In permanganate titrations, potassium pemanganats is used an an oxidizing agent in acidic medium.
The medium is maintained acidic by the use of dilute sulphuric acid. Potassium permanganate acts as self indicator . The potential equation, when potassium permanganate acts as an oxidizing agent is
`2KMnO_(4) + 3H_(2)SO_(4) to K_(2)SO_(4) + 2MnSO_(4) + 3H_(2)O + 5[O]`
or `MnO_(4)^(-) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_(2)O`
Before the end point, the solution remains colourless but after teh equivalence point only one extra drop of `KMnO_(4)` solution imparts pink colour , i.e. appearance of pink colour indicates end point. These titrations are used for estimation of ferrous salts, oxalic acid, oxalates, hydrogen peroxide, `As_(2)O_(3)` etc.
For decolourization of 1 mol of `KMnO_(4)` , the number of moles of `H_(2)O_(2)` required is

Options

A. `1/2`
B. `3/2`
C. `5/2`
D. `7/2`

Answer

To determine the number of moles of \( H_2O_2 \) required to decolorize 1 mole of \( KMnO_4 \), we need to analyze the redox reaction that occurs between potassium permanganate and hydrogen peroxide in an acidic medium.

### Step-by-Step Solution:

1. **Identify the Redox Reaction**:
The reaction between potassium permanganate (\( KMnO_4 \)) and hydrogen peroxide (\( H_2O_2 \)) can be represented as:
\[ 2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO_4 + 2MnSO_4 + 3H_2O + 5O_2 \]
In this reaction, \( KMnO_4 \) is reduced, and \( H_2O_2 \) is oxidized.

2. **Determine Oxidation States**:
- In \( KMnO_4 \), manganese (\( Mn \)) is in the +7 oxidation state.
- In the product \( Mn^{2+} \), manganese is in the +2 oxidation state.
- The change in oxidation state for manganese is \( 7 - 2 = 5 \).

For \( H_2O_2 \):
- The oxidation state of oxygen in \( H_2O_2 \) is -1.
- In molecular oxygen (\( O_2 \)), the oxidation state of oxygen is 0.
- The change in oxidation state for each oxygen atom is \( -1 \) to \( 0 \), which means it changes by +1. Since there are 2 oxygen atoms in \( H_2O_2 \), the total change is \( 2 \).

3. **Calculate the Valency Factors**:
- The valency factor for \( KMnO_4 \) is 5 (since it undergoes a change of 5 in oxidation state).
- The valency factor for \( H_2O_2 \) is 2 (since it undergoes a change of 2 in oxidation state).

4. **Set Up the Equivalence Equation**:
The equivalence of \( KMnO_4 \) must equal the equivalence of \( H_2O_2 \):
\[ \text{Equivalence of } KMnO_4 = \text{Equivalence of } H_2O_2 \]
This can be expressed as:
\[ \text{moles of } KMnO_4 \times \text{valency factor of } KMnO_4 = \text{moles of } H_2O_2 \times \text{valency factor of } H_2O_2 \]

5. **Substitute the Known Values**:
Let \( x \) be the number of moles of \( H_2O_2 \) required. Substituting the known values:
\[ 1 \text{ mole} \times 5 = x \times 2 \]

6. **Solve for \( x \)**:
\[ 5 = 2x \implies x = \frac{5}{2} = 2.5 \]

Thus, **the number of moles of \( H_2O_2 \) required to decolorize 1 mole of \( KMnO_4 \) is 2.5 moles**.