Question

For the complete redox reactions given here, (i) break down each reaction into its half-reactions; (ii) identify the oxidizing agent; (iii) identify the reducing agent. (a) \(2 \mathrm{Sr}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{SrO}\) (b) \(2 \mathrm{Li}+\mathrm{H}_{2} \longrightarrow 2 \mathrm{LiH}\) (c) \(2 \mathrm{Cs}+\mathrm{Br}_{2} \longrightarrow 2 \mathrm{CsBr}\) (d) \(3 \mathrm{Mg}+\mathrm{N}_{2} \longrightarrow \mathrm{Mg}_{3} \mathrm{~N}_{2}\)

Short answer

The oxidizing agents are \(O_2\), \(H_2\), \(Br_2\), and \(N_2\) for reactions (a), (b), (c), and (d) respectively. The reducing agents are \(Sr\), \(Li\), \(Cs\), and \(Mg\) for reactions (a), (b), (c), and (d) respectively.

Step-by-step solution

Determine the Half-Reactions (Step 1)

The first step is to break down the reactions into half-reactions. For each compound, identify which atoms gain and lose electrons. The half-reactions for each are: (a) \(2 Sr \rightarrow 2 Sr^{2+} + 4 e^-\) (Oxidation half-reaction) and \(O_2 + 4 e^- \rightarrow 2 O^{2-}\) (Reduction half-reaction) (b) \(2 Li \rightarrow 2 Li^+ + 2 e^-\) (Oxidation half-reaction)and \(H_2 + 2 e^- \rightarrow 2 H^-\) (Reduction half-reaction)(c) \(2 Cs \rightarrow 2 Cs^+ + 2 e^-\) (Oxidation half-reaction)and \(Br_2 + 2 e^- \rightarrow 2 Br^-\) (Reduction half-reaction)(d) \(3 Mg \rightarrow 3 Mg^{2+} + 6 e^-\) (Oxidation half-reaction)and \(N_2 + 6 e^- \rightarrow 2 N^{3-}\) (Reduction half-reaction)

Identify the Oxidizing Agents (Step 2)

Oxidizing agents are the species gaining electrons in a redox process. So, from our half-reactions, the oxidizing agents are:(a) \(O_2\) in \(2 Sr + O_2 \rightarrow 2 SrO\),(b) \(H_2\) in \(2 Li + H_2 \rightarrow 2 LiH\),(c) \(Br_2\) in \(2 Cs + Br_2 \rightarrow 2 CsBr\), and (d) \(N_2\) in \(3 Mg + N_2 \rightarrow Mg_3N_2\).

Identify the Reducing Agents (Step 3)

Reducing agents are the species that lose electrons in a redox process. So, from our half-reactions, the reducing agents are:(a) \(Sr\) in \(2 Sr + O_2 \rightarrow 2 SrO\),(b) \(Li\) in \(2 Li + H_2 \rightarrow 2 LiH\),(c) \(Cs\) in \(2 Cs + Br_2 \rightarrow 2 CsBr\), and (d) \(Mg\) in \(3 Mg + N_2 \rightarrow Mg_3N_2\).

Key concepts

Oxidation and Reduction Half-Reactions

Understanding redox reactions is pivotal in the field of chemistry, where the heart of these reactions lies in the oxidation and reduction processes. In any redox reaction, there are always two simultaneous changes occurring—a loss of electrons, known as oxidation, and a gain of electrons, termed reduction.

Let's break down half-reactions using a simple example from the textbook exercise: For reaction (a), the oxidation half-reaction is, \(2 Sr \rightarrow 2 Sr^{2+} + 4 e^-\). Here, strontium (Sr) atoms lose electrons becoming positively charged ions. Conversely, the reduction half-reaction is \(O_2 + 4 e^- \rightarrow 2 O^{2-}\). Oxygen (O2) gains electrons to form oxide ions. It's important to recognize that the electrons lost in the oxidation process are the same ones gained in the reduction process.

Reading Redox Reactions

Just like in the given solutions, dividing the overall equation into half-reactions allows students to identify precisely what gets oxidized and what gets reduced. This approach simplifies complex reactions into manageable parts, helping students identify electron flow and understand the reaction's direction. Remember, oxidation and reduction always occur together; you cannot have one without the other in a redox reaction.

Oxidizing Agents

An oxidizing agent, also known as an oxidant, is a substance that promotes oxidation by accepting electrons from another substance. In doing so, it gets reduced itself. From the textbook solutions, we see examples of oxidizing agents in various reactions. For instance, in reaction (a), \(O_2\) acts as the oxidizing agent because it accepts electrons during the process.

Identifying Oxidizing Agents

When looking at half-reactions, the species that gains electrons (and is therefore reduced) is your oxidizing agent. It could be a simple element like oxygen or a compound. In all the given reactions from the solutions, the oxidizing agents are the non-metals or the elements that are not in their elemental state: \(O_2\), \(H_2\), \(Br_2\), and \(N_2\). By accepting electrons, these agents facilitate the oxidation of the other species in the reaction, which is why understanding their role is crucial in analyzing redox reactions.

Reducing Agents

In contrast to oxidizing agents, a reducing agent (or reductant) donates electrons to another substance, causing the other substance to be reduced while being oxidized itself. From the exercise above, strontium (Sr) in reaction (a) is the reducing agent as it donates electrons, becoming oxidized to Sr2+ ions in the process.

Role of Reducing Agents

The reducing agents seen in the solutions are all metals that lose electrons and form cations. They include \(Sr\), \(Li\), \(Cs\), and \(Mg\). Each metal gives away electrons readily, making them excellent reducing agents. By donating electrons, they are indispensable in driving redox reactions; without them, other substances wouldn’t be able to gain electrons and reduce. Knowing the characteristic of reducing agents to lose electrons will enhance a student's ability to maneuver through complicated redox equations and better predict reaction outcomes.