The reduction half equation of the following reaction is:
$Zn_{(s)} + CuSO_{4(aq)} → ZnSO_{4(aq)} + Cu_{(s)}$

$Zn_{(s)} → Zn^{2+}_{(aq)} + 2e^-$

$CuSO_{4(s)} + H_2O_{(l)}$ → $Cu^{2+}_{(aq)}$ + $SO_{4(aq)}^{2-}$

$Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}$

$Cu^{2+}_{(aq)} + e^- → Cu_{(s)}$

Correct answer is C

The reduction half equation for the given reaction is:

$Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}$

In the overall reaction, copper(II) ions ( $Cu^{2+})$ gain two electrons (2e^-) and get reduced to form solid copper (Cu) metal. This is the reduction half-reaction in the redox reaction between zinc (Zn) and copper(II) sulfate ( $CuSO_4)$ .

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The reduction half equation of the following reaction is:
$Zn_{(s)} + CuSO_{4(aq)} → ZnSO_{4(aq)} + Cu_{(s)}$

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The reduction half equation of the following reaction is: Zn(s)+CuSO4(aq)→ZnSO4(aq)+Cu(s)Zn_{(s)} + CuSO_{4(aq)} → ZnSO_{4(aq)} + Cu_{(s)}

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The reduction half equation of the following reaction is:
$Zn_{(s)} + CuSO_{4(aq)} → ZnSO_{4(aq)} + Cu_{(s)}$