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`
Among the following, identify the correct statement.
(a)Chloride ion is oxidized by `O_(2)`.
(b)`Fe^(2+)` is oxidized by iodine.
(c)Iodide ion is oxidized by chlorine
(d)`Mn^(2+)` is oxidized by chlorine.

Options

A. Chloride ion is oxidized by `O_(2)`.
B. `Fe^(2+)` is oxidized by iodine.
C. Iodide ion is oxidized by chlorine
D. `Mn^(2+)` is oxidized by chlorine.

Explanation

To solve the problem, we need to analyze the given half-cell reactions and their standard reduction potentials (E°) to determine which statements about oxidation and reduction are correct.

### Step-by-Step Solution:

1. **List the Given Half-Cell Reactions and Their E° Values:**
- \( I_2 + 2e^- \rightarrow 2I^- \), \( E° = 0.54 \, V \)
- \( Cl_2 + 2e^- \rightarrow 2Cl^- \), \( E° = 1.36 \, V \)
- \( Mn^{3+} + e^- \rightarrow Mn^{2+} \), \( E° = 1.50 \, V \)
- \( Fe^{3+} + e^- \rightarrow Fe^{2+} \), \( E° = 0.77 \, V \)
- \( O_2 + 4H^+ + 4e^- \rightarrow 2H_2O \), \( E° = 1.23 \, V \)

2. **Understand the Concept of Oxidation and Reduction:**
- A species with a higher reduction potential will act as a stronger oxidizing agent and can oxidize a species with a lower reduction potential.
- The reduction potential indicates the tendency of a species to gain electrons (be reduced).

3. **Analyze Each Statement:**
- **(a) Chloride ion is oxidized by \( O_2 \):**
- \( E°(Cl_2) = 1.36 \, V \) (for reduction to \( Cl^- \))
- \( E°(O_2) = 1.23 \, V \) (for reduction to \( H_2O \))
- Since \( E°(Cl_2) > E°(O_2) \), \( O_2 \) cannot oxidize \( Cl^- \). **This statement is incorrect.**

- **(b) \( Fe^{2+} \) is oxidized by iodine:**
- \( E°(I_2) = 0.54 \, V \)
- \( E°(Fe^{3+}) = 0.77 \, V \)
- Since \( E°(Fe^{3+}) > E°(I_2) \), \( I_2 \) cannot oxidize \( Fe^{2+} \). **This statement is incorrect.**

- **(c) Iodide ion is oxidized by chlorine:**
- \( E°(Cl_2) = 1.36 \, V \) (for reduction to \( Cl^- \))
- \( E°(I_2) = 0.54 \, V \)
- Since \( E°(Cl_2) > E°(I_2) \), \( Cl_2 \) can oxidize \( I^- \). **This statement is correct.**

- **(d) \( Mn^{2+} \) is oxidized by chlorine:**
- \( E°(Cl_2) = 1.36 \, V \)
- \( E°(Mn^{3+}) = 1.50 \, V \)
- Since \( E°(Mn^{3+}) > E°(Cl_2) \), \( Cl_2 \) cannot oxidize \( Mn^{2+} \). **This statement is incorrect.**

4. **Conclusion:**
- The only correct statement is **(c)**: Iodide ion is oxidized by chlorine.