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 of what products are formed when \( S_2O_8^{2-} \) oxidizes \( Fe^{2+} \), we can follow these steps:

### Step 1: Identify the reactants and their oxidation states
- The reactants are \( S_2O_8^{2-} \) (persulfate ion) and \( Fe^{2+} \) (ferrous ion).
- In \( Fe^{2+} \), the oxidation state of iron is +2.

### Step 2: Determine the oxidation state of sulfur in \( S_2O_8^{2-} \)
- The formula for the persulfate ion is \( S_2O_8^{2-} \).
- Each oxygen has an oxidation state of -2.
- There are 8 oxygen atoms, contributing a total of \( 8 \times (-2) = -16 \).
- Let the oxidation state of sulfur be \( x \). Since there are 2 sulfur atoms, the equation becomes:
\[ 2x - 16 = -2 \]
- Solving for \( x \):
\[ 2x = 14 \implies x = +7 \]
- Therefore, the oxidation state of sulfur in \( S_2O_8^{2-} \) is +7.

### Step 3: Determine the oxidation state of sulfur in the product \( SO_4^{2-} \)
- In the sulfate ion \( SO_4^{2-} \):
- Let the oxidation state of sulfur be \( y \). The equation is:
\[ y + 4(-2) = -2 \]
- Solving for \( y \):
\[ y - 8 = -2 \implies y = +6 \]
- Thus, the oxidation state of sulfur in \( SO_4^{2-} \) is +6.

### Step 4: Identify the changes in oxidation states
- The oxidation state of sulfur decreases from +7 in \( S_2O_8^{2-} \) to +6 in \( SO_4^{2-} \), indicating that sulfur is reduced.
- The iron \( Fe^{2+} \) is oxidized to \( Fe^{3+} \) as its oxidation state increases from +2 to +3.

### Step 5: Write the balanced equation for the reaction
- The balanced redox reaction can be represented as:
\[ S_2O_8^{2-} + 2Fe^{2+} \rightarrow 2Fe^{3+} + 2SO_4^{2-} \]

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

### Final Answer
The product formed is \( SO_4^{2-} \).

---