Problem 83 - Chemistry Matter and Change

Question

The following equations show redox reactions that are sometimes used in the laboratory to generate pure nitrogen gas and pure dinitrogen monoxide gas (nitrous oxide, \(\left.\mathrm{N}_{2} \mathrm{O}\right)\). \(\mathrm{NH}_{4} \mathrm{NO}_{2}(\mathrm{~s}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{~s}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) a. Determine the oxidation number of each element in the two equations, and then make diagrams showing the changes in oxidation numbers that occur in each reaction. b. Identify the atom that is oxidized and the atom that is reduced in each of the two reactions. c. Identify the oxidizing and reducing agents in each of the two reactions. d. Write a sentence telling how the electron transfer taking place in these two reactions differs from that taking place here. \(2 \mathrm{AgNO}_{3}+\mathrm{Zn} \rightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{Ag}\)

Step-by-Step Solution

Answer

In \(\mathrm{NH}_4 \mathrm{NO}_2\), \(\mathrm{N}\) in ammonium is oxidized, and \(\mathrm{N}\) in nitrite is reduced. In \(\mathrm{NH}_4 \mathrm{NO}_3\), \(\mathrm{N}\) in ammonium is oxidized, and \(\mathrm{N}\) in nitrate is reduced. Electron transfer differs as it involves metals in the last reaction.

Step 1: Determine Oxidation Numbers in Reaction 1

In the reaction: \(\mathrm{NH}_{4} \mathrm{NO}_{2}(\mathrm{~s}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), assign oxidation numbers. For \(\mathrm{NH}_4^+\): \(\mathrm{N} = -3\) and \(\mathrm{H} = +1\). For \(\mathrm{NO}_2^-\): \(\mathrm{N} = +3\) and \(\mathrm{O} = -2\). In \(\mathrm{N}_2\): \(\mathrm{N} = 0\). The \(\mathrm{H}_2\mathrm{O}\) oxidation numbers remain: \(\mathrm{H} = +1\) and \(\mathrm{O} = -2\).

Step 2: Changes in Oxidation Numbers for Reaction 1

Draw diagrams showing \(\mathrm{N}\) in \(\mathrm{NH}_4^+\) goes from \(-3\) to \(0\), oxidized by losing electrons. \(\mathrm{N}\) in \(\mathrm{NO}_2^-\) goes from \(+3\) to \(0\), reduced by gaining electrons.

Step 3: Determine Oxidized and Reduced Elements in Reaction 1

In \(\mathrm{NH}_{4} \mathrm{NO}_{2}\), the \(\mathrm{N}\) in \(\mathrm{NH}_4^+\) gets oxidized, and the \(\mathrm{N}\) in \(\mathrm{NO}_2^-\) gets reduced.

Step 4: Identify Agents in Reaction 1

\(\mathrm{NO}_2^-\) is the oxidizing agent (it gets reduced), and \(\mathrm{NH}_4^+\) is the reducing agent (it gets oxidized).

Step 5: Determine Oxidation Numbers in Reaction 2

In the reaction: \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{~s}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), assign oxidation numbers. \(\mathrm{NH}_4^+\): \(\mathrm{N} = -3\), \(\mathrm{H} = +1\). \(\mathrm{NO}_3^-\): \(\mathrm{N} = +5\), \(\mathrm{O} = -2\). \(\mathrm{N}_2\mathrm{O}\): \(\mathrm{N} = +1\) in one part, and \(\mathrm{N} = 0\) in the other. \(\mathrm{H}_2\mathrm{O}\) oxidation numbers remain the same.

Step 6: Changes in Oxidation Numbers for Reaction 2

Draw diagrams showing \(\mathrm{N}\) in \(\mathrm{NH}_4^+\) goes from \(-3\) to \(+1\), oxidized by losing electrons. \(\mathrm{N}\) in \(\mathrm{NO}_3^-\) goes from \(+5\) to \(0\), reduced by gaining electrons.

