Question

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 minutes. 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.

• A. $7.95\times {10}^{-5}M$
• B. $7.95\times {10}^{-4}M$
• C. $7.95\times {10}^{-3}M$
• D. $7.95\times {10}^{-2}M$

Answer 1

The correct option is A $7.95\times {10}^{-5}M$

$Totalno.offaradayspassed=\frac{2\times {10}^{-3}\times 16\times 60}{96500}=1.98\times {10}^{-5}$

Moles of $C{u}^{2+}deposited=\frac{1.98\times {10}^{-5}}{2}$

Since absorbance was reduced to 50% of its original value the initial moles of $C{u}^{2+}$would be two times the moles of $C{u}^{2+}$ reduced.

Therefore initial moles of $C{u}^{2+}$$=\frac{1.98\times {10}^{-5}}{2}\times 2$ $=1.98\times {10}^{-5}$.

The concentration of $CuS{O}_4$ in the solution $=\frac{1.98\times {10}^{-5}\times 1000}{250}=7.95\times {10}^{-5}M$