times 0.5=1.0$
Moles of $Ni^{2+}$ ion remaining after electrolysis $=1.0-\frac{0.828}{2}$
$= 0.586 $
$\Rightarrow \text { Molarity of } Ni^{2+} \text { in final solution }=\frac{0.586}{0.50}=1.172 M$
BETA
