A volume of $500 mL$ of $0.1 M C u S O_{4}$ solution is electrolysed for $5 min$ at a current of $0.161 A$ . If $C u$ is produced at one electrode and oxygen at the other, the approximate pH of the final solution is:

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Solution

Cathode: $C u^{2 +} + 2 e^{-} \to C u$
Anode: $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$
Moles of electrons used $= \frac{Q}{F} = \frac{0.161 \times 5 \times 60}{96500} ≃ 5 \times 10^{- 4}$
Equivalents of $C u^{2 +}$ present = $\frac{500 \times 0.1}{1000} \times 2 = 0.1 (e x c e s s)$
Since, $C u^{2 +}$ is in excess, $H^{+}$ will not be consumed at cathode.
$∴ Moles of H^{+} produced=Moles of electrons = 5 \times 10^{- 4}$
$∴ {[H^{+}]}_{f i n a l} = \frac{5 \times 10^{- 4}}{500} \times 1000 = 10^{- 3} M$
$p H = - l o g [H^{+}]$
$\Rightarrow p H = 3.0$

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