Question

Copper sulphate solution $\left(250mL\right)$ was electrolysed using a platinum anode and a copper cathode. A constant current of $2mA$ was passed for $16$ minutes. If 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 being with.

Answer 1

Number of faraday passed through the aqueous solution of $CuS{O}_4$
$=\frac{I\times t}{96500}=\frac{(2\times {10}^{-3})\times 16\times 60}{96500}=1.98\times {10}^{-5}$.
Thus, number of equivalents of $CuS{O}_4$ involved electrolysis from $1$ litre solution
$=4\times 1.98\times {10}^{-5}=7.92\times {10}^{-5}$
Since, after electrolysis, the absorbance reduces to $50$% hence initial number of equivalent of $CuS{O}_4$ per litre
$=2\times 7.92\times {10}^{-5}$
$=1.584\times {10}^{-4}$
or Normality of $CuS{O}_4$ solution $=1.584\times {10}^{-4}N$.