The standard reduction potentials of some half cell reactions are given below:
`PbO_2+4H^++2e^(-)hArrPb^(2+)+2H_(2)O , E^@=1.455 V`
`MnO_4^(-)+8H^++5e^(-)hArrMn^(2+)+4H_(2)O , E^@=1.51 V`
`Ce^(4+)+e^(-)hArrCe^(3+), E^@=1.61 V`
`H_2O_2+2H^++2e^(-)hArr 2H_2O, E^@=1.71 V`
Pick out the correct statement:

Options

A. `Ce^(+4)` will oxidise `Pb^(2+)` to `PbO_2`
B. `MnO_(4)^(-)` will oxidise `Pb^(2+)` to `PbO_2`
C. `H_2O_2` will oxidise `Mn^(+2)` to `MnO_4^(-)`
D. `PbO_2` will oxidise `Mn^(+2)` to `MnO_4^(-)`

Explanation

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

### Given Half-Cell Reactions and Their Standard Reduction Potentials:
1. \( \text{PbO}_2 + 4\text{H}^+ + 2e^- \rightarrow \text{Pb}^{2+} + 2\text{H}_2\text{O} \), \( E^\circ = 1.455 \, \text{V} \)
2. \( \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \), \( E^\circ = 1.51 \, \text{V} \)
3. \( \text{Ce}^{4+} + e^- \rightarrow \text{Ce}^{3+} \), \( E^\circ = 1.61 \, \text{V} \)
4. \( \text{H}_2\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow 2\text{H}_2\text{O} \), \( E^\circ = 1.71 \, \text{V} \)

### Step-by-Step Analysis of Each Statement:

**Statement 1:** \( \text{Ce}^{4+} \) will oxidize \( \text{Pb}^{2+} \) to \( \text{PbO}_2 \).

- **Analysis:**
- The standard reduction potential of \( \text{Ce}^{4+} \) is \( 1.61 \, \text{V} \) (higher).
- The standard reduction potential of \( \text{PbO}_2 \) is \( 1.455 \, \text{V} \) (lower).
- Since \( \text{Ce}^{4+} \) has a higher potential, it will be reduced, and \( \text{Pb}^{2+} \) will be oxidized to \( \text{PbO}_2 \).

- **Conclusion:** This statement is **correct**.

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**Statement 2:** \( \text{MnO}_4^- \) will oxidize \( \text{Pb}^{2+} \) to \( \text{PbO}_2 \).

- **Analysis:**
- The standard reduction potential of \( \text{MnO}_4^- \) is \( 1.51 \, \text{V} \) (higher).
- The standard reduction potential of \( \text{PbO}_2 \) is \( 1.455 \, \text{V} \) (lower).
- Since \( \text{MnO}_4^- \) has a higher potential, it will be reduced, and \( \text{Pb}^{2+} \) will be oxidized to \( \text{PbO}_2 \).

- **Conclusion:** This statement is **correct**.

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**Statement 3:** \( \text{H}_2\text{O}_2 \) will oxidize \( \text{Mn}^{2+} \) to \( \text{MnO}_4^- \).

- **Analysis:**
- The standard reduction potential of \( \text{H}_2\text{O}_2 \) is \( 1.71 \, \text{V} \) (higher).
- The standard reduction potential of \( \text{MnO}_4^- \) is \( 1.51 \, \text{V} \) (lower).
- Since \( \text{H}_2\text{O}_2 \) has a higher potential, it will be reduced, and \( \text{Mn}^{2+} \) will be oxidized to \( \text{MnO}_4^- \).

- **Conclusion:** This statement is **correct**.

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**Statement 4:** \( \text{PbO}_2 \) will oxidize \( \text{Mn}^{2+} \) to \( \text{MnO}_4^- \).

- **Analysis:**
- The standard reduction potential of \( \text{PbO}_2 \) is \( 1.455 \, \text{V} \) (lower).
- The standard reduction potential of \( \text{MnO}_4^- \) is \( 1.51 \, \text{V} \) (higher).
- Since \( \text{PbO}_2 \) has a lower potential, it cannot oxidize \( \text{Mn}^{2+} \) to \( \text{MnO}_4^- \).

- **Conclusion:** This statement is **incorrect**.

### Final Conclusion:
The correct statements are **1, 2, and 3**.

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