Chapter 17, Problem 8ST

Interpretation Introduction

(a)

Interpretation:

The oxidation half-reaction for the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is to be stated.

Concept introduction:

The study of the conversion of chemical energy to electrical energy and vice versa is known as electrochemistry. In an electrochemical cell, two electrodes are connected by a wire and electrons are free to move between two compartments. Each compartment is known as a half-cell.

Answer to Problem 8ST

The oxidation half-reaction for the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is shown below.

${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{NO}}_{\text{3}}{}^{-}\left(aq\right)+\text{2}{\text{H}}^{+}\left(aq\right)+\text{2}{\text{e}}^{-}$

Explanation of Solution

The given redox reaction is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$

The reduction is defined as a process in which the gain of electrons takes place, whereas the oxidation is defined as the process in which the loss of electrons occurs. The substance that is oxidized is a reducing agent and the substance that is reduced is an oxidizing agent.

In the above redox reaction, the ${\text{NO}}_{\text{2}}{}^{-}$ oxidized to ${\text{NO}}_{\text{3}}{}^{-}$. In an electrochemical cell, oxidation occurs at anode electrode and reduction occurs at cathode electrode.

The oxidation half-reaction is shown below.

${\text{NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{NO}}_{\text{3}}{}^{-}\left(aq\right)+\text{2}{\text{e}}^{-}$

The number of atom is balanced by adding ${\text{H}}_{\text{2}}\text{O}$ molecule on the left hand side and ${\text{2H}}^{+}$ on the right hand side. The oxidation half-reaction is shown below.

${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{NO}}_{\text{3}}{}^{-}\left(aq\right)+\text{2}{\text{H}}^{+}\left(aq\right)+\text{2}{\text{e}}^{-}$

Conclusion

The oxidation half-reaction for the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is shown below.

${\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{NO}}_{\text{3}}{}^{-}\left(aq\right)+\text{2}{\text{H}}^{+}\left(aq\right)+\text{2}{\text{e}}^{-}$

Interpretation Introduction

(b)

Interpretation:

The reduction half-reaction for the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is to be stated.

Concept introduction:

The study of the conversion of chemical energy to electrical energy and vice versa is known as electrochemistry. In an electrochemical cell, two electrodes are connected by a wire and electrons are free to move between two compartments. Each compartment is known as a half-cell.

Answer to Problem 8ST

The reduction half-reaction for the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+2{\text{e}}^{-}+{\text{2H}}^{+}\left(aq\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The given redox reaction is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$

The reduction is defined as a process in which the gain of electrons takes place, whereas the oxidation is defined as the process in which the loss of electrons occurs. The substance that is oxidized is a reducing agent and the substance that is reduced is an oxidizing agent.

In the above redox reaction, the ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ reduced to ${\text{H}}_{\text{2}}\text{O}$. In an electrochemical cell, oxidation occurs at anode electrode and reduction occurs at cathode electrode.

Therefore, the reduction half-reaction is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+2{\text{e}}^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number of atoms and charge are balanced by adding ${\text{2H}}^{+}$ on the left hand side. Therefore, the reduction half-reaction is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+2{\text{e}}^{-}+{\text{2H}}^{+}\left(aq\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The reduction half-reaction for the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+2{\text{e}}^{-}+{\text{2H}}^{+}\left(aq\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Interpretation Introduction

(c)

Interpretation:

The substance that is oxidized in the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is to be stated.

Concept introduction:

The study of the conversion of chemical energy to electrical energy and vice versa is known as electrochemistry. In an electrochemical cell, two electrodes are connected by a wire and electrons are free to move between two compartments. Each compartment is known as a half-cell.

Answer to Problem 8ST

The substance that is oxidized in the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is ${\text{NO}}_{\text{2}}{}^{-}$.

Explanation of Solution

The given redox reaction is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$

The reduction is defined as a process in which the gain of electrons takes place, whereas the oxidation is defined as the process in which the loss of electrons occurs. The substance that is oxidized is a reducing agent and the substance that is reduced is an oxidizing agent.

In the above redox reaction, the ${\text{NO}}_{\text{2}}{}^{-}$ oxidized to ${\text{NO}}_{\text{3}}{}^{-}$. Therefore, the substance that is oxidized in the given redox reaction is ${\text{NO}}_{\text{2}}{}^{-}$.

Conclusion

The substance that is oxidized in the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is ${\text{NO}}_{\text{2}}{}^{-}$.

Interpretation Introduction

(d)

Interpretation:

The substance that is reduced in the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is to be stated

Concept introduction:

The study of the conversion of chemical energy to electrical energy and vice versa is known as electrochemistry. In an electrochemical cell, two electrodes are connected by a wire and electrons are free to move between two compartments. Each compartment is known as a half-cell.

Answer to Problem 8ST

The substance that is reduced in the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$.

Explanation of Solution

The given redox reaction is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$

The reduction is defined as a process in which the gain of electrons takes place, whereas the oxidation is defined as the process in which the loss of electrons occurs. The substance that is oxidized is a reducing agent and the substance that is reduced is an oxidizing agent.

In the above redox reaction, the ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ reduced to ${\text{H}}_{\text{2}}\text{O}$. Therefore, the substance that is reduced in the given redox reaction is ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$.

Conclusion

The substance that is reduced in the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$.

Interpretation Introduction

(e)

Interpretation:

The oxidizing agent in the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is to be stated.

Concept introduction:

The study of the conversion of chemical energy to electrical energy and vice versa is known as electrochemistry. In an electrochemical cell, two electrodes are connected by a wire and electrons are free to move between two compartments. Each compartment is known as a half-cell.

Answer to Problem 8ST

The oxidizing agent in the reaction ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$.

Explanation of Solution

The given redox reaction is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$

The reduction is defined as a process in which the gain of electrons takes place, whereas the oxidation is defined as the process in which the loss of electrons occurs. The substance that is oxidized is a reducing agent and the substance that is reduced is an oxidizing agent.

In the above redox reaction, the ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ reduced to ${\text{H}}_{\text{2}}\text{O}$. Therefore, the ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is an oxidizing agent.

Conclusion

The oxidizing agent in the reaction ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$.

Interpretation Introduction

(f)

Interpretation:

The reducing agent in the redox reaction

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is to be stated.

Concept introduction:

The study of the conversion of chemical energy to electrical energy and vice versa is known as electrochemistry. In an electrochemical cell, two electrodes are connected by a wire and electrons are free to move between two compartments. Each compartment is known as a half-cell.

Answer to Problem 8ST

The reducing agent in the reaction ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is ${\text{NO}}_{\text{2}}{}^{-}$.

Explanation of Solution

The given redox reaction is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$

The reduction is defined as a process in which the gain of electrons takes place, whereas the oxidation is defined as the process in which the loss of electrons occurs. The substance that is oxidized is a reducing agent and the substance that is reduced is an oxidizing agent.

In the above redox reaction, the ${\text{NO}}_{\text{2}}{}^{-}$ oxidized to ${\text{NO}}_{\text{3}}{}^{-}$. Therefore, the ${\text{NO}}_{\text{2}}{}^{-}$ is a reducing agent.

Conclusion

The reducing agent in the reaction ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+{\text{ NO}}_{\text{2}}{}^{-}\left(aq\right)\to {\text{ H}}_{\text{2}}\text{O}\left(l\right)+{\text{ NO}}_3{}^{-}\left(aq\right)$ is ${\text{NO}}_{\text{2}}{}^{-}$.