`2Na_(2)S_(2)O_(3)+I_(2)toNa_(2)S_(4)O_(6)+2NaI`
How many equivalents of Hypo is oxidised by one mole of Iodine?

Answer

To determine how many equivalents of sodium thiosulfate (hypo, Na2S2O3) are oxidized by one mole of iodine (I2), we can follow these steps:

### Step 1: Write the Balanced Chemical Equation
The balanced chemical equation given is:
\[ 2 \text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2 \text{NaI} \]

### Step 2: Identify the Oxidation States
In sodium thiosulfate (Na2S2O3), the oxidation state of sulfur is +2. In the product sodium tetrathionate (Na2S4O6), the oxidation state of sulfur is +2.5. This indicates that sulfur is being oxidized.

### Step 3: Calculate the Change in Oxidation State
The change in oxidation state for sulfur can be calculated as follows:
- Initial oxidation state of sulfur in Na2S2O3 = +2
- Final oxidation state of sulfur in Na2S4O6 = +2.5

The change in oxidation state per sulfur atom is:
\[ +2.5 - +2 = +0.5 \]

### Step 4: Determine the N Factor
The n-factor for the reaction can be calculated based on the change in oxidation state:
- Since there are 2 moles of Na2S2O3, the total change for 2 moles of sulfur is:
\[ 2 \text{ moles} \times 0.5 = 1 \text{ equivalent} \]

### Step 5: Relate Iodine to Sodium Thiosulfate
From the balanced equation, we see that 1 mole of I2 reacts with 2 moles of Na2S2O3. Therefore, for 1 mole of iodine, the equivalents of sodium thiosulfate oxidized can be calculated as follows:
- 1 mole of I2 oxidizes 2 equivalents of Na2S2O3.

### Conclusion
Thus, **2 equivalents of sodium thiosulfate (hypo) are oxidized by one mole of iodine (I2)**.

---