The reduction half equation of the following reaction is:
\(Zn_{(s)} + CuSO_{4(aq)} → ZnSO_{4(aq)} + Cu_{(s)}\)
A.
\(Zn_{(s)} → Zn^{2+}_{(aq)} + 2e^-\)
B.
\(CuSO_{4(s)} + H_2O_{(l)}\) → \(Cu^{2+}_{(aq)}\) + \(SO_{4(aq)}^{2-}\)
C.
\(Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}\)
D.
\(Cu^{2+}_{(aq)} + e^- → Cu_{(s)}\)
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 (\(CuSO_4)\).
