Consider the redox reaction
`2S_(2)O_(3)^(2-)+I_(2)rarrS_(4)O_(6)^(2-)+2I^(ө)`

Options

A. `S_(2)O_(3)^(2-)` gets reduced to `S_(4)O_(6)^(2-)`
B. `S_(2)O_(3)^(2-)` gets oxidised to `S_(4)O_(6)^(2-)`
C. `I_(2)` gets reduced to `I^(ө)`
D. `I_(2)` gets oxidised to `I^(ө)`

Explanation

To solve the redox reaction given by:

\[ 2S_{2}O_{3}^{2-} + I_{2} \rightarrow S_{4}O_{6}^{2-} + 2I^{-} \]

we will follow these steps:

### Step 1: Identify Oxidation and Reduction
In a redox reaction, one species gets oxidized (loses electrons) and another gets reduced (gains electrons).

- **Oxidation**: The species that loses electrons.
- **Reduction**: The species that gains electrons.

### Step 2: Write the Half-Reactions
We will write the half-reactions for both oxidation and reduction.

**Oxidation Half-Reaction:**
The thiosulfate ion \( S_{2}O_{3}^{2-} \) is oxidized to the tetrathionate ion \( S_{4}O_{6}^{2-} \).

\[ 2S_{2}O_{3}^{2-} \rightarrow S_{4}O_{6}^{2-} + 2e^{-} \]

**Reduction Half-Reaction:**
The iodine molecule \( I_{2} \) is reduced to iodide ions \( I^{-} \).

\[ I_{2} + 2e^{-} \rightarrow 2I^{-} \]

### Step 3: Combine the Half-Reactions
Now, we combine the two half-reactions to get the overall balanced redox reaction.

\[ 2S_{2}O_{3}^{2-} + I_{2} \rightarrow S_{4}O_{6}^{2-} + 2I^{-} \]

### Step 4: Identify the Oxidized and Reduced Species
From our half-reactions, we can conclude:
- \( S_{2}O_{3}^{2-} \) is oxidized to \( S_{4}O_{6}^{2-} \).
- \( I_{2} \) is reduced to \( I^{-} \).

### Conclusion
In this reaction:
- The thiosulfate ion \( S_{2}O_{3}^{2-} \) is the reducing agent as it is oxidized.
- The iodine molecule \( I_{2} \) is the oxidizing agent as it is reduced.

### Final Answer
The correct identification of the oxidation and reduction in the reaction is:
- \( S_{2}O_{3}^{2-} \) is oxidized to \( S_{4}O_{6}^{2-} \).
- \( I_{2} \) is reduced to \( I^{-} \).