Break down the following redox reaction into its half reactions: 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu.

• A. Oxidation: 3Cu²⁺ → 3Cu + 6e^-; Reduction: 2Al³⁺ + 6e^- → 2Al
• B. Oxidation: 2Al + 6e^- → 2Al³⁺; Reduction: 3Cu²⁺ → 3Cu + 6e^-
• C. Oxidation: 2Al → 2Al³⁺ + 6e^-; Reduction: 3Cu²⁺ + 6e^- → 3Cu
• D. Oxidation: 2Al³⁺ + 6e^- → 2Al; Reduction: 3Cu → 3Cu²⁺ + 6e^-

Answer: (C)