In this reaction: `S_(2)O_(8)^(2-)+2I^(-) to 2SO_(4)^(2-)+I_(2)`

Options

A. Oxidation of iodide into iodine takes place
B. Reduction of iodine into iodide takes place
C. Both oxidation and reduction of iodine takes place
D. All of these

Answer

To solve the question regarding the reaction \( S_2O_8^{2-} + 2I^{-} \rightarrow 2SO_4^{2-} + I_2 \), we need to analyze the oxidation and reduction processes involved.

### Step-by-Step Solution:

**Step 1: Identify the oxidation states of the elements involved.**
- In the reactants, the oxidation state of iodine in \( I^{-} \) is -1.
- In the products, the oxidation state of iodine in \( I_2 \) is 0.
- The oxidation state of sulfur in \( S_2O_8^{2-} \) is +7, and in \( SO_4^{2-} \), it is +6.

**Step 2: Determine the changes in oxidation states.**
- For iodine:
- The change from -1 (in \( I^{-} \)) to 0 (in \( I_2 \)) indicates that iodine is losing electrons.
- This is an oxidation process.

- For sulfur:
- The change from +7 (in \( S_2O_8^{2-} \)) to +6 (in \( SO_4^{2-} \)) indicates that sulfur is gaining electrons.
- This is a reduction process.

**Step 3: Conclude the type of reactions occurring.**
- Since iodine is losing electrons, it is being oxidized.
- Since sulfur is gaining electrons, it is being reduced.

**Step 4: Identify the correct answer.**
- The question asks about the oxidation and reduction processes taking place. The correct option is that oxidation of iodine into iodine takes place.

### Final Answer:
The correct option is: **Oxidation of iodine into iodine takes place.**