1 mole of `FeSO_4` (atomic weight of Fe is `55.84 g mol^(-1)` ) is oxidized to `Fe_2(SO_4)_3` . Calculate the equivalent weight of ferrous ion.

Options

A. 55.84
B. 27.92
C. 18.61
D. 111.68

Answer

To calculate the equivalent weight of the ferrous ion (Fe²⁺) when it is oxidized to ferric ion (Fe³⁺), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The oxidation reaction involves the conversion of ferrous ion (Fe²⁺) to ferric ion (Fe³⁺). The half-reaction can be represented as:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]

2. **Determine the Change in Oxidation State**:
In this reaction, the ferrous ion (Fe²⁺) is oxidized to ferric ion (Fe³⁺), which means it loses one electron. Therefore, the change in oxidation state is from +2 to +3.

3. **Calculate the Valence Factor (n)**:
The valence factor (n) is defined as the number of electrons lost or gained in the reaction. Since Fe²⁺ loses one electron to become Fe³⁺, we have:
\[ n = 1 \]

4. **Calculate the Molar Mass of Fe²⁺**:
The atomic weight of iron (Fe) is given as 55.84 g/mol. Therefore, the molar mass of the ferrous ion (Fe²⁺) is also 55.84 g/mol.

5. **Calculate the Equivalent Weight**:
The equivalent weight of an ion or molecule can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n} \]
Substituting the values we have:
\[ \text{Equivalent Weight of Fe}^{2+} = \frac{55.84 \, \text{g/mol}}{1} = 55.84 \, \text{g/equiv} \]

### Final Answer:
The equivalent weight of the ferrous ion (Fe²⁺) is **55.84 g/equiv**.