Problem 59 - General Chemistry: Principles and Modern Applications

Question

Attempts to make \(\mathrm{CuI}_{2}\) by the reaction of \(\mathrm{Cu}^{2+}(\mathrm{aq})\) and \(\mathrm{I}^{-}(\text {aq })\) produce \(\mathrm{CuI}(\mathrm{s})\) and \(\mathrm{I}_{3}^{-}(\mathrm{aq})\) instead. Without performing detailed calculations, show why this reaction should occur. $$2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{I}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{CuI}(\mathrm{s})+\mathrm{I}_{3}^{-}(\mathrm{aq})$$

Step-by-Step Solution

Answer

The reaction \[2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{I}^{-}(\mathrm{aq})\longrightarrow 2 \mathrm{CuI}(\mathrm{s})+\mathrm{I}_{3}^{-}(\mathrm{aq})\] produces \(\mathrm{CuI}(\mathrm{s})\) and \(\mathrm{I}_{3}^{-}(\mathrm{aq})\) instead of \(\mathrm{CuI}_{2}\) because these are the more thermodynamically stable compounds. Iodides are less stable than other halides which explains why \( \mathrm{I}_{3}^{-}(\mathrm{aq}) \) is formed.

Step 1: Analyze the given reaction

Looking at the reaction, \[2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{I}^{-}(\mathrm{aq})\longrightarrow 2 \mathrm{CuI}(\mathrm{s})+\mathrm{I}_{3}^{-}(\mathrm{aq})\], we see that when \(\mathrm{Cu}^{2+}(\mathrm{aq})\) and \(\mathrm{I}^{-}(\text {aq })\) are combined, they produce \(\mathrm{CuI}(\mathrm{s})\) and \(\mathrm{I}_{3}^{-}(\mathrm{aq})\), not \(\mathrm{CuI}_{2}\).

Step 2: Identify thermodynamically stable products

In a chemical reaction, the products that are formed are generally the ones that have lower energy and are more stable. In this case, \(\mathrm{CuI}(\mathrm{s})\) and \(\mathrm{I}_{3}^{-}(\mathrm{aq})\) are more stable than \(\mathrm{CuI}_{2}\). Therefore, \(\mathrm{CuI}(\mathrm{s})\) and \(\mathrm{I}_{3}^{-}(\mathrm{aq})\) are produced instead of \(\mathrm{CuI}_{2}\).

Step 3: Justification based on Periodic trends

In the periodic table, as we go down the group, the halides become less stable. That means iodides (I-) are less stable when compared to the upper halides which justifies why \(\mathrm{I}_{3}^{-}\) is produced.

Key Concepts

Copper Iodide Formation

The formation of Copper Iodide, specifically CuI, relates to how copper ions (\(\text{Cu}^{2+}\)) interact with iodide ions (\(\text{I}^{-}\)). When copper(II) ions mix with iodide ions in an aqueous solution, they don't form the expected \(\text{CuI}_2\). Instead, the reaction yields CuI precipitate and \(\text{I}_3^{-}\) ions. CuI is a solid and forms due to the

• High stability of the CuI compound,
• The limited solubility of CuI in water.

These factors drive the reaction to produce a different set of products than one might initially expect. Thus, understanding the chemical environment and properties of these ions is key to predicting the reaction outcome.

Thermodynamic Stability

In chemistry, products that are more stable have lower energies. This concept helps to predict which products form in a reaction. In the case of copper iodide reactions,

• CuI is more stable than CuI₂,
• \(\text{I}_3^{-}\) is more stable than free iodide ions.

By achieving a lower-energy state, the system naturally favors forming these stable products. These aspects of chemical stability guide us to understand why certain products, like CuI and \(\text{I}_3^{-}\), are preferred over less stable alternatives even without running calculations.

Periodic Trends

Periodic trends offer great insight into element behavior. As you move down a group in the periodic table, such as the halogen group, the stability of halides decreases. This trend means:

• Iodides (\(\text{I}^{-}\)) are less stable compared to lighter halides like chlorides (\(\text{Cl}^{-}\)).
• This reduction in stability is one reason the \(\text{I}_3^{-}\) ion forms instead of simple iodine compounds.

Understanding these trends helps make sense of why certain reactions, like the formation of copper iodide and polyiodide ions, occur. It shows how intrinsic periodic properties influence chemical reactions and stability choices.