Which of the following represents the balanced chemical equation for the cell diagram: Al(s) | Al^{3+}(aq) || Cd^{2+}(aq) | Cd(s)?

• A. Al(s) + Cd^{2+}(aq) → Al^{3+}(aq) + Cd(s)
• B. Al(s) + 2Cd^{2+}(aq) → Al^{3+}(aq) + 2Cd(s)
• C. 3Al(s) + 2Cd^{2+}(aq) → 3Al^{3+}(aq) + 2Cd(s)
• D. 2Al(s) + 3Cd^{2+}(aq) → 2Al^{3+}(aq) + 3Cd(s)

Answer: (D)

Verified step by step guidance

1. Identify the oxidation states and half-reactions for each electrode in the cell diagram: Al(s) | Al^{3+}(aq) || Cd^{2+}(aq) | Cd(s). Aluminum (Al) is oxidized from 0 to +3, and cadmium (Cd) is reduced from +2 to 0.
2. Write the oxidation half-reaction: \(\mathrm{Al(s) \rightarrow Al^{3+}(aq) + 3e^-}\), and the reduction half-reaction: \(\mathrm{Cd^{2+}(aq) + 2e^- \rightarrow Cd(s)}\).
3. Balance the electrons lost in oxidation with the electrons gained in reduction by finding the least common multiple of electrons: 3 electrons from Al and 2 electrons from Cd. The least common multiple is 6.
4. Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3 to balance the electrons: \$2 \times (\mathrm{Al(s) \rightarrow Al^{3+}(aq) + 3e^-})\( and \)3 \times (\mathrm{Cd^{2+}(aq) + 2e^- \rightarrow Cd(s)})$.
5. Combine the balanced half-reactions to write the overall balanced chemical equation: \(\mathrm{2Al(s) + 3Cd^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Cd(s)}\).