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Which correctly identifies oxidation and reduction in the following reaction?
\(\mathrm{Zn + CuSO_4 \rightarrow ZnSO_4 + Cu}\)

Answer and explanation

Correct answer: Zinc is oxidised and \(\mathrm{Cu^{2+}}\) is reduced

The oxidation number of zinc changes from \(0\) to \(+2\), so it loses two electrons and is oxidised: \(\mathrm{Zn \rightarrow Zn^{2+}+2e^-}\). The oxidation number of \(\mathrm{Cu^{2+}}\) changes from \(+2\) to \(0\); it gains electrons and is reduced: \(\mathrm{Cu^{2+}+2e^- \rightarrow Cu}\). Therefore, option A is correct. Option B reverses both changes.

Related tags

Redox ReactionsOxidationReductionOxidation NumberChemical EquationsClass 10 Science

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