Electrochemistry - Result Question 85
####33. Copper sulphate solution $(250 mL)$ was electrolysed using a platinum anode and a copper cathode. A constant current of $2 mA$ was passed for $16 min$. 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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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 $Cu^{2+}$ deposited
$$ =1.99 \times 10^{-5} $$
$\Rightarrow$ number of moles of $Cu^{2+}$ deposited $=\frac{1.99}{2} \times 10^{-5} \approx 10^{-5}$
Absorbance is directly proportional to $\left[Cu^{2+}\right]$. Therefore, if ’ $C$ ’ be the initial molarity, $0.5 C$ will be the final molarity.
$\Rightarrow \quad 0.5 C \times 0.25=10^{-5} \Rightarrow C=8 \times 10^{-5} M$
