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) \(\mathrm{IO}_{3}{\underline{\phantom{xx}}}^{-}\)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

(d) all are true.

Step 1: Identify the Oxidizing Agent

An oxidizing agent is a substance that causes another substance to lose electrons and itself gains electrons, undergoing a reduction. In the given reaction, we identify that \(\mathrm{IO}_{3}^{-}\) is reduced to \(\mathrm{I}^{-}\) while \(\mathrm{Cr}( ext{OH})_{3}\) is oxidized. Therefore, \(\mathrm{IO}_{3}^{-}\) is the oxidizing agent because it gains electrons (is reduced).

Step 2: Determine the Substance Being Oxidized

In the reaction, \(\mathrm{Cr}( ext{OH})_{3}\) is transformed into \(\mathrm{CrO}_{4}^{2-}\). Since \(\mathrm{Cr}( ext{OH})_{3}\) loses electrons during this transformation, it is being oxidized. Thus, the statement that \(\mathrm{Cr}( ext{OH})_{3}\) is oxidized is correct.

Step 3: Evaluate Electron Transfer

To determine how many electrons are transferred, examine the change in oxidation state. In \(\mathrm{IO}_{3}^{-}\) to \(\mathrm{I}^{-}\), the iodine goes from a +5 to -1 oxidation state, indicating it gains 6 electrons. Similarly, consider the chromium change: each Cr goes from an oxidation state in \(\mathrm{Cr}( ext{OH})_{3}\) to +6 in \(\mathrm{CrO}_{4}^{2-}\), losing 3 electrons. Thus the reaction balances with 6 electrons being exchanged in total.

Step 4: Compare Statements with Analysis

Evaluate the correctness of each given statement with the analysis: - (a) True: \(\mathrm{IO}_{3}^{-}\) is indeed the oxidizing agent.- (b) True: \(\mathrm{Cr}( ext{OH})_{3}\) is oxidized.- (c) True: 6 electrons are transferred in total per atom (considering the stoichiometric ratios).Thus, all statements are true.

Key Concepts

Oxidizing Agent

In an oxidation-reduction reaction, the oxidizing agent plays a crucial role. It is a substance that gains electrons and gets reduced in the process, causing another substance to lose electrons and become oxidized. In our reaction of interest, the species \[\text{IO}_3^-\] acts as the oxidizing agent. This is because it is reduced to \[\text{I}^-\] as it gains electrons. To remember, you can think of the oxidizing agent as the electron hog. It "hogs" the electrons from the substance it's interacting with, leading to that substance's oxidation. The oxidizing agent is always reduced itself in the reaction. So, next time you face an oxidation-reduction reaction, look for changes in the oxidation states: the substance undergoing a decrease in oxidation state is your oxidizing agent.

Electron Transfer

Electron transfer is at the heart of any oxidation-reduction (redox) reaction. It's the movement of electrons from one reactant to another. In the given reaction:\[\text{Cr(OH)}_3 + \text{OH}^- + \text{IO}_3^- \rightarrow \text{CrO}_4^{2-} + \text{H}_2\text{O} + \text{I}^-\]electrons are transferred during the change of oxidation states of the elements involved. Let's break it down:

• The reduction of \[\text{IO}_3^-\] to \[\text{I}^-\] involves a gain of 6 electrons by iodine, evident by the change in its oxidation state.
• On the other hand, \[\text{Cr(OH)}_3\] is oxidized to \[\text{CrO}_4^{2-}\] by losing 3 electrons per chromium atom.

The electrons lost by chromium are then gained by iodine, thereby balancing the electron transfer in the reaction. These movements are key for understanding redox reactions, as they highlight the interaction between substances undergoing oxidation and reduction.

Oxidation State

The concept of oxidation state is pivotal in understanding and predicting the behavior of compounds in reactions. Oxidation state refers to the degree of oxidation of an atom in a chemical compound, which can be thought of as an indicator of the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. It gives us a way to keep track of electron transfer in oxidation-reduction reactions.

For example, in the transition of \[\text{IO}_3^-\] to \[\text{I}^-\], the oxidation state of iodine changes from +5 to -1. This significant shift showcases a gain of 6 electrons. Meanwhile, in the transformation of \[\text{Cr(OH)}_3\] to \[\text{CrO}_4^{2-}\], the chromium oxidation state shifts from +3 to +6, indicating a loss of 3 electrons. By comparing these oxidation states before and after the reaction, one can determine which elements are oxidized and reduced. Learning to calculate and interpret oxidation states is essential for mastering redox reactions.