If `{:(Sn^(2+) + 2e^(-) rarr Sn(s), E^(o) = - 0.14 V),(Sn^(4+)+2e^(-) rarr Sn^(2+),E^(o)= + 0.13 V):}` then which of these is true?

Options

A. `Sn^(2+)` is unstable and disproportionates to form `Sn^(4+)` and Sn
B. `Sn^(2+)` is stable and disproportionation reaction is not spontaneous
C. `Sn^(4+)` is easily reduced to Sn
D. None of these is correct

Answer

To solve the problem, we need to analyze the given half-reactions and their standard electrode potentials (E° values) to determine the stability of the species involved and the spontaneity of the reactions.

### Step-by-Step Solution:

1. **Identify the Half-Reactions and Their E° Values**:
- The first half-reaction is:
\[ \text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}(s), \quad E^\circ = -0.14 \, \text{V} \]
- The second half-reaction is:
\[ \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+}, \quad E^\circ = +0.13 \, \text{V} \]

2. **Analyze the Stability of Sn²⁺**:
- The negative E° value for the first half-reaction indicates that the reduction of Sn²⁺ to Sn(s) is not favorable. Therefore, Sn²⁺ is less stable than Sn(s).
- The positive E° value for the second half-reaction indicates that the reduction of Sn⁴⁺ to Sn²⁺ is favorable, suggesting that Sn²⁺ is more stable than Sn⁴⁺.

3. **Examine the Disproportionation of Sn²⁺**:
- Disproportionation involves a species undergoing both oxidation and reduction. For Sn²⁺ to disproportionate, it would convert to Sn(s) and Sn⁴⁺:
\[ 2 \text{Sn}^{2+} \rightarrow \text{Sn}(s) + \text{Sn}^{4+} \]
- Since Sn²⁺ is more stable than both Sn(s) and Sn⁴⁺, this disproportionation reaction is not spontaneous.

4. **Evaluate the Spontaneity of the Reactions**:
- The spontaneity of a reaction can be determined by the sign of the overall E° value. If E° is positive, the reaction is spontaneous; if negative, it is non-spontaneous.
- For the reaction of Sn⁴⁺ being reduced to Sn, we can combine the two half-reactions:
\[ \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+} \quad (E^\circ = +0.13 \, \text{V}) \]
\[ \text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}(s) \quad (E^\circ = -0.14 \, \text{V}) \]
- Adding these reactions gives:
\[ \text{Sn}^{4+} + 2e^- + \text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}^{2+} + \text{Sn}(s) \]
- The overall E° value is:
\[ E^\circ_{\text{overall}} = +0.13 \, \text{V} - 0.14 \, \text{V} = -0.01 \, \text{V} \]
- Since this value is negative, the reduction of Sn⁴⁺ to Sn is also non-spontaneous.

5. **Conclusion**:
- Based on the analysis, the correct statement is that Sn²⁺ is stable and does not undergo spontaneous disproportionation. Therefore, the correct option is that Sn²⁺ is stable and the disproportionation reaction is not spontaneous.

### Final Answer:
The correct statement is: **Sn²⁺ is stable and the disproportionation reaction is not spontaneous.**