Which of the following reaction occurs at anode during the electrolysis of copper sulphate solution using copper electrodes?

C u_{(s)} \to C u^{+ 2}_{(a q)} + 2 e^{-}

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C u^{+ 2} \to C u (s) - 2 e^{-}

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2 H^{+} + 2 e^{-} \Leftrightarrow H_{2}

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S O {_{4}}^{2 -} (a q) + 2 H^{+} \Leftrightarrow H_{2} S O_{4}

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Solution

The correct option is A $C u_{(s)} \to C u^{+ 2}_{(a q)} + 2 e^{-}$
During the electrolysis of copper sulphate,

At anode oxidation occurs, Cu acts as an active electrode and gets oxidize into $C u^{2 +}$ and moves into solution.

$C u (s) \to C u^{2 +} + 2 e^{-}$

So, the correct option is ( A )

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C u S O_{4}

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

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

C u S O_{4}

The oxidation and reduction half reactions of the reaction

2Al_(s) + 3Cu²⁺_(aq) → 2Al³⁺_(aq) + 3Cu_(s) are

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V

C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{0} = 0.34 V

A g^{+} (a q) + e^{-} \to A g (s); E^{0} = 0.80 V

2 H^{+} (a q) + 2 e^{-} \to H_{2} (g); E^{0} = 0 V

N i^{2 +} (a q) + 2 e^{-} \to N i (s); E^{0} = - 0.25 V

[C u^{2 +}, O H^{-}, S O_{4}^{2 -}, H^{+}]

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