In a reaction, `H_(2)O +C rarr CO +H_(2)`

Options

A. `H_(2)O` is the reducing agent
B. `H_(2)O` is the oxidising agent
C. carbon is the oxidising agent
D. oxidation -reduction does not occur

Answer

To analyze the reaction \( H_2O + C \rightarrow CO + H_2 \) and identify the oxidizing and reducing agents, we can follow these steps:

### Step 1: Assign Oxidation States
- In \( H_2O \):
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

- In elemental carbon (C), the oxidation state is 0.

- In \( CO \):
- Carbon (C) has an oxidation state of +2.
- Oxygen (O) has an oxidation state of -2.

- In \( H_2 \), the oxidation state of hydrogen is 0.

### Step 2: Determine Changes in Oxidation States
- In the reactants:
- Oxygen remains at -2.
- Carbon changes from 0 (in elemental form) to +2 (in CO).

- In the products:
- Hydrogen changes from +1 (in \( H_2O \)) to 0 (in \( H_2 \)).

### Step 3: Identify Oxidation and Reduction
- **Oxidation**: An increase in oxidation state.
- Carbon is oxidized from 0 to +2.

- **Reduction**: A decrease in oxidation state.
- Hydrogen is reduced from +1 to 0.

### Step 4: Identify the Oxidizing and Reducing Agents
- The substance that gets oxidized (loses electrons) is the reducing agent.
- The substance that gets reduced (gains electrons) is the oxidizing agent.

- Here:
- Carbon is the reducing agent because it is oxidized.
- Water (\( H_2O \)) acts as the oxidizing agent because it is reduced.

### Conclusion
In this reaction:
- Carbon is oxidized (reducing agent).
- Water is reduced (oxidizing agent).