Electrochemistry - Result Question 93
####41. A current of $1.70 A$ is passed through $300.0 mL$ of $0.160 M$ solution of a $ZnSO _4$ for $230 s$ with a current efficiency of $90 %$. Find out the molarity of $Zn^{2+}$ after the deposition $Zn$. Assume the volume of the solution to remain constant during the electrolysis.
(1991, 4M)
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Answer:
Correct Answer: 41. $(0.154 M)$
Solution:
- Faraday’s passed $=\frac{1.7 \times 230}{96500}=4.052 \times 10^{-3} F$
Faradays used for reduction of $Zn^{2+}=4.052 \times 10^{-3} \times 0.9$
$$ =3.65 \times 10^{-3} $$
$\Rightarrow$ Meq. of $Zn^{2+}$ reduced $=3.65$
Initial meq. of $Zn^{2+}=300 \times 0.16 \times 2=96$
$\Rightarrow$ Meq. of $Zn^{2+}$ remaining $=96-3.65=92.35$
$\Rightarrow$ Molarity of $Zn^{2+}=\frac{92.35}{2} \times \frac{1}{300}=0.154 M$
