Question

The equivalent weight of \(\text{Na}_2\text{S}_2\text{O}_3\) (Gram molecular weight = \(M\)) in the given reaction is:
\[\text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6\]

• A. M/2
• B. M
• C. 2M
• D. M/4

Answer 1

The Correct Option is B

The chemical process is: \( \text{I}_2 + 2\text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2\text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \)

Each \( \text{Na}_2\text{S}_2\text{O}_3 \) molecule donates one electron, resulting in a total transfer of two electrons per \( \text{I}_2 \) molecule.

Sulfur's Oxidation States: In \( \text{Na}_2\text{S}_2\text{O}_3 \), the average sulfur oxidation state is +2. In the reaction, sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) is partially oxidized to \( \text{Na}_2\text{S}_4\text{O}_6 \), with an average oxidation state of +2.5.

Equivalent Weight Definition: The equivalent weight is calculated as: \( \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n} \). Here, \( n \) signifies the number of electrons exchanged per molecule during the reaction.

Calculation: For \( \text{Na}_2\text{S}_2\text{O}_3 \), \( n = 1 \), since each molecule transfers one electron. Consequently, the equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) equals its molecular weight, \( M \). Therefore, the equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) in this reaction is \( M \).