Problem 132 - A Complete Resource Book in Chemistry for JEE Main

Question

Excess of \(\mathrm{KI}\) react with \(\mathrm{CuSO}_{4}\) solution and then \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution is added to it. Which of the following statements is incorrect for this reaction? [2004] (a) \(\mathrm{Cu}_{2} \mathrm{I}_{2}\) is formed (b) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) is oxidized (c) \(\mathrm{CuI}_{2}\) is formed (d) evolved \(\mathrm{I}\), is reduced

Step-by-Step Solution

Answer

The incorrect statement is (c) CuI2 is formed.

Step 1: Understand the Reaction Between KI and CuSO4

When excess KI reacts with CuSO4, a reaction occurs where copper(II) ions are reduced to copper(I) ions, resulting in the formation of copper(I) iodide, Cu2I2, as a precipitate. The iodide ions are oxidized to iodine, I2. This is a redox reaction.

Step 2: Analyze the Role of Na2S2O3

The Na2S2O3 reacts with the iodine produced in the first step. Thiosulfate ions (S2O3^2-) are strong reducing agents and will reduce I2 back to iodide ions (I-), themselves getting oxidized in the process. This means that thiosulfate acts as a reducer here.

Step 3: Evaluate Each Statement

(a) Cu2I2 is the correct product of the reaction between KI and CuSO4, so this statement is correct. (b) Na2S2O3 indeed gets oxidized when it reduces I2, so this statement is correct. (c) CuI2 does not form; only Cu2I2 forms, making this statement incorrect. (d) Iodine (I2) is reduced back to I- ions by Na2S2O3, completing the conversion, thus this statement is correct.

Key Concepts

Copper(I) Iodide Formation

The formation of Copper(I) iodide, represented as \(\mathrm{Cu}_2\mathrm{I}_2\), is an essential part of redox reactions involving copper sulfate (\(\mathrm{CuSO}_4\)) and potassium iodide (\(\mathrm{KI}\)).
When \(\mathrm{KI}\) is added in excess to \(\mathrm{CuSO}_4\), the copper(II) ions (\(\mathrm{Cu}^{2+}\)) in the solution undergo a reduction process.
They receive electrons, transitioning from a +2 oxidation state to a +1 oxidation state, creating the copper(I) species.
Consequently, the iodide ions (\(\mathrm{I}^-\)) in \(\mathrm{KI}\) are oxidized to form iodine (\(\mathrm{I}_2\)). This means the iodide ions lose electrons, evident in the formation of \(\mathrm{Cu}_2\mathrm{I}_2\) as a white precipitate.

• This is also why the solution may turn brown, owing to the liberation of iodine.
• This step highlights the reciprocal nature of redox reactions, marked by simultaneous oxidation and reduction processes.

Thiosulfate Reduction

Thiosulfate ions (\(\mathrm{S}_2\mathrm{O}_3^{2-}\)) play a critical role as reducing agents in this chemical reaction.
As the reaction progresses, thiosulfate comes into contact with iodine (\(\mathrm{I}_2\)), where it functions to reduce iodine back to iodide ions (\(\mathrm{I}^-\)).
During this time, thiosulfate ions are oxidized themselves to form tetrathionate (\(\mathrm{S}_4\mathrm{O}_6^{2-}\)).
This transition can be understood from the equation: \[\mathrm{2S}_2\mathrm{O}_3^{2-} + \mathrm{I}_2 \rightarrow \mathrm{2I}^- + \mathrm{S}_4\mathrm{O}_6^{2-}\]

• This reduction of iodine ensures that the original product of copper iodide can remain stable without copious free iodine disrupting the reaction conditions.
• It exemplifies the versatile nature of thiosulfate as agents that can revert oxidized species back to a reduced form.

Iodine Reduction

The reduction of iodine is a pivotal aspect of maintaining the integrity of redox reactions.
When iodine molecules (\(\mathrm{I}_2\)) are formed initially during the copper iodide formation, it is crucial to counteract their oxidative potential.
Adding sodium thiosulfate (\(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\)) serves this purpose efficiently.

• The resulting reaction of thiosulfate with iodine not only rescinds iodine to iodide ions (\(\mathrm{2I}^-\)) but sustains the continuity of the redox cycle.
• This conversion is part of many reactions where iodine is a by-product and needs to be managed to prevent its interference with desired outcomes.

In the context of the original solution, acknowledging the reduction of iodine showcases an understanding of dynamic chemical principles and necessity of controlling intermediate products in synthesis reactions.