Write the oxidation product when `XeO_(3)` oxidises `I^(-)` in acidic medium.

Answer

To determine the oxidation product when \( \text{XeO}_3 \) oxidizes \( \text{I}^- \) in acidic medium, we can follow these steps:

### Step 1: Write the balanced reaction
In acidic medium, \( \text{XeO}_3 \) acts as an oxidizing agent. The oxidation of iodide ions (\( \text{I}^- \)) can produce iodine (\( \text{I}_2 \)) or triiodide ions (\( \text{I}_3^- \)).

### Step 2: Identify the reactants and products
The reactants in this reaction are:
- \( \text{XeO}_3 \) (xenon trioxide)
- \( \text{I}^- \) (iodide ion)
- \( \text{H}^+ \) (from the acidic medium)

The products can be either:
- \( \text{I}_2 \) (iodine)
- \( \text{I}_3^- \) (triiodide ion)
- Water (\( \text{H}_2\text{O} \))
- Xenon (\( \text{Xe} \))

### Step 3: Write the half-reactions
1. For the oxidation of iodide:
- \( 2 \text{I}^- \rightarrow \text{I}_2 + 2 \text{e}^- \) (for producing iodine)
- or \( 3 \text{I}^- \rightarrow \text{I}_3^- + 2 \text{e}^- \) (for producing triiodide)

2. For the reduction of xenon trioxide:
- \( \text{XeO}_3 + 6 \text{H}^+ + 6 \text{e}^- \rightarrow \text{Xe} + 3 \text{H}_2\text{O} \)

### Step 4: Combine the half-reactions
To balance the electrons transferred, we can use the first reaction with \( 9 \text{I}^- \) to produce \( 3 \text{I}_2 \):
\[ \text{XeO}_3 + 6 \text{H}^+ + 6 \text{I}^- \rightarrow \text{Xe} + 3 \text{H}_2\text{O} + 3 \text{I}_2 \]

### Step 5: Final balanced equation
Thus, the final balanced equation for the oxidation of \( \text{I}^- \) by \( \text{XeO}_3 \) in acidic medium is:
\[ \text{XeO}_3 + 6 \text{H}^+ + 6 \text{I}^- \rightarrow \text{Xe} + 3 \text{H}_2\text{O} + 3 \text{I}_2 \]

### Conclusion
The oxidation product when \( \text{XeO}_3 \) oxidizes \( \text{I}^- \) in acidic medium is \( \text{I}_2 \) (iodine).

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