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Chapter 18: Problem 18

The Ostwald process for the commercial production of nitric acid involves the following three steps: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ a. Which reactions in the Ostwald process are oxidationreduction reactions? b. Identify each oxidizing agent and reducing agent.

Short Answer

Expert verified
a. All three reactions in the Ostwald process are oxidation-reduction reactions. b. - In Reaction 1, NH3 is the reducing agent, and O2 is the oxidizing agent. - In Reaction 2, NO is the reducing agent, and O2 is the oxidizing agent. - In Reaction 3, NO2 acts as both the reducing agent (for the +4 to +5 change) and the oxidizing agent (for the +4 to +2 change).

Step by step solution

01

Assign oxidation states

For each reaction, we will assign oxidation states to each atom in the reactants and products. Remember the following rules for assigning oxidation states: - The oxidation state of an element in its pure form is 0. - The oxidation state of group 1 elements is +1, and group 2 elements is +2. - The oxidation state of oxygen is usually -2, except in peroxides where it is -1. - The oxidation state of hydrogen is +1 in most of its compounds. - The sum of the oxidation states of all atoms in a compound is 0. Reaction 1: \(4NH_3(g) + 5O_2(g) \longrightarrow 4NO(g) + 6H_2O(g)\) N: -3 (in NH3), +2 (in NO) H: +1 O: 0 (in O2), -2 (in H2O) Reaction 2: \(2NO(g) + O_2(g) \longrightarrow 2NO_2(g)\) N: +2 (in NO), +4 (in NO2) O: 0 (in O2), -2 (in NO2) Reaction 3: \(3NO_2(g) + H_2O(l) \longrightarrow 2HNO_3(aq) + NO(g)\) N: +4 (in NO2), +5 (in HNO3), +2 (in NO) O: -2 (in H2O and HNO3) H: +1
02

Identify redox reactions

Now we will compare the oxidation states of atoms in reactants and products to identify if any redox reaction has taken place. a. Reaction 1: N: -3 (NH3) to +2 (NO), which represents an increase in oxidation state, i.e., oxidation O: 0 (O2) to -2 (H2O), which represents a decrease in oxidation state, i.e., reduction So, Reaction 1 is a redox reaction. b. Reaction 2: N: +2 (NO) to +4 (NO2), which represents an increase in oxidation state, i.e., oxidation O: 0 (O2) to -2 (NO2), which represents a decrease in oxidation state, i.e., reduction So, Reaction 2 is a redox reaction. c. Reaction 3: N: +4 (NO2) to +5 (HNO3) and +2 (NO), where +4 to +5 represents an increase in oxidation state, i.e., oxidation and +4 to +2 represents a decrease in oxidation state, i.e., reduction. So, Reaction 3 is a redox reaction.
03

Identify oxidizing and reducing agents

Now, we will identify the oxidizing and reducing agents in the above redox reactions: a. Reaction 1: NH3 is the reducing agent as it is getting oxidized, and O2 is the oxidizing agent as it is getting reduced. b. Reaction 2: NO is the reducing agent as it is getting oxidized, and O2 is the oxidizing agent as it is getting reduced. c. Reaction 3: NO2 is both the reducing agent (for the +4 to +5 change) and the oxidizing agent (for the +4 to +2 change).

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Key Concepts

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

Redox Reactions
Redox reactions, or oxidation-reduction reactions, are a fundamental class of chemical reactions that involve the transfer of electrons between two substances. The term 'redox' comes from 'reduction' and 'oxidation,' which are the two complementary processes that occur during such reactions. In reduction, an atom or molecule gains electrons, thereby decreasing its oxidation state. Conversely, in oxidation, an atom or molecule loses electrons, which increases its oxidation state.

