200 ml of 0.01 M `KMnO_(4)` oxidise 20 ml of `H_(2)O_(2)` sample in acidic medium. The volume strength of `H_(2)O_(2)` is

Options

A. 2.8 volume
B. 5.6 volume
C. 0.5 volume
D. 0.25 volume

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced half-reactions
In acidic medium, the half-reaction for the reduction of permanganate ion (MnO4^-) to manganese ion (Mn^2+) is:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

The half-reaction for the oxidation of hydrogen peroxide (H2O2) to oxygen (O2) is:
\[ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2 \text{H}^+ + 2 \text{e}^- \]

### Step 2: Determine the valency factors
- For MnO4^-, the valency factor (n1) is 5 (it gains 5 electrons).
- For H2O2, the valency factor (n2) is 2 (it loses 2 electrons).

### Step 3: Use the stoichiometric relationship
Using the equation for redox reactions:
\[ N_1V_1 = N_2V_2 \]
Where:
- \( N_1 = n1 = 5 \)
- \( V_1 = 200 \, \text{ml} \)
- \( N_2 = n2 = 2 \)
- \( V_2 = 20 \, \text{ml} \)

### Step 4: Calculate the moles of KMnO4 used
Given the molarity of KMnO4:
\[ \text{Molarity} = 0.01 \, \text{M} \]
\[ \text{Volume} = 200 \, \text{ml} = 0.2 \, \text{L} \]
Moles of KMnO4:
\[ \text{Moles} = \text{Molarity} \times \text{Volume} = 0.01 \times 0.2 = 0.002 \, \text{moles} \]

### Step 5: Calculate the moles of electrons transferred
Since each mole of KMnO4 corresponds to 5 moles of electrons:
\[ \text{Moles of electrons} = 0.002 \times 5 = 0.01 \, \text{moles} \]

### Step 6: Relate moles of H2O2 to moles of electrons
From the half-reaction of H2O2, we see that 2 moles of electrons correspond to 1 mole of H2O2:
\[ \text{Moles of H2O2} = \frac{0.01}{2} = 0.005 \, \text{moles} \]

### Step 7: Calculate the molarity of H2O2
The volume of H2O2 used is 20 ml or 0.02 L:
\[ \text{Molarity of H2O2} = \frac{\text{Moles}}{\text{Volume}} = \frac{0.005}{0.02} = 0.25 \, \text{M} \]

### Step 8: Convert molarity to volume strength
The relationship between molarity and volume strength is given by:
\[ \text{Volume strength} = \text{Molarity} \times 11.2 \]
\[ \text{Volume strength} = 0.25 \times 11.2 = 2.8 \, \text{volume} \]

### Final Answer
The volume strength of H2O2 is **2.8 volume**.

---