What product is formed when `H_2S` gas is passed through acidified `KMnO_4` solution ?

Options

A. `MnO_2`
B. S
C. `K_2SO_3`
D. `K_2S`

Answer

To determine the product formed when hydrogen sulfide (H₂S) gas is passed through acidified potassium permanganate (KMnO₄) solution, we can analyze the redox reaction that occurs.

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in this reaction are hydrogen sulfide (H₂S) and acidified potassium permanganate (KMnO₄). The acidification usually involves adding sulfuric acid (H₂SO₄) to provide H⁺ ions.

2. **Understand the Redox Reaction**: KMnO₄ acts as an oxidizing agent in acidic medium. H₂S will be oxidized to sulfur (S), while KMnO₄ will be reduced to manganese ions (Mn²⁺).

3. **Write the Half-Reactions**:
- **Oxidation Half-Reaction**:
\[ H_2S \rightarrow S + 2H^+ + 2e^- \]
- **Reduction Half-Reaction**:
\[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \]

4. **Balance the Half-Reactions**:
- The oxidation half-reaction is already balanced.
- For the reduction half-reaction, we need to ensure that the number of electrons lost in the oxidation half-reaction matches the number of electrons gained in the reduction half-reaction.

5. **Multiply the Half-Reactions to Equalize Electrons**:
- Multiply the oxidation half-reaction by 5:
\[ 5H_2S \rightarrow 5S + 10H^+ + 10e^- \]
- The reduction half-reaction remains the same.

6. **Combine the Half-Reactions**:
- Add the balanced half-reactions together:
\[ 5H_2S + 8H^+ + MnO_4^- \rightarrow 5S + Mn^{2+} + 4H_2O \]

7. **Identify the Products**: From the combined reaction, we can see that the products formed are sulfur (S) and manganese ions (Mn²⁺).

8. **Conclusion**: Therefore, when H₂S gas is passed through acidified KMnO₄ solution, the primary product formed is sulfur (S).

### Final Answer:
The product formed is sulfur (S).