`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 H₂S in the given redox reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Half-Reactions:**
The overall reaction is:
\[ 2KMnO_4 + 5H_2S + 6H^+ \rightarrow 2Mn^{2+} + 2K^+ + 5S + 8H_2O \]
The two half-reactions can be identified as:
- Reduction: \( KMnO_4 \rightarrow Mn^{2+} + K^+ \)
- Oxidation: \( H_2S \rightarrow S \)

2. **Determine Oxidation States:**
- For \( KMnO_4 \): Manganese (Mn) has an oxidation state of +7.
- For \( H_2S \): Sulfur (S) has an oxidation state of -2.

3. **Calculate Change in Oxidation States:**
- In the reduction half-reaction, Mn changes from +7 to +2, which is a change of 5 (7 - 2 = 5).
- In the oxidation half-reaction, S changes from -2 to 0, which is a change of 2 (0 - (-2) = 2).

4. **Balance the Half-Reactions:**
- For the reduction half-reaction:
\[ KMnO_4 + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \]
- For the oxidation half-reaction:
\[ H_2S \rightarrow S + 2e^- \]

5. **Balance Electrons:**
To combine the half-reactions, we need to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction. The reduction half-reaction involves 5 electrons, while the oxidation half-reaction involves 2 electrons. To balance, we can multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:
- Reduction: \( 2KMnO_4 + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O \)
- Oxidation: \( 5H_2S \rightarrow 5S + 10e^- \)

6. **Determine Electrons for 1 Mole of Oxidation:**
From the balanced oxidation half-reaction, we see that 5 moles of \( H_2S \) require 10 electrons. Therefore, for 1 mole of \( H_2S \):
\[ \text{Electrons required} = \frac{10 \text{ electrons}}{5 \text{ moles}} = 2 \text{ electrons} \]

### Final Answer:
For the oxidation of 1 mole of \( H_2S \), **2 electrons** are involved.

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