When the redox (disproportionation) reaction
`StoSO_(3)^(--)+S^(--)` is balanced by ion electron method in basic medium.
What is correct co-efficient of `S^(--)`.

Answer

To balance the redox reaction \( \text{S} + \text{SO}_3^{2-} \rightarrow \text{S}^{2-} \) using the ion-electron method in a basic medium, we will follow these steps:

### Step 1: Identify Oxidation States
- The oxidation state of elemental sulfur (S) is 0.
- In \( \text{SO}_3^{2-} \), sulfur has an oxidation state of +4.
- In \( \text{S}^{2-} \), sulfur has an oxidation state of -2.

### Step 2: Determine Oxidation and Reduction
- The oxidation half-reaction involves the conversion of elemental sulfur (0) to \( \text{SO}_3^{2-} \) (+4), which means sulfur is oxidized.
- The reduction half-reaction involves the conversion of sulfur (0) to \( \text{S}^{2-} \) (-2), which means sulfur is reduced.

### Step 3: Write the Half-Reactions
1. **Oxidation Half-Reaction**:
\[ \text{S} + 6 \text{OH}^- \rightarrow \text{SO}_3^{2-} + 3 \text{H}_2\text{O} + 4 \text{e}^- \]
2. **Reduction Half-Reaction**:
\[ \text{S} + 2 \text{e}^- \rightarrow \text{S}^{2-} \]

### Step 4: Balance Electrons
- To balance the electrons, multiply the reduction half-reaction by 2:
\[ 2 \text{S} + 4 \text{e}^- \rightarrow 2 \text{S}^{2-} \]

### Step 5: Combine the Half-Reactions
Now we can add the oxidation and reduction half-reactions together:
\[ \text{S} + 6 \text{OH}^- + 2 \text{S} + 4 \text{e}^- \rightarrow \text{SO}_3^{2-} + 3 \text{H}_2\text{O} + 2 \text{S}^{2-} + 4 \text{e}^- \]

### Step 6: Cancel Out Electrons
Cancel the 4 electrons from both sides:
\[ 3 \text{S} + 6 \text{OH}^- \rightarrow \text{SO}_3^{2-} + 3 \text{H}_2\text{O} + 2 \text{S}^{2-} \]

### Step 7: Final Balanced Equation
The final balanced equation is:
\[ 3 \text{S} + 6 \text{OH}^- \rightarrow 2 \text{S}^{2-} + \text{SO}_3^{2-} + 3 \text{H}_2\text{O} \]

### Conclusion
The coefficient of \( \text{S}^{2-} \) in the balanced equation is **2**.

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