Problem 20 - A Complete Resource Book in Chemistry for JEE Main

Question

In the following reaction \(\mathrm{Cr}(\mathrm{OH})_{3}+\mathrm{OH}^{-}+\mathrm{IO}_{3}^{-} \longrightarrow \mathrm{CrO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O}+\mathrm{I}^{-}\) (a) \(\mathrm{IO}_{3}^{-}\)is oxidizing agent (b) \(\mathrm{Cr}(\mathrm{OH})_{3}\) is oxidized (c) \(6 \mathrm{e}^{-}\)are being taken per 1 atom (d) all are true

Step-by-Step Solution

Answer

Both statements (a) and (b) are true.

Step 1: Determine Oxidation States

Assign oxidation states to each element in the reaction for both reactants and products to identify changes. For \( ext{IO}_3^-\), Iodine ( ext{I}) is in the +5 state, and for \( ext{I}^-\), it's in -1. Chromium ( ext{Cr}) in \( ext{Cr}( ext{OH})_3\) is +3, and in \( ext{CrO}_4^{2-}\) it's +6.

Step 2: Analyze Oxidation and Reduction

Compare the oxidation states from Step 1. \( ext{IO}_3^-\) is reduced as the oxidation state of I decreases from +5 to -1 (gains electrons). \( ext{Cr}( ext{OH})_3\) is oxidized as Cr goes from +3 to +6 (loses electrons).

Step 3: Calculate Electrons Transferred

Iodine's reduction involves the gain of \( (5 - (-1)) = 6 \) electrons per atom of iodine. Chromium's oxidation involves the loss of \( (6 - 3) = 3 \) electrons per atom of chromium. The balanced reaction shows the same molecules of iodine and chromium being oxidized and reduced, ensuring the stoichiometry is maintained with 6 electrons transfer.

Step 4: Evaluate Given Statements

(a) is true: \(\text{IO}_3^-\) acts as an oxidizing agent since it gains electrons and is reduced. (b) is also true: \(\text{Cr}( ext{OH})_3\) is oxidized. (c) is false as only 6 electrons are involved in reduction per molecule of \(\text{IO}_3^-\), not per atom. (d) is false since not all statements are true.

Key Concepts

Oxidation States

Oxidation states are a helpful concept in chemistry to track the transfer of electrons in chemical reactions, especially in redox reactions. They are assigned to atoms to indicate the degree of oxidation or reduction they have undergone. For every element in a compound, an oxidation state is a hypothetical charge on the atom if all bonds were ionic. In the reaction presented, we find the oxidation states by considering known rules, such as:

• The oxidation state of hydrogen in most compounds is +1.
• Oxygen generally holds an oxidation state of -2.
• The sum of oxidation states for all atoms in a molecule or polyatomic ion equals its charge.

For Iodine in \( \text{IO}_3^- \), the state is +5, decreasing to -1 in \( \text{I}^- \), indicating a gain of electrons. Chromium in \( \text{Cr(OH)}_3 \) starts at +3, increasing to +6 in \( \text{CrO}_4^{2-} \), showing a loss of electrons. By assigning these states, we can succinctly discern which species are oxidized and which ones are reduced.

Oxidizing Agent

An oxidizing agent is a substance that has the capability to oxidize other substances by accepting electrons from them, and in redox reactions, this is the substance that gets reduced. In the reaction provided, \( \text{IO}_3^- \) acts as an oxidizing agent. Initially, Iodine is in a +5 oxidation state and transforms into \( \text{I}^- \) with an oxidation state of -1. This transformation means Iodine gains electrons, hence is reduced.
An important point to grasp about oxidizing agents is that they drive the oxidation process by becoming reduced themselves. This is crucial in context: while Chromate (\( \text{CrO}_4^{2-} \)) shows Chromium's increase in oxidation state (evidence of oxidation), Iodine's decrease clearly identifies its role as the oxidizing agent in this process.

Reduction and Oxidation

Reduction and Oxidation are two halves of a complete redox reaction where electrons are exchanged between participating substances. **Reduction** is defined as the gain of electrons, and **oxidation** is the loss of electrons. Redox reactions are characterized by two main changes: one species undergoes oxidation (losing electrons), while another undergoes reduction (gaining electrons).
In this exercise: - **Chromium (\( \text{Cr} \))** is oxidized, losing electrons as its oxidation state increases from +3 in \( \text{Cr(OH)}_3 \) to +6 in \( \text{CrO}_4^{2-} \). - **Iodine (\( \text{I} \))** is reduced, gaining electrons as its oxidation state decreases from +5 in \( \text{IO}_3^- \) to -1 in \( \text{I}^- \).These transformations are part of balanced and interconnected processes that uphold the principle that electrons lost are always equal to electrons gained. This maintains charge neutrality across the reaction.