1. **State the problem:** Balance the redox reaction involving tin and chromium ions: $\text{Sn}^{2+} + 2 \text{Cr}^{2+} + \text{Sn}^{2+} \rightarrow 2 \text{Cr}^{3+} + \text{Sn}(s)$ 2. **Identify the species and their oxidation states:** - Tin ion $\text{Sn}^{2+}$ has oxidation state +2. - Chromium ion $\text{Cr}^{2+}$ has oxidation state +2. - Chromium ion $\text{Cr}^{3+}$ has oxidation state +3. - Tin solid $\text{Sn}(s)$ has oxidation state 0. 3. **Determine oxidation and reduction:** - $\text{Cr}^{2+}$ is oxidized to $\text{Cr}^{3+}$ (oxidation state increases from +2 to +3). - $\text{Sn}^{2+}$ is reduced to $\text{Sn}(s)$ (oxidation state decreases from +2 to 0). 4. **Write half-reactions:** - Oxidation: $\text{Cr}^{2+} \rightarrow \text{Cr}^{3+} + e^-$ - Reduction: $\text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}(s)$ 5. **Balance electrons:** - Multiply oxidation half-reaction by 2 to balance electrons: $$2 \text{Cr}^{2+} \rightarrow 2 \text{Cr}^{3+} + 2e^-$$ - Reduction half-reaction already has 2 electrons. 6. **Add half-reactions:** $$2 \text{Cr}^{2+} \rightarrow 2 \text{Cr}^{3+} + 2e^-$$ $$\text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}(s)$$ Sum: $$2 \text{Cr}^{2+} + \text{Sn}^{2+} \rightarrow 2 \text{Cr}^{3+} + \text{Sn}(s)$$ 7. **Final balanced equation:** $$2 \text{Cr}^{2+} + \text{Sn}^{2+} \rightarrow 2 \text{Cr}^{3+} + \text{Sn}(s)$$ This matches the given reaction except the original had an extra $\text{Sn}^{2+}$ on the reactant side which is redundant. **Answer:** The balanced redox reaction is: $$2 \text{Cr}^{2+} + \text{Sn}^{2+} \rightarrow 2 \text{Cr}^{3+} + \text{Sn}(s)$$