Electrochemistry 1 Question 42
41. A current of $1.70 \mathrm{~A}$ is passed through $300.0 \mathrm{~mL}$ of $0.160 \mathrm{M}$ solution of a $\mathrm{ZnSO}_{4}$ for $230 \mathrm{~s}$ with a current efficiency of $90 %$. Find out the molarity of $\mathrm{Zn}^{2+}$ after the deposition $\mathrm{Zn}$. Assume the volume of the solution to remain constant during the electrolysis.
(1991, 4M)
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Answer:
Correct Answer: 41. $(0.154 \mathrm{M})$
Solution:
- Faraday’s passed $=\frac{1.7 \times 230}{96500}=4.052 \times 10^{-3} \mathrm{~F}$
Faradays used for reduction of $\mathrm{Zn}^{2+}=4.052 \times 10^{-3} \times 0.9$
$$ =3.65 \times 10^{-3} $$
$\Rightarrow$ Meq. of $\mathrm{Zn}^{2+}$ reduced $=3.65$
Initial meq. of $\mathrm{Zn}^{2+}=300 \times 0.16 \times 2=96$
$\Rightarrow$ Meq. of $\mathrm{Zn}^{2+}$ remaining $=96-3.65=92.35$
$\Rightarrow$ Molarity of $\mathrm{Zn}^{2+}=\frac{92.35}{2} \times \frac{1}{300}=0.154 \mathrm{M}$
