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 analyze the redox reaction given by:

\[ 2 \text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2 \text{I}^{-} \]

we will follow these steps:

### Step 1: Identify the oxidation states
First, we need to determine the oxidation states of the elements involved in the reaction.

- In \( \text{S}_2\text{O}_3^{2-} \), sulfur (S) has an oxidation state of +2.
- In \( \text{S}_4\text{O}_6^{2-} \), sulfur (S) has an oxidation state of +4.
- In \( \text{I}_2 \), iodine (I) has an oxidation state of 0.
- In \( \text{I}^{-} \), iodine (I) has an oxidation state of -1.

### Step 2: Determine oxidation and reduction
Now, we can identify which species is oxidized and which is reduced:

- **Oxidation**: The oxidation state of sulfur increases from +2 in \( \text{S}_2\text{O}_3^{2-} \) to +4 in \( \text{S}_4\text{O}_6^{2-} \). Therefore, \( \text{S}_2\text{O}_3^{2-} \) is oxidized.
- **Reduction**: The oxidation state of iodine decreases from 0 in \( \text{I}_2 \) to -1 in \( \text{I}^{-} \). Therefore, \( \text{I}_2 \) is reduced.

### Step 3: Write half-reactions
Next, we can write the half-reactions for oxidation and reduction:

- **Oxidation half-reaction**:
\[ 2 \text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} + 2 \text{e}^- \]

- **Reduction half-reaction**:
\[ \text{I}_2 + 2 \text{e}^- \rightarrow 2 \text{I}^- \]

### Step 4: Combine half-reactions
The overall balanced redox reaction can be obtained by combining the oxidation and reduction half-reactions:

\[ 2 \text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2 \text{I}^- \]

### Step 5: Conclusion
From the analysis, we conclude:
- \( \text{S}_2\text{O}_3^{2-} \) is oxidized (loses electrons).
- \( \text{I}_2 \) is reduced (gains electrons).