Which of the following correctly balances the chemical equation for the reaction between Fe^{3+}(aq) and Sn(s) to produce Fe(s) and Sn^{2+}(aq)?

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

Answer: (C)

Verified step by step guidance

1. Identify the oxidation states of the elements involved in the reaction. Fe^{3+} is reduced to Fe(s), and Sn(s) is oxidized to Sn^{2+}.
2. Write the half-reactions for oxidation and reduction separately: Reduction: \(\mathrm{Fe^{3+} + 3e^- \rightarrow Fe}\) Oxidation: \(\mathrm{Sn \rightarrow Sn^{2+} + 2e^-}\)
3. Balance the electrons lost and gained by finding the least common multiple of electrons in both half-reactions. Here, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 3 to balance electrons:
4. \$2 \times (\mathrm{Fe^{3+} + 3e^- \rightarrow Fe})\( and \)3 \times (\mathrm{Sn \rightarrow Sn^{2+} + 2e^-})$
5. Combine the balanced half-reactions to write the overall balanced equation: \(\mathrm{2Fe^{3+} + 3Sn \rightarrow 2Fe + 3Sn^{2+}}\).