Problem 13 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction, \(\mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2}\), equivalent weight of iodine will be equal to (a) twice its molecular weight (b) its molecular weight (c) \(1 / 2\) its moleculear weight (d) \(1 / 4\) its molecular weight

Step-by-Step Solution

Answer

The equivalent weight of iodine is equal to its molecular weight.

Step 1: Determine the Reaction Type

Given the reaction \[\mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\]we need to find the equivalent weight of iodine, \(\mathrm{I}_2\). This reaction involves the reduction of iodine, indicating it's a redox reaction.

Step 2: Calculate Change in Oxidation State

In the reaction, iodine \((\mathrm{I}_2)\) is reduced to iodide \((2\mathrm{I}^-)\). The oxidation state of \(\mathrm{I}_2\) changes from 0 to -1 for each iodine atom, totaling a change of 2 electrons per \(\mathrm{I}_2\) molecule.

Step 3: Determine the Equivalent Weight Formula

The equivalent weight of a substance in a redox reaction is calculated using the formula:\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{n-factor}} \]where the n-factor is the number of electrons exchanged per molecule in the balanced equation.

Step 4: Apply the Formula to Iodine

For iodine \((\mathrm{I}_2)\), the molecular weight is 254 g/mol, and since 2 electrons are exchanged per molecule, the n-factor is 2. Thus, the equivalent weight is:\[ \frac{254}{2} = 127 \text{ g/equiv} \]

Key Concepts

Equivalent Weight

When studying redox reactions, one important concept is the equivalent weight of a substance involved in the reaction. Equivalent weight is particularly significant because it helps to understand how much of a material reacts with a fixed quantity of another reactant.
In a redox reaction, equivalent weight can be calculated using the formula:

• \[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{n-factor}} \]

The n-factor in this context is the number of moles of electrons exchanged per mole of the substance in the chemical equation.
For iodine \(\mathrm{I}_2\), the molar mass is given as 254 g/mol. In the provided chemical reaction, iodine exchanges 2 electrons per molecule. Thus, the equivalent weight of iodine is 127 g/equiv.
This value provides insight into how iodine participates in other chemical reactions, particularly when reacting stoichiometrically with thiosulfate anions relevant to the given chemical equation.

Oxidation State

The concept of oxidation state is central to redox reactions. It denotes the degree of oxidation of an atom within a compound. By assigning oxidation states, we can follow the transfer of electrons in a chemical reaction.
In the reaction between iodine \(\mathrm{I}_2\) and thiosulfate \(\mathrm{S}_2\mathrm{O}_3^{2-}\), iodine serves as the oxidizing agent. Initially, each iodine atom in \(\mathrm{I}_2\) has an oxidation state of 0. As the reaction proceeds, each iodine atom is reduced to iodide \(\mathrm{I}^-\), where its oxidation state becomes -1.
This change in oxidation state demonstrates a gain of electrons by iodine atoms, crucial for identifying redox behavior:

• Iodine oxidation state change: 0 to -1
• Number of electrons gained per iodine molecule: 2

Understanding the shifts in oxidation states allows chemists to balance redox reactions and determine the stoichiometry accurately.

Iodine Reduction

In the outlined reaction, iodine undergoes a reduction process. Reduction refers to the gain of electrons by a molecule, atom, or ion. This particular reaction highlights the interaction between iodine \(\mathrm{I}_2\) and thiosulfate ions \(\mathrm{S}_2\mathrm{O}_3^{2-}\), wherein iodine is reduced to iodide ions \(\mathrm{I}^-\).
The reduction of iodine is depicted by these key changes:

• The oxidation state shifts from 0 to -1, meaning iodine atoms gain electrons.
• A total of 2 electrons are accepted per molecular unit of iodine \(\mathrm{I}_2\).

The reaction showcases a typical feature of reduction – iodine, initially elemental and neutral, gains electrons to become negatively charged iodide ions. This is an essential step in many chemical analyses and industrial applications, particularly in redox titrations where iodine's role is critical in determining concentrations of reducing agents.