Problem 46 - Chemistry Matter and Change

Question

Identify each of these half-reactions as either oxidation or reduction. $$ \begin{array}{l}{\text { a. } \mathrm{Al} \rightarrow \mathrm{Al}^{3+}+3 \mathrm{e}^{-}} \\ {\text { b. } \mathrm{Cu}^{2+}+\mathrm{e}^{-} \rightarrow \mathrm{Cu}^{+}}\end{array} $$

Step-by-Step Solution

Answer

Half-reaction a is oxidation; half-reaction b is reduction.

Step 1: Analyze Half-Reaction a

Let's examine the half-reaction \( \mathrm{Al} \rightarrow \mathrm{Al}^{3+} + 3\mathrm{e}^{-} \). Here, an aluminum atom (Al) loses three electrons to become \( \mathrm{Al}^{3+} \). Since oxidation involves the loss of electrons, this half-reaction represents oxidation.

Step 2: Identify Type of Reaction for a

Since the aluminum atom is losing electrons, half-reaction a is classified as an oxidation process.

Step 1: Analyze Half-Reaction b

Consider the half-reaction \( \mathrm{Cu}^{2+} + \mathrm{e}^{-} \rightarrow \mathrm{Cu}^{+} \). In this reaction, copper ion \( \mathrm{Cu}^{2+} \) is gaining an electron to form \( \mathrm{Cu}^{+} \). Gaining electrons corresponds to reduction.

Step 2: Identify Type of Reaction for b

Since the copper ion is gaining electrons, half-reaction b is classified as a reduction process.

Key Concepts

Understanding Oxidation

Oxidation is a key concept in redox reactions and is defined as the process where an atom, ion, or molecule loses electrons. This loss of electrons results in an increase in oxidation state. In redox reactions, oxidation is always accompanied by reduction. It's vital to recognize oxidation in half-reactions. Consider the exercise's example:

• The half-reaction \( \mathrm{Al} \rightarrow \mathrm{Al}^{3+} + 3\mathrm{e}^{-} \) involves aluminum losing three electrons, transforming into \( \mathrm{Al}^{3+} \).
• Since there's a loss of electrons, aluminum's oxidation state increases from 0 to +3.

This transformation reveals that aluminum is undergoing oxidation. When remembering oxidation, think of the mnemonic "OIL RIG": Oxidation Is Loss (of electrons). This helps keep the concept simple and helps distinguish between oxidation and its complementary process, reduction.

Reduction Explained

Reduction is the opposite of oxidation in redox reactions. It describes the process where an atom, ion, or molecule gains electrons. This gain in electrons leads to a decrease in oxidation state.Referring to the given exercise:

• The half-reaction \( \mathrm{Cu}^{2+} + \mathrm{e}^{-} \rightarrow \mathrm{Cu}^{+} \) demonstrates copper gaining an electron.
• The oxidation state of copper decreases from +2 to +1, indicating reduction.

Reduction is crucial as it always occurs alongside oxidation. The combined effect of oxidation and reduction allows for the flow of electrons, which is the foundation of energy transfer in many biochemical and industrial processes.To help remember, use "OIL RIG": Reduction Is Gain (of electrons). This can quickly remind you that reduction involves gaining electrons, setting it apart from oxidation.

Half-Reaction Analysis

Half-reaction analysis is a method used to determine the type of reaction—either oxidation or reduction—involved in a redox process.To analyze a half-reaction, follow these steps:

• Identify the species in the reaction that is losing or gaining electrons.
• Look at the change in oxidation states to determine if the species is undergoing oxidation or reduction.

In the given examples:1. **For Aluminum:** In the half-reaction \( \mathrm{Al} \rightarrow \mathrm{Al}^{3+} + 3\mathrm{e}^{-} \), the aluminum loses electrons, indicating oxidation.2. **For Copper:** In the half-reaction \( \mathrm{Cu}^{2+} + \mathrm{e}^{-} \rightarrow \mathrm{Cu}^{+} \), the copper ion gains an electron, which is characteristic of reduction.By systematically breaking down each half-reaction, you can easily determine the flow of electrons and how each reactant is transformed. This approach not only clarifies individual reactions but also helps in balancing redox equations, ensuring a deep understanding of electron transfer in chemistry.