The equivalent weight of \(\text{Na}_2\text{S}_2\text{O}_3\) (Gram molecular weight = \(M\)) in the given reaction is: \[\text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6\]
Here, each molecule of \( \text{Na}_2\text{S}_2\text{O}_3 \) donates 1 electron, leading to a total of 2 electrons being transferred per molecule of \( \text{I}_2 \).
Oxidation states of sulfur: In \( \text{Na}_2\text{S}_2\text{O}_3 \), the average oxidation state of sulfur is +2. During the reaction, sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) is partially oxidized to \( \text{Na}_2\text{S}_4\text{O}_6 \), where the average oxidation state becomes +2.5.
Definition of equivalent weight: The equivalent weight of a substance is given by: \( \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n} \) Here, \( n \) represents the number of electrons transferred per molecule in the reaction.
Calculation: For \( \text{Na}_2\text{S}_2\text{O}_3 \), \( n = 1 \), because each molecule transfers 1 electron. Therefore, the equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) is equal to its molecular weight, \( M \). Thus, the equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) in this reaction is \( M \).
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Approach Solution -2
Correct Answer:
Option 2: M
Explanation:
1. Analyze the Reaction:
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
2. Determine the Change in Oxidation State:
In Na2S2O3, the average oxidation state of sulfur is +2.
In Na2S4O6, the average oxidation state of sulfur is +2.5.
3. Calculate the Change per Molecule:
Each S atom in Na2S2O3 changes its oxidation state from +2 to +2.5, a change of 0.5.
There are two sulfur atoms in Na2S2O3, so the total change in oxidation state per molecule of Na2S2O3 is 2 * 0.5 = 1.
4. Determine the n-factor:
The n-factor (number of electrons transferred per molecule) is equal to the total change in oxidation state, which is 1.
