Given`:`
Oxidation`H_(2)O_(2) rarr O_(2)+2H^(o+)+2e^(-)" "E^(c-)=-0.69V,`
`2F^(c-)rarr F_(2)+2e^(-)" "E^(c-)=-287V,`
Reduction `: H_(2)O_(2)+2H^(o+)+2e^(-) rarr 2H_(2)O" "E^(c-)=1.77V,`
2I^(c-)rarr I_(2)+2e^(-)" "E^(c-)=-0.54V,`
Which of the following statements is `//` are correct ?

Options

A. `H_(2)O_(2)` behaves as an oxidant for `I_(2)//I^(c-)`
B. `H_(2)O_(2)` behaves as an reductant for `I_(2)//I^(c-)`
C. `I^(c-)//I_(2)` behaves as an reductant for `H_(2)O_(2)`
D. None of these is correct

Explanation

To solve the given electrochemistry problem, we need to analyze the provided half-reactions and their standard electrode potentials (E°). We will determine which reactions are spontaneous and which statements regarding these reactions are correct.

### Step-by-Step Solution:

1. **Identify the Half-Reactions and Their Potentials:**
- Oxidation:
\[ H_2O_2 \rightarrow O_2 + 2H^+ + 2e^- \quad E^\circ = -0.69 \, V \]
- Reduction (for H2O2):
\[ H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O \quad E^\circ = 1.77 \, V \]
- Oxidation of Fluorine:
\[ 2F^- \rightarrow F_2 + 2e^- \quad E^\circ = -287 \, V \]
- Reduction of Iodine:
\[ 2I^- \rightarrow I_2 + 2e^- \quad E^\circ = -0.54 \, V \]

2. **Determine the Spontaneity of Each Reaction:**
- A reaction is spontaneous if the overall cell potential (E°cell) is positive.
- For the oxidation of \(H_2O_2\), the potential is negative, indicating it is not spontaneous.
- For the reduction of \(H_2O_2\), the potential is positive, indicating it is spontaneous.
- The oxidation of \(F^-\) is highly negative, indicating it is not spontaneous.
- The reduction of \(I^-\) is also negative, indicating it is not spontaneous.

3. **Calculate the Overall Cell Potentials:**
- For the reaction involving \(H_2O_2\):
\[ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} = 1.77 \, V - (-0.69 \, V) = 1.77 \, V + 0.69 \, V = 2.46 \, V \]
- This indicates that the reaction involving \(H_2O_2\) is spontaneous.

4. **Evaluate the Other Reactions:**
- For fluorine and iodine, since both reactions have negative potentials, they do not contribute to a spontaneous reaction.

5. **Conclusion:**
- The only spontaneous reaction is the reduction of \(H_2O_2\) to \(H_2O\).
- Therefore, any statement claiming that the oxidation of \(H_2O_2\) or the oxidation of \(F^-\) is correct would be false.

### Final Answer:
The correct statement is that the reduction of \(H_2O_2\) is spontaneous, while the oxidation of \(H_2O_2\) and \(F^-\) is not spontaneous.