The eq.wt of iodine in`I_(2)+2S_(2)O_(3)^(2-) to 2I^(-)+S_(4)O_(6)^(2-)` is equal to its

Options

A. mol.wt
B. mol.wt/2
C. mol.wt/4
D. none of these

Answer

To find the equivalent weight of iodine in the reaction \( I_2 + 2S_2O_3^{2-} \rightarrow 2I^{-} + S_4O_6^{2-} \), we can follow these steps:

### Step 1: Determine the oxidation states
In the reaction, iodine changes from an oxidation state of 0 in \( I_2 \) to -1 in \( 2I^{-} \).

**Hint:** Look for the oxidation states of the elements involved in the reaction to identify whether they are being oxidized or reduced.

### Step 2: Identify the change in oxidation state
The change in oxidation state for iodine is from 0 to -1. This indicates that iodine is being reduced, as it is gaining electrons.

**Hint:** Remember that reduction is the gain of electrons, which can be identified by a decrease in oxidation state.

### Step 3: Calculate the number of electrons gained
Since there are 2 iodine atoms in \( I_2 \) and each atom goes from 0 to -1, the total number of electrons gained is 2.

**Hint:** Count the number of atoms involved in the change to determine how many electrons are transferred.

### Step 4: Determine the n-factor
The n-factor for iodine in this reaction is equal to the total number of electrons gained per molecule of \( I_2 \). Since 2 electrons are gained for 1 molecule of \( I_2 \), the n-factor is 2.

**Hint:** The n-factor can be thought of as the number of moles of electrons transferred per mole of substance.

### Step 5: Calculate the molecular weight of iodine
The molecular weight of iodine (\( I_2 \)) is approximately 253.8 g/mol (since the atomic weight of iodine is about 126.9 g/mol).

**Hint:** Use the periodic table to find the atomic weights of the elements involved to calculate the molecular weight.

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

**Hint:** Remember that the equivalent weight is a measure of how much of a substance will react with or supply one mole of hydrogen ions (H⁺) or electrons in a reaction.

### Final Answer
The equivalent weight of iodine in the given reaction is 126.9 g/equiv.

---