The equivalent weight of`KMnO_(4)` (formula weight M) when it is used as an oxidant in neural medium is

Options

A. M
B. `M//2`
C. `M//3`
D. `M//5`

Answer

To find the equivalent weight of KMnO₄ when it is used as an oxidant in a neutral medium, we can follow these steps:

### Step 1: Determine the oxidation state of manganese in KMnO₄
- In KMnO₄, potassium (K) has an oxidation state of +1, and each oxygen (O) has an oxidation state of -2.
- Let the oxidation state of manganese (Mn) be x.
- The overall charge of the compound is neutral (0).
- The equation can be set up as follows:
\[ +1 + x + 4(-2) = 0 \]
- Simplifying this gives:
\[ +1 + x - 8 = 0 \implies x - 7 = 0 \implies x = +7 \]
- Therefore, the oxidation state of Mn in KMnO₄ is +7.

### Step 2: Identify the reduction that occurs in neutral medium
- In a neutral medium, KMnO₄ is reduced from Mn in the +7 oxidation state to Mn in the +4 oxidation state (as MnO₂).
- The half-reaction can be represented as:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{MnO}_2 + 4\text{H}_2\text{O} \]
- However, in neutral medium, the reaction can be simplified to:
\[ \text{MnO}_4^- + 3\text{e}^- + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{OH}^- \]
- Here, 3 electrons are involved in the reduction of Mn from +7 to +4.

### Step 3: Calculate the equivalent weight
- The formula for equivalent weight is given by:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{n-factor}} \]
- The n-factor in this case is the number of electrons transferred, which we found to be 3.
- Therefore, the equivalent weight of KMnO₄ when used as an oxidant in neutral medium is:
\[ \text{Equivalent Weight} = \frac{M}{3} \]

### Conclusion
- The equivalent weight of KMnO₄ when used as an oxidant in neutral medium is \( \frac{M}{3} \).

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