For example, in the Ostwald process, ammonia (NH3) is oxidized to form nitric oxide (NO) while oxygen molecules are reduced to form water (H2O), signifying a redox reaction. The oxidation state of nitrogen increases from -3 to +2, and the oxygen is reduced as it gains electrons to go from an oxidation state of 0 to -2. To determine if a redox reaction has taken place, we must compare the oxidation states of all atoms in the reactants and products, as done in the step-by-step solution of the exercise.
Oxidizing Agent
An oxidizing agent, also known as an oxidant, is a substance that promotes oxidation in a redox reaction by receiving electrons from another substance. It is called an 'agent' because it causes another substance to oxidize. In the context of the Ostwald process, oxygen (O2) is the oxidizing agent in the first reaction, as it gains electrons from ammonia (NH3), leading to the formation of water (H2O).

An effective way to identify the oxidizing agent in a redox reaction is by looking for the element that undergoes a reduction in its oxidation state. In the reactions of the Ostwald process, we see oxygen transforming from a free diatomic molecule with an oxidation state of 0 to becoming part of water with an oxidation state of -2, clearly indicating it has been reduced and hence is the oxidizing agent.
Reducing Agent
In contrast to an oxidizing agent, a reducing agent, or reductant, is a substance that donates electrons to another substance during a redox reaction, causing the other substance to be reduced. As it donates electrons, the reducing agent itself undergoes oxidation. For example, in the Ostwald process, ammonia (NH3) acts as the reducing agent in the first reaction because it loses electrons to oxygen and is oxidized as a result, transforming into nitric oxide (NO).

A helpful approach to pinpoint the reducing agent is to identify which species is increasing in oxidation state in the course of the reaction. In the Ostwald process's first step, ammonia's oxidation state increases from -3 to +2, demonstrating that it has lost electrons and thus is the reducing agent. Note that sometimes one species can be both an oxidizing and reducing agent in different steps of a process, as seen with NO2 in the Ostwald process.

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Most popular questions from this chapter

The free energy change for a reaction, \(\Delta G\), is an extensive property. What is an extensive property? Surprisingly, one can calculate \(\Delta G\) from the cell potential, \(\mathscr{E}\), for the reaction. This is surprising because \(\mathscr{E}\) is an intensive property. How can the extensive property \(\Delta G\) be calculated from the intensive property \(\mathscr{E}\) ?

A galvanic cell is based on the following half-reactions: $$ \begin{aligned} \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ} &=0.34 \mathrm{~V} \\ \mathrm{~V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) & \mathscr{E}^{\circ} &=-1.20 \mathrm{~V} \end{aligned} $$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M\), and the vanadium compartment contains a vanadium electrode and \(\mathrm{V}^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{~L}\) of solution) was titrated with \(0.0800 M \mathrm{H}_{2} \mathrm{EDTA}^{2-}\), resulting in the reaction $$ \begin{array}{r} \mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \\ K=? \end{array} $$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of \(500.0 \mathrm{~mL} \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell }}\) was observed to be \(1.98 \mathrm{~V}\). The solution was buffered at a pH of \(10.00\). a. Calculate \(\mathscr{E}_{\text {cell }}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K\), for the titration reaction. c. Calculate \(\mathscr{B}_{\text {cell }}\) at the halfway point in the titration.

Under standard conditions, what reaction occurs, if any, when each of the following operations is performed? a. Crystals of \(\mathrm{I}_{2}\) are added to a solution of \(\mathrm{NaCl}\). b. \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of NaI. c. A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\). d. An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air. For the reactions that occur, write a balanced equation and calculate \(\mathscr{8}^{\circ}, \Delta G^{\circ}\), and \(K\) at \(25^{\circ} \mathrm{C}\).

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? a. molten KF b. molten \(\mathrm{CuCl}_{2}\) c. molten \(\mathrm{MgI}_{2}\)

Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the \(\mathrm{AuCl}_{4}^{-}\) ion and gaseous \(\mathrm{NO}\). Write a balanced equation for the dissolution of gold in aqua regia.
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