During the electrolysis of copper sulphate aqueous solution using copper electrode, the reaction taking place at the cathode is:

C u ⟶ {C u}_{(a q)}^{2 +} + 2 e^{-}

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{C u}_{(a q)}^{2 +} + 2 e^{-} ⟶ {C u}_{(s)}

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H_{(a q)}^{+} + e^{-} ⟶ \frac{1}{2} H_{2_{(g)}}

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{S O}_{4 (a q)}^{2 -} ⟶ {S O}_{3_{(g)}} + \frac{1}{2} O_{2_{(g)}} + 2 e^{-}

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Solution

The correct option is B ${C u}_{(a q)}^{2 +} + 2 e^{-} ⟶ {C u}_{(s)}$
During the electrolysis of copper sulphate aqueous solution using copper electrode, the reaction taking place at the cathode is

${C u^{2 +}}_{(a q)} + 2 e^{-} \to {C u}_{(s)}$
$C u (I I)$ ions having lower discharge potential than proton so will be liberated at the cathode in preference to protons.

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C u S O_{4}

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

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C u_{(s)} \to C u_{(a q)}^{2 +} + 2 e^{-}; E^{o} = - 0.325 V

C u_{(a q)}^{+} + e^{-} \to C u_{(s)}; E^{o} = 0.50 V

E^{o}

2 C u_{(a q)}^{+} \to C u_{(a q)}^{2 +} + C u_{(s)}

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