Problem 2 - Chemistry Matter and Change

Question

Identify what is oxidized and what is reduced in the following processes. $$ \begin{array}{l}{\text { a. } 2 \mathrm{Br}^{-}+\mathrm{Cl}_{2} \rightarrow \mathrm{Br}_{2}+2 \mathrm{Cl}^{-}} \\ {\text { b. } 2 \mathrm{Ce}+3 \mathrm{Cu}^{2+} \rightarrow 3 \mathrm{Cu}+2 \mathrm{Ce}^{3+}} \\ {\text { c. } 2 \mathrm{zn}+\mathrm{O}_{2} \rightarrow 2 \mathrm{nO}} \\ {\text { d. } 2 \mathrm{Na}+2 \mathrm{H}^{+} \rightarrow 2 \mathrm{Na}^{+}+\mathrm{H}_{2}}\end{array} $$

Step-by-Step Solution

Answer

a) \( \mathrm{Br}^{-} \) oxidized, \( \mathrm{Cl}_{2} \) reduced. b) \( \mathrm{Ce} \) oxidized, \( \mathrm{Cu}^{2+} \) reduced. c) \( \mathrm{Zn} \) oxidized, \( \mathrm{O}_{2} \) reduced. d) \( \mathrm{Na} \) oxidized, \( \mathrm{H}^{+} \) reduced.

Step 1: Understanding Redox Reaction in Part (a)

For the reaction \( 2 \mathrm{Br}^{-} + \mathrm{Cl}_{2} \rightarrow \mathrm{Br}_{2} + 2 \mathrm{Cl}^{-}\), identify the oxidation states. \( \mathrm{Br}^{-} \) is at -1 and \( \mathrm{Cl}_{2} \) at 0. \( \mathrm{Br}_{2} \) is 0, and \( \mathrm{Cl}^{-} \) is -1. \( \mathrm{Br}^{-} \) is oxidized to \( \mathrm{Br}_{2} \), and \( \mathrm{Cl}_{2} \) is reduced to \( \mathrm{Cl}^{-} \).

Step 2: Understanding Redox Reaction in Part (b)

For the reaction \( 2 \mathrm{Ce} + 3 \mathrm{Cu}^{2+} \rightarrow 3 \mathrm{Cu} + 2 \mathrm{Ce}^{3+}\), identify oxidation states. \( \mathrm{Ce} \) is 0 and changes to +3 in \( \mathrm{Ce}^{3+} \). \( \mathrm{Cu}^{2+} \) is +2 and changes to 0 in \( \mathrm{Cu} \). \( \mathrm{Ce} \) is oxidized, and \( \mathrm{Cu}^{2+} \) is reduced.

Step 3: Understanding Redox Reaction in Part (c)

For the reaction \( 2 \mathrm{Zn} + \mathrm{O}_{2} \rightarrow 2 \mathrm{ZnO} \), identify oxidation states. \( \mathrm{Zn} \) is 0 and changes to +2 in \( \mathrm{ZnO} \). \( \mathrm{O}_{2} \) is 0 and changes to -2 in \( \mathrm{ZnO} \). \( \mathrm{Zn} \) is oxidized, and \( \mathrm{O}_{2} \) is reduced.

Step 4: Understanding Redox Reaction in Part (d)

For the reaction \( 2 \mathrm{Na} + 2 \mathrm{H}^{+} \rightarrow 2 \mathrm{Na}^{+} + \mathrm{H}_{2} \), identify oxidation states. \( \mathrm{Na} \) is 0 and changes to +1 in \( \mathrm{Na}^{+} \). \( \mathrm{H}^{+} \) is +1 and changes to 0 in \( \mathrm{H}_{2} \). \( \mathrm{Na} \) is oxidized, and \( \mathrm{H}^{+} \) is reduced.

Key Concepts

Oxidation States

In chemistry, oxidation states help us understand how electrons are distributed in molecules and ions during a chemical reaction. They are hypothetical charges assigned to atoms in a chemical compound. To analyze redox reactions, we assign oxidation states to all elements involved. The basic rule is that the sum of the oxidation states in a neutral molecule is zero, and in a polyatomic ion, it equals the ion's charge. For instance:

• In a simple ion such as Br^-, the oxidation state is -1 because it has gained an electron.
• In Cl₂, consisting of neutral molecules, each chlorine atom has an oxidation state of 0, as they are equally sharing electrons.

Understanding oxidation states is essential for identifying which substance is oxidized and which is reduced in a reaction.

Oxidation

Oxidation is a process where a substance loses electrons, leading to an increase in its oxidation state. This happens during redox reactions, with the oxidized species acting as the reducing agent, which means it donates electrons to another substance. For example, in the reaction \[2 \text{Br}^{-} + \text{Cl}_2 \rightarrow \text{Br}_2 + 2 \text{Cl}^{-}\] we observe:

• Each Br^- ion starts with an oxidation state of -1. It loses an electron to form neutral Br₂ with an oxidation state of 0. Hence, Br^- is oxidized.

This loss of electrons is a hallmark of oxidation, emphasizing its role in electron transfer reactions.

Reduction

Reduction is essentially the opposite of oxidation. It involves gaining electrons, which results in a decrease in oxidation state. In redox reactions, the reduced substance is the oxidizing agent, as it accepts electrons from another substance. For instance, in the reaction \[2 \text{Ce} + 3 \text{Cu}^{2+} \rightarrow 3 \text{Cu} + 2 \text{Ce}^{3+}\] we observe:

• The Cu²⁺ ions begin with an oxidation state of +2 and gain enough electrons to become neutral Cu, with an oxidation state of 0. Thus, the Cu²⁺ ions are reduced.

This process underscores the importance of electron acceptance in establishing reduction.

Chemical Reactions

Chemical reactions are the transformative processes where substances, known as reactants, convert into different substances called products. These transformations often involve breaking and forming bonds with the exchange or sharing of electrons. In redox reactions, the hallmark of chemical change involves electron transfer between reactants. Let's explore with an example from our original exercise: \[2 \text{Na} + 2 \text{H}^{+} \rightarrow 2 \text{Na}^{+} + \text{H}_2\] In this reaction:

• Two Na atoms (with oxidation states starting at 0) lose electrons to form Na⁺, signifying oxidation.
• Two H⁺ ions (starting at +1) gain electrons to form H₂, indicating reduction.

This showcases the dynamic interplay of electron transfer, highlighting the principles of oxidation and reduction which are core to many chemical reactions.