Peroxodisulphate salts, `(e.g., Na_(2)S_(2)O_(8))` are strong oxidizing agents used be bleaching agents for fats, oils, ets.
Given `:`
`O_(2)(g)+4H^(o+)(aq)+4e^(-)rarr 2H_(2)O(l)" "E^(c-)=1.23V`
`S_(2)O_(8)^(2-)(aq)+2e^(-)rarr 2SO_(4)^(2-)(aq)" "E^(C-)=2.01V`
Which of the following statement is `(` are `)` correct ?

Options

A. Oxygen gas can oxidize sulphate ion to per`-` oxo disulphate ion `(S_(2)O_(8)^(2-))` in acidic solution.
B. `O_(2)(g)` is reduced to water.
C. Water is oxidized to `O_(2)`
D. `S_(2)O_(8)^(2-)` ions are reduced to `SO_(4)^(2-)` ions.

Explanation

To solve the question regarding the peroxodisulphate salts and their oxidizing properties, we will analyze the given half-reactions and their standard electrode potentials (E°) to determine which statements are correct.

### Step-by-Step Solution:

1. **Identify the Half-Reactions and Their Potentials**:
- The first half-reaction is:
\[ O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l) \quad E° = 1.23 \, V \]
- The second half-reaction is:
\[ S_2O_8^{2-}(aq) + 2e^- \rightarrow 2SO_4^{2-}(aq) \quad E° = 2.01 \, V \]

2. **Determine the Oxidation and Reduction Processes**:
- In the first half-reaction, oxygen is reduced to water.
- In the second half-reaction, peroxodisulphate ions are reduced to sulphate ions.

3. **Calculate the Standard Cell Potential (E°cell)**:
- The overall cell reaction can be determined by combining the two half-reactions. To balance the electrons, we need to multiply the first reaction by 2:
\[ 2(O_2(g) + 4H^+ + 4e^- \rightarrow 2H_2O) \quad E° = 1.23 \, V \]
- The second half-reaction remains as is:
\[ S_2O_8^{2-} + 2e^- \rightarrow 2SO_4^{2-} \quad E° = 2.01 \, V \]
- The overall reaction becomes:
\[ 2O_2(g) + 8H^+ + 4SO_4^{2-} \rightarrow 4H_2O + 2S_2O_8^{2-} \]

4. **Calculate the E°cell**:
- The standard cell potential is calculated as:
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]
- Here, the cathode reaction (reduction) is the one with oxygen:
\[ E°_{cathode} = 1.23 \, V \]
- The anode reaction (oxidation) is the one with peroxodisulphate:
\[ E°_{anode} = 2.01 \, V \]
- Therefore:
\[ E°_{cell} = 1.23 \, V - 2.01 \, V = -0.78 \, V \]

5. **Interpret the Result**:
- Since \( E°_{cell} < 0 \), the reaction is non-spontaneous under standard conditions.

6. **Evaluate the Statements**:
- **Statement 1**: "Oxygen gas can oxidize sulphate ion to peroxodisulphate ion in acidic solution." - **Incorrect** (as the reaction is not spontaneous).
- **Statement 2**: "Oxygen is reduced to water." - **Correct** (this is the cathode reaction).
- **Statement 3**: "Water is oxidized to oxygen." - **Incorrect** (the reverse reaction is not spontaneous).
- **Statement 4**: "S2O8^{2-} ions are reduced to SO4^{2-} ions." - **Correct** (this is the anode reaction).

### Conclusion:
The correct statements are **2 and 4**.