`Sn^(2+)` and `Fe^(3+)` cannot exist in the same solution.

Explanation

To solve the question of why \( \text{Sn}^{2+} \) and \( \text{Fe}^{3+} \) cannot exist in the same solution, we will analyze the redox reaction that occurs between these two ions.

### Step-by-Step Solution:

1. **Identify the Ions**:
We have two ions in question: \( \text{Sn}^{2+} \) (tin ion) and \( \text{Fe}^{3+} \) (iron ion).

2. **Determine Possible Reactions**:
When \( \text{Sn}^{2+} \) and \( \text{Fe}^{3+} \) are present in the same solution, they can undergo a redox reaction.

3. **Write the Half-Reactions**:
- **Oxidation**: \( \text{Sn}^{2+} \) can be oxidized to \( \text{Sn}^{4+} \):
\[ \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \]
- **Reduction**: \( \text{Fe}^{3+} \) can be reduced to \( \text{Fe}^{2+} \):
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \]

4. **Combine the Half-Reactions**:
To balance the electrons transferred, we can multiply the reduction half-reaction by 2:
\[ 2\text{Fe}^{3+} + 2e^- \rightarrow 2\text{Fe}^{2+} \]
Now, we can combine the oxidation and reduction reactions:
\[ \text{Sn}^{2+} + 2\text{Fe}^{3+} \rightarrow \text{Sn}^{4+} + 2\text{Fe}^{2+} \]

5. **Conclusion**:
In this reaction, \( \text{Sn}^{2+} \) is oxidized to \( \text{Sn}^{4+} \) (loss of electrons), and \( \text{Fe}^{3+} \) is reduced to \( \text{Fe}^{2+} \) (gain of electrons). Since a redox reaction occurs between \( \text{Sn}^{2+} \) and \( \text{Fe}^{3+} \), they cannot coexist in the same solution without reacting.

### Final Statement:
Thus, the statement that \( \text{Sn}^{2+} \) and \( \text{Fe}^{3+} \) cannot exist in the same solution is true due to the occurrence of a redox reaction between them.

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