In the reaction : `Cl_2+2OH^(-) rarr OCl^(-) +Cl^(-) +H_2O`

Options

A. `OH^(-)` is oxidising and `Cl^(-)` is reducing agent
B. `Cl_2` is oxidising and `OH^(-)` is reducing agent
C. `OH^(-)` is both oxidising and reducing agent .
D. `Cl_2` is both oxidising and reducing agent

Answer

To determine the oxidizing and reducing agents in the reaction \( \text{Cl}_2 + 2\text{OH}^- \rightarrow \text{OCl}^- + \text{Cl}^- + \text{H}_2\text{O} \), we will follow these steps:

### Step 1: Assign Oxidation States
We need to assign oxidation states to all the elements in the reaction.

1. **For \( \text{Cl}_2 \)**: Chlorine in its elemental form has an oxidation state of 0.
2. **For \( \text{OH}^- \)**:
- Oxygen (O) has an oxidation state of -2.
- Hydrogen (H) has an oxidation state of +1.
3. **For \( \text{OCl}^- \)**:
- Let the oxidation state of Cl be \( x \).
- The overall charge is -1, so we have:
\[ x + (-2) = -1 \implies x = +1 \]
- Therefore, the oxidation state of Cl in \( \text{OCl}^- \) is +1.
4. **For \( \text{Cl}^- \)**: Chlorine has an oxidation state of -1.
5. **For \( \text{H}_2\text{O} \)**:
- Oxygen has an oxidation state of -2.
- Each Hydrogen has an oxidation state of +1.

### Step 2: Identify Changes in Oxidation States
Now we will summarize the oxidation states:

- In \( \text{Cl}_2 \): Cl = 0
- In \( \text{OCl}^- \): Cl = +1
- In \( \text{Cl}^- \): Cl = -1

Now let's analyze the changes:
- The chlorine in \( \text{Cl}_2 \) (0) is oxidized to \( \text{OCl}^- \) (+1).
- The chlorine in \( \text{Cl}_2 \) (0) is reduced to \( \text{Cl}^- \) (-1).

### Step 3: Determine the Oxidizing and Reducing Agents
- **Oxidizing Agent**: The species that gets reduced (gains electrons) is the oxidizing agent. Here, \( \text{Cl} \) in \( \text{OCl}^- \) is the oxidizing agent since it goes from 0 to +1.
- **Reducing Agent**: The species that gets oxidized (loses electrons) is the reducing agent. Here, \( \text{Cl} \) in \( \text{Cl}^- \) is the reducing agent since it goes from 0 to -1.

### Conclusion
In this reaction:
- \( \text{Cl}_2 \) acts as both an oxidizing agent (for the formation of \( \text{OCl}^- \)) and a reducing agent (for the formation of \( \text{Cl}^- \)).

### Final Answer
Thus, \( \text{Cl}_2 \) is both the oxidizing agent and the reducing agent in this reaction.

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