Step 7: Determine Oxidized and Reduced Elements in Reaction 2

In \(\mathrm{NH}_{4} \mathrm{NO}_{3}\), the \(\mathrm{N}\) in \(\mathrm{NH}_4^+\) gets oxidized, and the \(\mathrm{N}\) in \(\mathrm{NO}_3^-\) gets reduced.

Step 8: Identify Agents in Reaction 2

\(\mathrm{NO}_3^-\) is the oxidizing agent (it gets reduced), and \(\mathrm{NH}_4^+\) is the reducing agent (it gets oxidized).

Step 9: Electron Transfer Explanation

In reactions 1 and 2, electron transfer involves the oxidation of nitrogen in \(\mathrm{NH}_4^+\) and the reduction of nitrogen in \(\mathrm{NO}_2^-\) or \(\mathrm{NO}_3^-\). In contrast, in \(2 \mathrm{AgNO}_{3} + \mathrm{Zn} \rightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} + 2 \mathrm{Ag}\), zinc oxidizes from 0 to +2, and silver reduces from +1 to 0.

Key Concepts

Oxidation Numbers

Oxidation numbers are crucial for understanding how redox reactions work. They represent the theoretical charges of atoms if all bonds were ionic. This helps us to track electron transfer in chemical reactions. In the reactions provided, we had to determine the oxidation numbers for nitrogen in different compounds to see how they change throughout the reaction. In both reactions, nitrogen experiences a change in oxidation state, which indicates a redox process is occurring.

• In the first reaction: \(\text{NH}_4^+\) has nitrogen with an oxidation number of \(-3\), and in \(\text{NO}_2^-\) nitrogen has an oxidation number of \(+3\).
• When these form nitrogen gas \(\text{N}_2\), the oxidation number becomes \(0\), showing that oxidation and reduction have happened.
• Similarly, in the second reaction: \(\text{NH}_4^+\) again has nitrogen with \(-3\), and \(\text{NO}_3^-\) has nitrogen with \(+5\).

These changes demonstrate how oxidation numbers help us recognize oxidation and reduction of elements in a reaction.

Oxidizing and Reducing Agents

In a redox reaction, one substance gets oxidized and another gets reduced. The oxidizing agent, or oxidant, is the substance that is reduced, meaning it gains electrons. The reducing agent, or reductant, is the one that donates electrons, so it is oxidized.
For the first reaction, the analysis is as follows:

• The nitrogen in \(\text{NH}_4^+\) is oxidized, changing from \(-3\) to \(0\), so \(\text{NH}_4^+\) is the reducing agent.
• The nitrogen in \(\text{NO}_2^-\) is reduced from \(+3\) to \(0\), making \(\text{NO}_2^-\) the oxidizing agent.

In the second reaction:

• Again, \(\text{NH}_4^+\) acts as the reducing agent because it is oxidized from \(-3\) to \(+1\).
• The \(\text{NO}_3^-\) acts as the oxidizing agent since the nitrogen within it is reduced from \(+5\) to \(0\).

Understanding these roles helps us identify the direction of electron flow in a reaction.

Electron Transfer

Electron transfer is the heart of redox reactions. It involves moving electrons from one atom to another. In the redox reactions provided, nitrogen atoms undergo both oxidation (loss of electrons) and reduction (gain of electrons), enabling them to change states.
The process is distinctively different in the third reaction presented in the exercise, which includes silver and zinc.

• In contrast, here, zinc acts as the reducing agent. It loses two electrons to become \(\text{Zn}^{2+}\). This occurs when zinc is oxidized from 0 to \(+2\).
• Silver gains electrons, reducing from \(+1\) to 0, thus, \(\text{Ag}^+\) is the oxidizing agent.

This kind of understanding of electron transfer allows us to visualize the electron flow and energy changes in chemical reactions, providing insights into why and how reactions occur.