In the electrolysis of aqueous solution of $C u S O_{4}$ using copper electrodes, the process taking place at the anode is:

S O_{4}^{2 -} \to S O 4 + 2 e^{-}

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C u \to C u^{+} + e^{-}

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2 O H^{-} \to H_{2} O + \frac{1}{2} O_{2} + 2 e^{-}

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C u \to C u^{2 +} + 2 e^{-}

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Solution

The correct option is C $2 O H^{-} \to H_{2} O + \frac{1}{2} O_{2} + 2 e^{-}$
When an aqueous solution of $C u S O_{4}$ is subjected to electrolysis using copper electrodes, the following reactions take place at cathode and anode.

$C u S O_{4} \to C u^{2 +} + S O_{4}^{2 -}$

Reduction at cathode:
$C u^{2 +} + 2 e^{-} \to C u$

Oxidation at anode:
$2 O H^{-} \to H_{2} O + \frac{1}{2} O_{2} + 2 e^{-}$ .

Hence, the correct answer is option $C$ .

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⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (c o r r e c t) C u (s) ⟶ & C u^{2 +} (a q) + 2 e^{-} E_{C u | C u^{2 +}}^{⊝} = - 0.34 V (w r o n g) 2 H_{2} O (l) ⟶ & O_{2} (g) + 4 H^{\oplus} (a q) + 4 e^{-} E_{H_{2} O | O_{2}}^{⊝} = - 1.23 V \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (c o r r e c t) N i (s) ⟶ & N i^{2 +} (a q) + 2 e^{-} E_{N i | N i^{2 +}}^{⊝} = - 0.25 V (w r o n g) 4 ⊝ O H (a q) ⟶ & O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} E_{⊝ O H | O_{2}}^{\oplus} = - 0.4 V \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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