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The commercial production of nitric acid involves the following chemical reactions: $$ \begin{gathered} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{gathered} $$ (a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction.

Short Answer

Expert verified
(a) All of the given reactions are redox reactions. (b) The element undergoing oxidation and reduction in each reaction are as follows: - In Reaction 1: Oxidation - N, Reduction - O - In Reaction 2: Oxidation - N, Reduction - O - In Reaction 3: Oxidation - N, Reduction - N.

Step by step solution

01

Reaction 1: Analysis

\(4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g) \) - Oxidation state of N in NH3: -3 - Oxidation state of H in NH3: +1 - Oxidation state of O in O2: 0 - Oxidation state of N in NO: +2 - Oxidation state of O in NO: -2 - Oxidation state of H in H2O: +1 - Oxidation state of O in H2O: -2 Here, the oxidation state of N increases from -3 to +2 (oxidation), and the oxidation state of O decreases from 0 to -2 (reduction). Hence, this is a redox reaction.
02

Reaction 2: Analysis

\(2 \mathrm{NO}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \) - Oxidation state of N in NO: +2 - Oxidation state of O in NO: -2 - Oxidation state of O in O2: 0 - Oxidation state of N in NO2: +4 - Oxidation state of O in NO2: -2 Here, the oxidation state of N increases from +2 to +4 (oxidation), and the oxidation state of O decreases from 0 to -2 (reduction). Hence, this is also a redox reaction.
03

Reaction 3: Analysis

\(3 \mathrm{NO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(aq) + \mathrm{NO}(g) \) - Oxidation state of N in NO2: +4 - Oxidation state of O in NO2: -2 - Oxidation state of H in H2O: +1 - Oxidation state of O in H2O: -2 - Oxidation state of H in HNO3: +1 - Oxidation state of N in HNO3: +5 - Oxidation state of O in HNO3: -2 - Oxidation state of N in NO: +2 - Oxidation state of O in NO: -2 Here, the oxidation state of N decreases from +4 to +2 (reduction), and the oxidation state of N increases from +4 to +5 (oxidation). Hence, this is also a redox reaction. Now that we have identified the redox reactions in each step, we can summarize the oxidizing and reducing elements for each reaction:
04

Reaction 1: Oxidizing and reducing elements

- Oxidizing element: O (oxidation state changes from 0 to -2) - Reducing element: N (oxidation state changes from -3 to +2)
05

Reaction 2: Oxidizing and reducing elements

- Oxidizing element: O (oxidation state changes from 0 to -2) - Reducing element: N (oxidation state changes from +2 to +4)
06

Reaction 3: Oxidizing and reducing elements

- Oxidizing element: N (oxidation state changes from +4 to +5) - Reducing element: N (oxidation state changes from +4 to +2) In conclusion: (a) All of the given reactions are redox reactions. (b) The element undergoing oxidation and reduction in each reaction are as follows: - In Reaction 1: Oxidation - N, Reduction - O - In Reaction 2: Oxidation - N, Reduction - O - In Reaction 3: Oxidation - N, Reduction - N.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation States
When examining redox reactions, the concept of oxidation states is crucial. The oxidation state, also referred to as the oxidation number, indicates the degree of oxidation of an atom in a substance. Think of it as a bookkeeping tool that helps track electrons during chemical reactions. It helps determine how many electrons are lost or gained during a reaction.

In the given chemical reactions, we analyze the oxidation states of elements such as nitrogen (N) and oxygen (O) to identify which elements are oxidized or reduced. For example, in Reaction 1, nitrogen in ammonia ( NH_3 ) starts with an oxidation state of -3 , and in nitric oxide ( NO ), it ends with an oxidation state of +2 . This change indicates that nitrogen is losing electrons, hence it is being oxidized.
Oxidizing Agents
An oxidizing agent is a substance that accepts electrons in a redox reaction, causing another substance to lose electrons. In simpler terms, it oxidizes something else while itself being reduced.

In the context of nitric acid production, oxygen ( O_2 ) is a common oxidizing agent. For instance, in both Reactions 1 and 2 in the provided exercise, oxygen starts with an oxidation state of 0 and ends up with -2 . This change shows that oxygen gains electrons, demonstrating its role as the oxidizing agent in these reactions. Through this process, the oxidizing agent lowers its oxidation state as it garners electrons from the other substance.
Reducing Agents
A reducing agent donates electrons in a chemical reaction, leading to the reduction of another substance. The reducing agent itself becomes oxidized, marking its characteristic role in redox reactions.

In nitric acid production, the role of the reducing agent varies across different reactions. For example, in Reaction 1, nitrogen in ammonia ( NH_3 ) acts as the reducing agent. Here, nitrogen's oxidation state changes from -3 to +2 , showing it leads the charge by passing electrons to oxygen. The release of these electrons is a clear indicator of its function as a reducing agent.
Chemical Reactions
Chemical reactions involve the transformation of one or more substances into others, and redox reactions are a specific type. In redox reactions, there is a simultaneous process of oxidation and reduction.

In these exercises concerning nitric acid production, each reaction represents a redox process. All reactions involve changes in oxidation states, highlighting the interplay between oxidization and reduction. For instance, in Reaction 3, nitrogen experiences both reduction (from +4 to +2 ) and oxidation (from +4 to +5 ). Such interactions are critical to the overall chemical transformations taking place during nitric acid synthesis.
Nitric Acid Production
Nitric acid production is an essential industrial process involving intricately orchestrated chemical reactions. These include a series of redox reactions where ammonia is converted into nitric acid, primarily through the Ostwald process.

The process begins with the oxidation of ammonia to form nitrogen monoxide and water. Subsequent reactions lead to the formation of nitrogen dioxide and ultimately, nitric acid. Each step requires precise control of the conditions and catalysts to ensure efficient conversion. Understanding the roles of oxidizing and reducing agents in these steps is vital for optimizing the production of this important chemical used in fertilizers, explosives, and more.

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Most popular questions from this chapter

Separate samples of a solution of an unknown ionic compound are treated with dilute \(\mathrm{AgNO}_{3}, \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\), and \(\mathrm{BaCl}_{2}\). Precipitates form in all three cases. Which of the following could be the anion of the unknown salt: \(\mathrm{Br}^{-} ; \mathrm{CO}_{3}^{2-} ; \mathrm{NO}_{3}^{-} ?\)

The labels have fallen off two bottles, one containing \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) and the other containing \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\). You have a bottle of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\). How could you use it to test a portion of each solution to identify which solution is which? [Section 4.2]

The metal cadmium tends to form \(\mathrm{Cd}^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q)\), cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)\), nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the observations made above. (b) What can you conclude about the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

We have seen that ions in aqueous solution are stabilized by the attractions between the ions and the water molecules. Why then do some pairs of ions in solution form precipitates? [Section 4.2]

The mass percentage of chloride ion in a 25.00-mL sample of seawater was determined by titrating the sample with silver nitrate, precipitating silver chloride. It took \(42.58 \mathrm{~mL}\) of \(0.2997 \mathrm{M}\) silver nitrate solution to reach the equivalence point in the titration. What is the mass percentage of chloride ion in the seawater if its density is \(1.025 \mathrm{~g} / \mathrm{mL} ?\)

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