Question

1 mole of $FeSO_4$ (atomic weight of Fe is $55.84\,\text{g mol}^{-1}$) is oxidized to $Fe_2(SO_4)_3$. Calculate the equivalent weight of ferrous ion

• A. 55.84
• B. 27.92
• C. 18.61
• D. 111.68
• E. 83.76

Hint:

For reactions where the valency change is 1, the equivalent weight is always equal to the atomic/molecular weight.

Answer 1

The Correct Option is A

Concept: Equivalent weight in a redox reaction is defined as: \[ \text{Equivalent Weight} = \frac{\text{Atomic Weight}}{\text{Change in Oxidation State } (n\text{-factor})} \]

Step 1: Determine the oxidation states.
In \(FeSO_4\) (Ferrous sulfate), iron is in the +2 oxidation state (\(Fe^{2+}\)). In \(Fe_2(SO_4)_3\) (Ferric sulfate), iron is in the +3 oxidation state (\(Fe^{3+}\)).

Step 2: Calculate the change (n-factor).
The reaction is: \[ Fe^{2+} \rightarrow Fe^{3+} + e^- \] The number of electrons lost per iron atom is 1. Therefore, \(n = 1\).

Step 3: Calculate equivalent weight.
\[ \text{Equivalent weight of } Fe^{2+} = \frac{55.84}{1} = 55.84 \]