In the reaction given below identify the species undergoing oxidation and reduction`H_2S+Cl_2rarr2HCl+S(s)`

Options

A. `H_2S` is oxidised and `Cl_2` is reduced
B. `H_2S` is reduced and `Cl_2` is oxidised
C. both `H_2S` and `Cl_2` are oxidised
D. both `H_2S` and `Cl_2` are reduced

Answer

To identify the species undergoing oxidation and reduction in the reaction \( H_2S + Cl_2 \rightarrow 2HCl + S(s) \), we can follow these steps:

### Step 1: Assign Oxidation States
First, we need to assign oxidation states to each element in the reactants and products.

- In \( H_2S \):
- Hydrogen (H) has an oxidation state of +1.
- Sulfur (S) has an oxidation state of -2.

- In \( Cl_2 \):
- Chlorine (Cl) has an oxidation state of 0 (as it is in its elemental form).

- In \( HCl \):
- Hydrogen (H) has an oxidation state of +1.
- Chlorine (Cl) has an oxidation state of -1.

- In elemental sulfur (S):
- Sulfur (S) has an oxidation state of 0.

### Step 2: Determine Changes in Oxidation States
Next, we analyze the changes in oxidation states:

- For sulfur in \( H_2S \) (from -2 to 0 in elemental sulfur):
- Oxidation state increases (from -2 to 0), indicating that sulfur is being oxidized.

- For chlorine in \( Cl_2 \) (from 0 to -1 in \( HCl \)):
- Oxidation state decreases (from 0 to -1), indicating that chlorine is being reduced.

### Step 3: Identify Oxidation and Reduction
Based on the changes in oxidation states:

- **Oxidation**: The species that undergoes oxidation is \( H_2S \) because sulfur is losing electrons (its oxidation state increases).
- **Reduction**: The species that undergoes reduction is \( Cl_2 \) because chlorine is gaining electrons (its oxidation state decreases).

### Conclusion
In the reaction \( H_2S + Cl_2 \rightarrow 2HCl + S(s) \):
- **Oxidation**: \( H_2S \) (sulfur is oxidized)
- **Reduction**: \( Cl_2 \) (chlorine is reduced)