`2KMnO_(4)+5H_(2)S+6H^(+)rarr2Mn^(2+)+2K^(+)+5S+8H_(2)O`.
In the above reaction, how many electrons would be involved in the oxidaton of 1 mole of reduction?

Options

A. Two
B. Five
C. Ten
D. One

Answer

To determine how many electrons are involved in the oxidation of 1 mole of reduction in the given redox reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Half-Reactions**:
- The reaction involves potassium permanganate (KMnO₄) being reduced to Mn²⁺ and hydrogen sulfide (H₂S) being oxidized to sulfur (S).
- The half-reactions can be written as:
- Reduction: \( \text{KMnO}_4 \rightarrow \text{Mn}^{2+} + \text{K}^+ + 4\text{H}_2O \)
- Oxidation: \( \text{H}_2S \rightarrow \text{S} + 2\text{H}^+ + 2e^- \)

2. **Balance the Reduction Half-Reaction**:
- For the reduction of KMnO₄ to Mn²⁺, we need to balance the oxygen and hydrogen:
- Add 4 water molecules to the right side:
\[ \text{KMnO}_4 + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + \text{K}^+ + 4\text{H}_2O \]
- This half-reaction shows that 5 electrons are gained.

3. **Balance the Oxidation Half-Reaction**:
- For the oxidation of H₂S to S, we balance the charges:
- The balanced half-reaction is:
\[ 5\text{H}_2S \rightarrow 5\text{S} + 10\text{H}^+ + 10e^- \]
- This indicates that 10 electrons are lost.

4. **Combine the Half-Reactions**:
- Since we have 2 moles of KMnO₄ and 5 moles of H₂S in the overall reaction, we can multiply the half-reactions accordingly:
- The reduction half-reaction (for 2 moles of KMnO₄) gives us 10 electrons.
- The oxidation half-reaction (for 5 moles of H₂S) also gives us 10 electrons.
- The total balanced reaction is:
\[ 2\text{KMnO}_4 + 5\text{H}_2S + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 2\text{K}^+ + 5\text{S} + 8\text{H}_2O \]

5. **Determine Electrons for 1 Mole of Reduction**:
- Since the reduction of 2 moles of KMnO₄ involves 10 electrons, the oxidation of 1 mole of reduction (which corresponds to 1 mole of KMnO₄) would involve:
\[ \frac{10 \text{ electrons}}{2} = 5 \text{ electrons} \]

### Final Answer:
The number of electrons involved in the oxidation of 1 mole of reduction is **5 electrons**.