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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

You have a concentration cell with Cu electrodes and [Cu^{2+} ] \(=1.00 M\) (right side) and \(1.0 \times 10^{-4} M\) (left side). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\). b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}\) by the following equation: \(\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)\) \(K=1.0 \times 10^{13}\) Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M}\)

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{~V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{~V}\)

Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\mathrm{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)\) c. \(\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)\)

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains \(1.0 M \mathrm{M}^{2+}\). Solution B in the other cell compartment has a volume of \(1.00 \mathrm{~L}\). At the beginning of the experiment \(0.0100\) mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.0100\) mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be \(0.44 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{~V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\text {sp }}\) for \(\operatorname{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\).

One of the few industrial-scale processes that produce organic compounds electrochemically is used by the Monsanto Company to produce 1,4 -dicyanobutane. The reduction reaction is $$ 2 \mathrm{CH}_{2}=\mathrm{CHCN}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN} $$ The \(\mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) is then chemically reduced using hydrogen gas to \(\mathrm{H}_{2} \mathrm{~N}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{NH}_{2}\), which is used in the production of nylon. What current must be used to produce 150. \(\mathrm{kg} \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) per hour?
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