Electrochemistry 1 Question 34
33. Copper sulphate solution $(250 \mathrm{mL})$ was electrolysed using a platinum anode and a copper cathode. A constant current of $2 \mathrm{mA}$ was passed for $16 \mathrm{min}$. It was found that after electrolysis the absorbance of the solution was reduced to $50 %$ of its original value. Calculate the concentration of copper sulphate in the solution to begin with.
(2000, 3M)
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Answer:
Correct Answer: 33. $8 \times 10^{-5} \mathrm{M}$
Solution:
- The number of Faraday’s passed $=\frac{2 \times 10^{-3} \times 16 \times 60}{96500}$
$ =1.99 \times 10^{-5} $
$\Rightarrow$ number of gram equivalent of $\mathrm{Cu}^{2+}$ deposited
$ =1.99 \times 10^{-5} $
$\Rightarrow$ number of moles of $\mathrm{Cu}^{2+}$ deposited $=\frac{1.99}{2} \times 10^{-5} \approx 10^{-5}$
Absorbance is directly proportional to $\left[\mathrm{Cu}^{2+}\right]$. Therefore, if ’ $C$ ’ be the initial molarity, $0.5 \mathrm{C}$ will be the final molarity.
$\Rightarrow \quad 0.5 \mathrm{C} \times 0.25=10^{-5} \Rightarrow C=8 \times 10^{-5} \mathrm{M}$
