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 answer is (c) \( 1/2 \) its molecular weight.

Step 1: Understanding the Reaction

First, let's break down what happens in the reaction. The balanced reaction shows that one mole of iodine, \( \mathrm{I}_2 \), reacts with two moles of \( \mathrm{S}_2 \mathrm{O}_3^{2-} \) to form two moles of \( \mathrm{I}^{-} \) and one mole of \( \mathrm{S}_4 \mathrm{O}_6^{2-} \). The iodine is being reduced from \( \mathrm{I}_2 \) to \( \mathrm{I}^{-} \).

Step 2: Determine Loss and Gain of Electrons

In this reaction, iodine changes from an oxidation state of 0 in \( \mathrm{I}_2 \) to -1 in \( \mathrm{I}^{-} \). Each iodine atom in \( \mathrm{I}_2 \) gains one electron (to become \( \mathrm{I}^{-} \)), so two electrons are gained for the two iodine atoms in one molecule of \( \mathrm{I}_2 \).

Step 3: Calculate Equivalent Weight of Iodine

Equivalent weight is calculated as the molar mass divided by the number of electrons transferred per molecule. In our reaction, iodine \( \mathrm{I}_2 \) gains 2 electrons. Therefore, the equivalent weight of iodine is its molecular weight divided by 2.

Step 4: Check the Options

Given that the equivalent weight of iodine is the molecular weight of \( \mathrm{I}_2 \) divided by 2, option (c) \( 1/2 \) its molecular weight is correct.

Key Concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are processes where electrons are transferred between substances. These reactions are fundamental to many chemical processes and are characterized by the movement of electrons from one reactant (which is oxidized) to another (which is reduced).
In our example, iodine () initially has an oxidation state of 0. During the reaction, it gains electrons and transitions to iodide (), reducing its oxidation state to -1. This gain of electrons is known as "reduction." On the other hand, the species that donates electrons undergoes "oxidation."

• Oxidation is the loss of electrons.
• Reduction is the gain of electrons.

Understanding these concepts is essential for analyzing the changes in oxidation states and electron flow during a redox reaction.

Iodine Chemistry

Iodine is a fascinating element, widely known for its role in both chemistry and biology. In the realm of chemistry, iodine can participate in various reactions, prominently redox reactions. In its elemental form, as in , iodine molecules consist of two iodine atoms bonded together.
When iodine acts as an oxidizing agent, it accepts electrons. For example, in the given reaction, iodine () accepts electrons from thiosulfate ions (0_3^{2-}), thus forming iodide ions (). This electron acceptance leads to its reduction. The versatile nature of iodine allows it to switch between multiple oxidation states, which is crucial for various applications in synthesis and analysis.

Electron Transfer

Electron transfer is a key component of redox reactions. It involves the movement of electrons from one chemical species to another during the reaction. Specifically, one substance donates electrons (undergoes oxidation), and another species accepts those electrons (undergoes reduction).
In our given reaction, the iodine molecules () gain electrons from the thiosulfate ions (0_3^{2-}). This transfer results in the formation of iodide ions () and rearranges the remaining atoms into tetrathionate (O_6^{2-}).

• Understanding the number of electrons transferred helps in calculating the equivalent weight of the reactants.
• The movement of electrons causes chemical changes essential for the synthesis and decomposition of compounds.

Electron transfer is thus a fundamental concept that not only plays a role in theoretical chemistry but also in practical applications like batteries and electrochemical cells.

Stoichiometry

Stoichiometry is a concept that involves the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict how much of each substance is involved or produced in a reaction based on the balanced chemical equation.
For the iodine reaction we've covered, stoichiometry helps us determine the exact proportions of  and 0_3^{2-} needed to achieve a complete reaction, leading to the formation of  and O_6^{2-}. This understanding is critical in determining yields and efficiencies of chemical processes.
Stoichiometry relies heavily on the concept of moles, which relates the masses of the reactants and products to their chemical quantities through Avogadro's number. This helps in:

• Balancing chemical equations
• Calculating reactant and product masses
• Determining limiting reactants in a reaction

Mastery of stoichiometry ensures that chemists can efficiently utilize resources and predict the outcomes of chemical reactions accurately.