Question

A volume of $500\text{mL}$ of $0.1MCuS{O}_4$ solution is electrolysed for $5\text{min}$ at a current of $0.161A$. If $Cu$ is produced at one electrode and oxygen at the other, the approximate pH of the final solution is:

• A. 3.0
• B. 3
• C. 3.00

Answer 1

Answer: (A), (B), (C)

Cathode: $C{u}^{2+}+2e^{-}\to Cu$
Anode: $2H_{2}O\to O_2+4H^{+}+4e^{-}$
Moles of electrons used$=\frac{Q}{F}=\frac{0.161\times 5\times 60}{96500}\simeq 5\times {10}^{-4}$
Equivalents of $C{u}^{2+}$ present =$\frac{500\times 0.1}{1000}\times 2=0.1\left(excess\right)$
Since, $C{u}^{2+}$ is in excess, $H^{+}$ will not be consumed at cathode.
$\therefore \text{Moles of}{H}^{+}\text{produced=Moles of electrons}=5\times {10}^{-4}$
$\therefore {\left[H^{+}\right]}_{final}=\frac{5\times {10}^{-4}}{500}\times 1000={10}^{-3}M$
$pH=-log\left[H^{+}\right]$
$\Rightarrow pH=3.0$