In the reaction,
`I_(2)+2S_(2)O_(3)^(2-) rarr 2I^(-)+S_(4)O_(6)^(2-)`.
Equivalent wieght of iodine will be equal to

Options

A. Its Mol. wt.
B. Mol. wt./2
C. Mol. wt./4
D. None of these

Answer

To find the equivalent weight of iodine in the given reaction, we can follow these steps:

### Step 1: Write down the reaction
The reaction provided is:
\[ I_2 + 2S_2O_3^{2-} \rightarrow 2I^{-} + S_4O_6^{2-} \]

### Step 2: Identify the oxidation states
- In \( I_2 \), the oxidation state of iodine (I) is 0.
- In \( I^{-} \), the oxidation state of iodine is -1.

### Step 3: Calculate the change in oxidation number
The change in oxidation number for iodine can be calculated as follows:
- Initial oxidation state = 0
- Final oxidation state = -1
- Change = Final - Initial = -1 - 0 = -1

### Step 4: Determine the n-factor
The n-factor is defined as the total change in oxidation number per mole of the substance. Since each iodine atom changes from 0 to -1, and there are 2 iodine atoms in \( I_2 \):
- Total change for 2 iodine atoms = 2 × (-1) = -2
- Therefore, the n-factor for iodine in this reaction is 2.

### Step 5: Calculate the molecular weight of iodine
The molecular weight of iodine (I2) can be calculated as:
- Atomic weight of iodine (I) = 126.9 g/mol
- Therefore, molecular weight of \( I_2 = 2 \times 126.9 = 253.8 \, \text{g/mol} \)

### Step 6: Calculate the equivalent weight
The formula for equivalent weight is:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} \]
Substituting the values we have:
\[ \text{Equivalent weight of iodine} = \frac{253.8 \, \text{g/mol}}{2} = 126.9 \, \text{g/equiv} \]

### Step 7: Conclusion
Thus, the equivalent weight of iodine in this reaction is 126.9 g/equiv.

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