When `S_(2)O_(8)^(2-)` oxidise `Fe^(2+)` then product formed is :-

Options

A. `s_(2)O_(6)^(2-)`
B. `SO_(4)^(2-)`
C. `S_(2)O_(8)^(2-)`
D. `SO_(5)^(2-)`

Answer

To solve the question regarding the oxidation of \( S_2O_8^{2-} \) (peroxydisulfate ion) by \( Fe^{2+} \) (ferrous ion), we need to identify the products formed during this redox reaction.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are \( S_2O_8^{2-} \) and \( Fe^{2+} \).

2. **Determine the Role of Each Reactant**:
- \( S_2O_8^{2-} \) acts as an oxidizing agent because it can accept electrons.
- \( Fe^{2+} \) acts as a reducing agent because it can donate electrons.

3. **Write the Half-Reactions**:
- The oxidation half-reaction for \( Fe^{2+} \) is:
\[ Fe^{2+} \rightarrow Fe^{3+} + e^- \]
- The reduction half-reaction for \( S_2O_8^{2-} \) is:
\[ S_2O_8^{2-} + 2e^- \rightarrow 2SO_4^{2-} \]

4. **Balance the Half-Reactions**:
- The oxidation half-reaction shows that one \( Fe^{2+} \) ion loses one electron to become \( Fe^{3+} \).
- The reduction half-reaction shows that \( S_2O_8^{2-} \) gains two electrons to form two \( SO_4^{2-} \) ions.

5. **Combine the Half-Reactions**:
- Since two moles of \( Fe^{2+} \) are needed to provide the two electrons required for the reduction of one mole of \( S_2O_8^{2-} \), we can write:
\[ 2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-} \]

6. **Write the Final Balanced Equation**:
- The final balanced equation for the reaction is:
\[ 2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-} \]

### Conclusion:
The products formed when \( S_2O_8^{2-} \) oxidizes \( Fe^{2+} \) are \( Fe^{3+} \) and \( SO_4^{2-} \).