Given the redox reaction: Zn + Cu^{2+} → Zn^{2+} + Cu, which of the following correctly separates the reaction into its oxidation and reduction half-reactions?

• A. Oxidation: Zn → Zn^{2+} + 2e^{-}; Reduction: Cu^{2+} + 2e^{-} → Cu
• B. Oxidation: Cu^{2+} + 2e^{-} → Cu; Reduction: Zn → Zn^{2+} + 2e^{-}
• C. Oxidation: Cu → Cu^{2+} + 2e^{-}; Reduction: Zn^{2+} + 2e^{-} → Zn
• D. Oxidation: Zn^{2+} + 2e^{-} → Zn; Reduction: Cu → Cu^{2+} + 2e^{-}

Answer: (A)

Verified step by step guidance

1. Identify the species that undergo oxidation and reduction by examining the changes in oxidation states in the reaction: \(\mathrm{Zn} + \mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+} + \mathrm{Cu}\).
2. Determine the oxidation half-reaction by finding the species that loses electrons. Here, zinc (Zn) goes from an oxidation state of 0 to +2, indicating it loses 2 electrons. Write this as: \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^{-}\).
3. Determine the reduction half-reaction by finding the species that gains electrons. Copper ion (\(\mathrm{Cu}^{2+}\)) goes from +2 to 0, indicating it gains 2 electrons. Write this as: \(\mathrm{Cu}^{2+} + 2e^{-} \rightarrow \mathrm{Cu}\).
4. Verify that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction to ensure electron balance.
5. Combine the two half-reactions to confirm they sum to the original overall reaction, ensuring the half-reactions are correctly separated.