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 ferrous ion (Fe²⁺) when it is oxidized to ferric ion (Fe³⁺), we can follow these steps:

### Step 1: Determine the oxidation states
- In `FeSO₄`, the oxidation state of iron (Fe) is +2.
- In `Fe₂(SO₄)₃`, the oxidation state of iron (Fe) is +3.

### Step 2: Calculate the change in oxidation state
- The change in oxidation state from Fe²⁺ to Fe³⁺ is:
\[ \text{Change} = +3 - (+2) = 1 \]

### Step 3: Use the formula for equivalent weight
- The formula for equivalent weight is given by:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n}} \]
where \( n \) is the number of electrons transferred in the redox reaction. Here, \( n = 1 \) (since the change in oxidation state is 1).

### Step 4: Calculate the molar mass of ferrous ion
- The atomic weight of iron (Fe) is given as 55.84 g/mol.

### Step 5: Calculate the equivalent weight
- Substituting the values into the equivalent weight formula:
\[ \text{Equivalent weight} = \frac{55.84 \text{ g/mol}}{1} = 55.84 \text{ g/equiv} \]

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

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