The equivalent mass of `KMnO_(4)` in the following reaction is:
`MnO_(4)^(-) + 5Fe^(2+) + 8H^(+) to Mn^(2+) + 5Fe^(2+) + 4H_(2)O`
(M = Molecular mass)

Options

A. `M/2`
B. `M/3`
C. `M/4`
D. `M/5`

Explanation

To find the equivalent mass of \( KMnO_4 \) in the given redox reaction, we will follow these steps:

### Step 1: Identify the oxidation states
In the reaction:
\[ MnO_4^{-} + 5Fe^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O \]

We need to determine the oxidation state of manganese (Mn) in \( MnO_4^{-} \).

### Step 2: Calculate the oxidation state of Mn
Let the oxidation state of Mn be \( x \). The oxidation state of oxygen is \(-2\). In \( MnO_4^{-} \):
\[ x + 4(-2) = -1 \]
\[ x - 8 = -1 \]
\[ x = +7 \]
Thus, the oxidation state of Mn in \( MnO_4^{-} \) is \( +7 \).

### Step 3: Determine the change in oxidation state
In the reaction, \( Mn \) changes from \( +7 \) in \( MnO_4^{-} \) to \( +2 \) in \( Mn^{2+} \). The change in oxidation state is:
\[ 7 - 2 = 5 \]
This means that \( MnO_4^{-} \) gains 5 electrons.

### Step 4: Calculate the molar mass of \( KMnO_4 \)
The molar mass of \( KMnO_4 \) can be calculated as follows:
- Potassium (K): \( 39.1 \, g/mol \)
- Manganese (Mn): \( 54.9 \, g/mol \)
- Oxygen (O): \( 16.0 \, g/mol \times 4 = 64.0 \, g/mol \)

Adding these together:
\[ Molar \, mass \, of \, KMnO_4 = 39.1 + 54.9 + 64.0 = 158.0 \, g/mol \]

### Step 5: Calculate the equivalent mass
The equivalent mass of a substance in a redox reaction can be calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{Number of electrons gained or lost}} \]
In this case, the number of electrons gained is 5. Therefore:
\[ \text{Equivalent mass of } KMnO_4 = \frac{158.0 \, g/mol}{5} = 31.6 \, g/equiv \]

### Final Answer
The equivalent mass of \( KMnO_4 \) in the given reaction is \( 31.6 \, g/equiv \).

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