Volumetric titrations involving `KMnO_(4)`, are carried out only in presence of dilute `H_(2)SO_(4)` but not in the presence of HCI or `HNO_(3)` This is because oxygen produced from `KMnO_(4)` + dil: `H_(2)SO_(4)`, is used only for oxidizing the reducing agent. Moreover, H2SO4 does not give any oxygen of its own to oxidize the reducing agent. In case HCI is used, the oxygen produced from `KMnO_(4) + HCI` is partly used up to oxidize HCI and in case `HNO_(3)` is used, it itself acts as oxidizing agent and partly oxidizes the reducing agent. `KMnO_(4)`, in various mediums gives following products`Mn^(7+)overset(H^(+))toMn^(2+)`
`Mn^(7+)overset(H_(2)O)toMn^(4+)`
`Mn^(7+)underset(OH^(-))toMn^(6+)`
QThe equivalent weight of `KMnO_(4)`, in the given reaction is`KMnO_(4)toK_(2)MnO_(4)+ MnO_(2) + O_(2)`

Options

A. `(M)/(6)`
B. `(M)/(2)`
C. M
D. `(M)/(3)`

Answer

To determine the equivalent weight of \( KMnO_4 \) in the reaction \( KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2 \), we can follow these steps:

### Step 1: Identify the oxidation states of manganese in the compounds involved.
- In \( KMnO_4 \), the oxidation state of manganese (Mn) is +7.
- In \( K_2MnO_4 \), the oxidation state of manganese is +6.
- In \( MnO_2 \), the oxidation state of manganese is +4.

### Step 2: Determine the change in oxidation states.
- The change from \( Mn^{+7} \) to \( Mn^{+6} \) represents a reduction of 1.
- The change from \( Mn^{+7} \) to \( Mn^{+4} \) represents a reduction of 3.

### Step 3: Calculate the total change in oxidation number.
- In total, for one mole of \( KMnO_4 \):
- 1 mole of \( Mn \) is reduced from +7 to +6 (1 unit decrease).
- 1 mole of \( Mn \) is reduced from +7 to +4 (3 units decrease).
- Therefore, the total change in oxidation number for one mole of \( KMnO_4 \) is \( 1 + 3 = 4 \).

### Step 4: Calculate the equivalent weight.
- The equivalent weight (EW) is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n} \]
where \( n \) is the total change in oxidation number.
- Since the total change is 4, we can express it as:
\[ \text{Equivalent weight} = \frac{M}{4} \]

### Step 5: Conclusion
Thus, the equivalent weight of \( KMnO_4 \) in the given reaction is \( \frac{M}{4} \), where \( M \) is the molecular weight of \( KMnO_4 \).

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