`K_(2)Cr_(2)O_(7), + l^(-) + H^(+ )to` Oxidized product. The product is

Options

A. `KIO_(3)`
B. `I_(2)`
C. `I_(3)^(-)`
D. `Cr_(2)O_(3)`

Answer

To determine the oxidized product from the reaction of potassium dichromate (K₂Cr₂O₇) with iodide ions (I⁻) in the presence of acid (H⁺), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are potassium dichromate (K₂Cr₂O₇), iodide ions (I⁻), and hydrogen ions (H⁺) from the acid.

2. **Write the Reaction**:
The reaction can be represented as:
\[ K_2Cr_2O_7 + I^- + H^+ \rightarrow \text{Oxidized Product} \]

3. **Determine the Role of Each Species**:
- Potassium dichromate (K₂Cr₂O₇) acts as an oxidizing agent.
- Iodide ions (I⁻) will be oxidized to iodine (I₂).
- The chromium in K₂Cr₂O₇ will be reduced from +6 oxidation state to +3 oxidation state.

4. **Write the Half-Reactions**:
- **Oxidation Half-Reaction**:
\[ 2 I^- \rightarrow I_2 + 2 e^- \]
- **Reduction Half-Reaction**:
\[ Cr_2O_7^{2-} + 14 H^+ + 6 e^- \rightarrow 2 Cr^{3+} + 7 H_2O \]

5. **Balance the Overall Reaction**:
To balance the overall reaction, we need to ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
- Multiply the oxidation half-reaction by 3:
\[ 6 I^- \rightarrow 3 I_2 + 6 e^- \]
- Combine the balanced half-reactions:
\[ Cr_2O_7^{2-} + 6 I^- + 14 H^+ \rightarrow 2 Cr^{3+} + 3 I_2 + 7 H_2O \]

6. **Identify the Oxidized Product**:
From the balanced reaction, we can see that the iodide ions (I⁻) are oxidized to iodine (I₂). Therefore, the oxidized product is:
\[ \text{I}_2 \]

### Final Answer:
The oxidized product is **I₂** (iodine).