Question

Excess of 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) \(\mathrm{CuI}\), is formed (d) Evolved \(\mathrm{I}_{2}\) is reduced

Short answer

The incorrect statement is (c) \( \mathrm{CuI} \) is formed.

Step-by-step solution

Understand the Reaction

When excess KI reacts with \( \mathrm{CuSO}_{4} \), it forms \( \mathrm{CuI} \) and \( \mathrm{I}_2 \). The reaction is: \[ 2\mathrm{CuSO}_{4} + 4\mathrm{KI} \rightarrow \mathrm{Cu}_2\mathrm{I}_2 + \mathrm{I}_2 + 2\mathrm{K}_2\mathrm{SO}_4 \]Here, \( \mathrm{CuI} \) and \( \mathrm{I}_2 \) are produced in the reaction.

Analyze the Role of \( \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \)

\( \mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \) is a reducing agent. When added to the reaction mixture, it reduces \( \mathrm{I}_2 \) back to \( \mathrm{I}^- \), while itself being oxidized to \( \mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6 \). The reaction is:\[ \mathrm{I}_2 + 2\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \rightarrow 2\mathrm{NaI} + \mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6 \]

Incorrect Statement Identification

Evaluate each statement:- (a) Correct: \( \mathrm{Cu}_2\mathrm{I}_2 \) is formed in the reaction with \( \mathrm{CuSO}_{4} \).- (b) Correct: \( \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \) acts as a reducing agent and gets oxidized to \( \mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6 \).- (c) Incorrect: \( \mathrm{CuI} \) is not formed directly; rather \( \mathrm{Cu}_2\mathrm{I}_2 \) is formed.- (d) Correct: \( \mathrm{I}_2 \) evolved is reduced back to \( \mathrm{I}^- \) by \( \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \).

Conclusion

Based on the evaluation of each statement, the incorrect statement about the reaction is (c) \"\( \mathrm{CuI} \) is formed.\" The correct compound formed is \( \mathrm{Cu}_2\mathrm{I}_2 \), not \( \mathrm{CuI} \).

Key concepts

Oxidation States

In the world of chemistry, oxidation states are a way of keeping track of electrons in a chemical reaction. They help us understand how electrons are transferred and which elements gain or lose them. When you look at a chemical compound, oxidation states can give you an idea of the composition and the behavior during reactions.

In our exercise, we see a redox reaction taking place, where KI reacts with \(\mathrm{CuSO}_4\). Here, copper in \(\mathrm{CuSO}_4\) has an oxidation state of +2. As the reaction proceeds, copper gets reduced to form \(\mathrm{Cu}_2\mathrm{I}_2\). This means the oxidation state of copper decreases. In contrast, iodine's oxidation state increases, as it is converted from \(I^-\) ion to \(\mathrm{I}_2\), which we recognize as elemental iodine.

This interplay of oxidation states allows us to see which atoms are oxidized and which are reduced. Keeping track of these changes tells us so much about the roles different species play in a reaction.

Reducing Agents

Reducing agents are substances that donate electrons in a chemical reaction. They themselves get oxidized in the process. In our reaction, \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\) acts as a reducing agent.

When it is added to the products of the initial reaction between KI and \(\mathrm{CuSO}_4\), it immediately starts reducing \(\mathrm{I}_2\) back to \(\mathrm{I}^-\). During this process, \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\) becomes oxidized to \(\mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6\).

• \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\) gives electrons to iodine, making it an excellent example of a reducing agent in action.
• Through this reaction, it neutralizes the iodine by converting it back to iodide ions, effectively reversing part of the initial chemical transformation.

This cycle of oxidation and reduction facilitated by reducing agents like \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\) is central to how many chemical processes occur.

Inorganic Chemistry

Inorganic chemistry is a broad field that studies compounds that generally do not contain carbon. This includes minerals, metals, and salts, many of which are key players in reactions studied through exercises like this.

The reaction between \(\mathrm{CuSO}_4\) and KI falls under inorganic chemistry because it involves ionic compounds and transition metals. These reactants lead to the formation of copper iodide, \(\mathrm{Cu}_2\mathrm{I}_2\), and iodine, \(\mathrm{I}_2\).

• \(\mathrm{CuSO}_4\) and KI are both inorganic compounds, commonly seen in laboratory work.
• Understanding the role these substances play highlights the importance of electronegativity, electron exchange, and oxidation states in the study of chemistry.

The behavior of inorganic compounds often involves processes like precipitation, acid-base reactions, and redox changes, making their study crucial for understanding chemical reactions across various applications and industries.