For decolourisation of `1 "mol of" KMnO_(4)`, the moles of `H_(2)O_(2)` required is

Options

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

Explanation

To determine the moles of \( H_2O_2 \) required for the decolorization of \( 1 \) mole of \( KMnO_4 \), we need to analyze the redox reaction involved. Here’s a step-by-step solution:

### Step 1: Identify the Reduction Reaction
The \( KMnO_4 \) (potassium permanganate) is reduced during the decolorization process. The \( MnO_4^- \) ion (permanganate ion) is reduced to \( Mn^{2+} \).

**Oxidation State Change:**
- The oxidation state of manganese in \( MnO_4^- \) is \( +7 \).
- The oxidation state of manganese in \( Mn^{2+} \) is \( +2 \).

**Change in Oxidation State:**
- The change in oxidation state is \( +7 \) to \( +2 \), which corresponds to a change of \( 5 \) (i.e., \( 7 - 2 = 5 \)).

### Step 2: Calculate the Moles of Electrons Required
Since \( 5 \) moles of electrons are required to reduce \( 1 \) mole of \( MnO_4^- \) to \( Mn^{2+} \), we can write:

\[ \text{Electrons required} = 5 \text{ moles of electrons} \]

### Step 3: Identify the Oxidation Reaction of \( H_2O_2 \)
The \( H_2O_2 \) (hydrogen peroxide) acts as a reducing agent and gets oxidized to \( O_2 \).

**Oxidation State Change:**
- The oxidation state of oxygen in \( H_2O_2 \) is \( -1 \).
- The oxidation state of oxygen in \( O_2 \) is \( 0 \).

**Change in Oxidation State:**
- The change in oxidation state is \( -1 \) to \( 0 \), which corresponds to a change of \( 1 \) for each oxygen atom. Since there are \( 2 \) oxygen atoms in \( H_2O_2 \), the total change is \( 2 \) (i.e., \( 2 \times 1 = 2 \)).

### Step 4: Calculate the Moles of Electrons Provided by \( H_2O_2 \)
From the oxidation of \( H_2O_2 \):
- \( 1 \) mole of \( H_2O_2 \) provides \( 2 \) moles of electrons.

### Step 5: Relate Electrons from \( H_2O_2 \) to \( KMnO_4 \)
To find out how many moles of \( H_2O_2 \) are needed to provide \( 5 \) moles of electrons:

\[ \text{Moles of } H_2O_2 = \frac{\text{Moles of electrons required}}{\text{Electrons provided by } H_2O_2} \]

Substituting the values:

\[ \text{Moles of } H_2O_2 = \frac{5 \text{ moles of electrons}}{2 \text{ moles of electrons per mole of } H_2O_2} = \frac{5}{2} = 2.5 \text{ moles of } H_2O_2 \]

### Final Answer
Thus, the moles of \( H_2O_2 \) required for the decolorization of \( 1 \) mole of \( KMnO_4 \) is \( 2.5 \) moles.

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