Given`:`
Oxidation`H_(2)O_(2) rarr O_(2)+2H^(o+)+2e^(-)" "E^(-)=-0.69V,`
`2F^(-)rarr F_(2)+2e^(-)" "E^(-)=-287V,`
Reduction `: H_(2)O_(2)+2H^(o+)+2e^(-) rarr 2H_(2)O" "E^(-)=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^(c-)`
B. `H_(2)O_(2)` behaves as a reductant for `1I_(2)`
C. `H_(2)O_(2)` behaves as an oxidant for `F^(-)`
D. `H_(2)O_(2)` behaves as a reductant for `F_(2)`.

Explanation

To solve the problem, we will analyze the given half-reactions and their standard reduction potentials (E°) to determine the correct statements regarding the behavior of H₂O₂ as an oxidant or reductant.

### Step-by-Step Solution:

1. **Identify the Half-Reactions and Their E° Values:**
- Oxidation of H₂O₂:
\[ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2e^- \quad E° = -0.69 \text{ V} \]
- Oxidation of F⁻:
\[ 2\text{F}^- \rightarrow \text{F}_2 + 2e^- \quad E° = -2.87 \text{ V} \]
- Reduction of H₂O₂:
\[ \text{H}_2\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow 2\text{H}_2\text{O} \quad E° = 1.77 \text{ V} \]
- Oxidation of I⁻:
\[ 2\text{I}^- \rightarrow \text{I}_2 + 2e^- \quad E° = -0.54 \text{ V} \]

2. **Analyze Each Statement:**
- **Statement A:** H₂O₂ behaves as an oxidant for I⁻.
- Since the reduction potential of H₂O₂ (1.77 V) is greater than that of I₂ (-0.54 V), H₂O₂ can oxidize I⁻.
- **Conclusion:** Statement A is **correct**.

- **Statement B:** H₂O₂ behaves as a reductant for I₂.
- For H₂O₂ to act as a reductant, it must have a lower oxidation potential than I₂. The oxidation potential of H₂O₂ is -0.69 V, while that of I₂ is -0.54 V.
- Since -0.69 V < -0.54 V, H₂O₂ cannot reduce I₂.
- **Conclusion:** Statement B is **incorrect**.

- **Statement C:** H₂O₂ behaves as an oxidant for F⁻.
- The reduction potential of H₂O₂ (1.77 V) is greater than that of F₂ (-2.87 V), indicating that H₂O₂ can oxidize F⁻.
- **Conclusion:** Statement C is **correct**.

- **Statement D:** H₂O₂ behaves as a reductant for F₂.
- For H₂O₂ to act as a reductant, its oxidation potential must be greater than that of F₂. However, the oxidation potential of H₂O₂ is -0.69 V, which is less than -2.87 V for F₂.
- **Conclusion:** Statement D is **incorrect**.

3. **Final Conclusion:**
- The correct statements are A and C.

### Summary of Correct Statements:
- **A:** H₂O₂ behaves as an oxidant for I⁻ (Correct)
- **B:** H₂O₂ behaves as a reductant for I₂ (Incorrect)
- **C:** H₂O₂ behaves as an oxidant for F⁻ (Correct)
- **D:** H₂O₂ behaves as a reductant for F₂ (Incorrect)

### Final Answer:
The correct statements are **A and C**.

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