Redox reactions play a pivotal role in chemistry and biology. The values standard redox potential `(E^(c-))` of two half cell reactions decided which way the reaction is expected to preceed. A simple example is a Daniell cell in which zinc goes into solution and copper sets deposited. Given below are a set of half cell reactions `(` acidic medium `)` along with their `E^(c-)(V` with respect to normal hydrogen electrode `)` values. Using this data, obtain correct explanations for Question.
`I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.54`
`Cl_(2)+2e^(-) rarr 2Cl^(c-), " "E^(c-)=1.36`
`Mn^(3+)+e^(-) rarr Mn^(2+), " "E^(c-)=1.50`
`Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.77`
`O_(2)+4H^(o+)+4e^(-) rarr 2H_(2)O, " "E^(c-)=1.23`
While `Fe^(3+)` is stable, `Mn^(3+)` is not stable in acid solution because

Options

A. `O_(2)` oxidizes `Mn^(2+)` to `Mn^(3+)`
B. `O_(2)` oxidizes both `Mn^(2+)` to `Mn^(3+)` and `Fe^(2+)` to `Fe^(3+)`
C. `Fe^(3+)` oxidizes `H_(2)O` to `O_(2)`
D. `Mn^(3+)` oxidized `H_(2)O` to `O_(2)`

Explanation

To determine why \( \text{Mn}^{3+} \) is not stable in acidic solution while \( \text{Fe}^{3+} \) is stable, we can analyze the standard reduction potentials given for the half-cell reactions. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Standard Reduction Potentials
The standard reduction potential \( E^\circ \) indicates the tendency of a species to gain electrons and be reduced. A higher \( E^\circ \) value means a greater tendency to be reduced.

### Step 2: List the Relevant Half-Reactions and Their Potentials
From the problem, we have the following half-cell reactions and their corresponding \( E^\circ \) values:

1. \( \text{I}_2 + 2e^- \rightarrow 2\text{I}^- \), \( E^\circ = 0.54 \, \text{V} \)
2. \( \text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \), \( E^\circ = 1.36 \, \text{V} \)
3. \( \text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+} \), \( E^\circ = 1.50 \, \text{V} \)
4. \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \), \( E^\circ = 0.77 \, \text{V} \)
5. \( \text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O} \), \( E^\circ = 1.23 \, \text{V} \)

### Step 3: Compare the Stability of \( \text{Mn}^{3+} \) and \( \text{Fe}^{3+} \)
- **Stability of \( \text{Fe}^{3+} \)**:
- The reduction potential for the half-reaction \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) is \( 0.77 \, \text{V} \). This indicates that \( \text{Fe}^{3+} \) can be reduced to \( \text{Fe}^{2+} \) but does not readily oxidize back to \( \text{Fe}^{2+} \) in acidic conditions, thus it remains stable.

- **Instability of \( \text{Mn}^{3+} \)**:
- The reduction potential for \( \text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+} \) is \( 1.50 \, \text{V} \). This is higher than the potential of \( \text{O}_2 \) (1.23 V) and indicates that \( \text{Mn}^{3+} \) can easily be reduced to \( \text{Mn}^{2+} \) in the presence of a stronger oxidizing agent like \( \text{O}_2 \). Thus, \( \text{Mn}^{3+} \) is not stable as it will tend to reduce back to \( \text{Mn}^{2+} \).

### Conclusion
Therefore, \( \text{Mn}^{3+} \) is not stable in acidic solution because it can be readily reduced to \( \text{Mn}^{2+} \) by \( \text{O}_2 \) or other reducing agents, while \( \text{Fe}^{3+} \) remains stable due to its lower tendency to be reduced back to \( \text{Fe}^{2+} \).