Chapter 4, Problem 78E

(a)

Interpretation Introduction

Interpretation: Whether the reaction, ${\text{CH}}_4\left(g\right)+{\text{2O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{2H}}_2\text{O}\left(g\right)$ , is oxidation-reduction reaction or not is to be identified. If the reaction is oxidation-reduction reaction then the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced are to be predicted.

Concept introduction: The oxidation-reduction reactions involve the changes with respect to the oxidation state of the species that are interacting in reaction. These reactions are helpful in the identification of a substance that is oxidized or reduced. Along with that these are also useful in predicting reaction species that act as an oxidizing agent/ reducing agent.

Answer to Problem 78E

The reaction, ${\text{CH}}_4\left(g\right)+{\text{2O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{2H}}_2\text{O}\left(g\right)$ , is an oxidation-reduction reaction. The oxidizing agent is ${\text{O}}_2\left(g\right)$ , reducing agent is ${\text{CH}}_4\left(g\right)$ , carbon is the substance that is being oxidized and oxygen is the substance that is being reduced.

Explanation of Solution

The given chemical equation is shown below.

  ${\text{CH}}_4\left(g\right)+{\text{2O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{2H}}_2\text{O}\left(g\right)$

The oxidation state of carbon in ${\text{CH}}_4\left(g\right)$ is $-4$ .

The oxidation state of each hydrogen atom in ${\text{CH}}_4\left(g\right)$ is $+1$ .

The oxidation state of oxygen in ${\text{O}}_2\left(g\right)$ is $0$ .

The oxidation state ofcarbon and each oxygen atom in ${\text{CO}}_2\left(g\right)$ is $+4$ and $-2$ .

The oxidation state of each hydrogen atom and oxygen in ${\text{H}}_2\text{O}\left(g\right)$ is $+1$ and $-2$ .

The oxidation state of carbon increases from $-4$ in ${\text{CH}}_4\left(g\right)$ to $+4$ in ${\text{CO}}_2\left(g\right)$ .

So, carbon is oxidized from $-4$ to $+4$ oxidation state.

Thus, ${\text{CH}}_4\left(g\right)$ is a reducing agent.

The oxidation state of oxygen decreases from $0$ in ${\text{O}}_2\left(g\right)$ to $-2$ in ${\text{CO}}_2\left(g\right)$ and ${\text{H}}_2\text{O}\left(g\right)$ .

So, oxygen is reduced from $0$ to $\text{}-2$ oxidation state.

Thus, ${\text{O}}_2\left(g\right)$ is an oxidizing agent.

(b)

Interpretation Introduction

Interpretation: Whether the reaction, $\text{Zn}\left(s\right)+\text{2HCl}\left(aq\right)\to {\text{ZnCl}}_2\left(aq\right)+{\text{H}}_2\left(g\right)$ , is oxidation-reduction reaction or not is to be identified. If the reaction is oxidation-reduction reaction then the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced are to be predicted.

Concept introduction: The oxidation-reduction reactions involve the changes with respect to the oxidation state of the species that are interacting in reaction. These reactions are helpful in the identification of a substance that is oxidized or reduced. Along with that these are also useful in predicting reaction species that act as an oxidizing agent/ reducing agent.

Answer to Problem 78E

The reaction, $\text{Zn}\left(s\right)+\text{2HCl}\left(aq\right)\to {\text{ZnCl}}_2\left(aq\right)+{\text{H}}_2\left(g\right)$ , is an oxidation-reduction reaction. The oxidizing agent is $\text{HCl}\left(aq\right)$ , reducing agent is $\text{Zn}\left(s\right)$ , zinc is the substance that is being oxidized and hydrogen is the substance that is being reduced.

Explanation of Solution

The given chemical equation is shown below.

  $\text{Zn}\left(s\right)+\text{2HCl}\left(aq\right)\to {\text{ZnCl}}_2\left(aq\right)+{\text{H}}_2\left(g\right)$

The oxidation state of zinc in $\text{Zn}\left(s\right)$ is $0$ .

The oxidation state of hydrogen in $\text{HCl}\left(aq\right)$ is $+1$ .

The oxidation state of chlorine in $\text{HCl}\left(aq\right)$ is $-1$ .

The oxidation state of zinc and each chlorine atom in ${\text{ZnCl}}_2\left(aq\right)$ is $+2$ and $-1$ .

The oxidation state of each hydrogen atom in ${\text{H}}_2\left(g\right)$ is $0$ .

The oxidation state of zinc increases from $0$ in $\text{Zn}\left(s\right)$ to $+2$ in ${\text{ZnCl}}_2\left(aq\right)$ .

So, zinc is oxidized from $0$ to $+2$ oxidation state.

Thus, $\text{Zn}\left(s\right)$ is a reducing agent.

The oxidation state of hydrogen decreases from $+1$ in $\text{HCl}\left(aq\right)$ to $0$ in ${\text{H}}_2\left(g\right)$ .

So, hydrogen is reduced from $+1$ to $0$ oxidation state.

Thus, $\text{HCl}\left(aq\right)$ is an oxidizing agent.

(c)

Interpretation Introduction

Interpretation: Whether the reaction, ${\text{Cr}}_2{\text{O}}_7{}^{2-}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)\to 2{\text{CrO}}_4{}^{2-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$ , is oxidation-reduction reaction or not is to be identified. If the reaction is oxidation-reduction reaction then the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced are to be predicted.

Concept introduction: The oxidation-reduction reactions involve the changes with respect to the oxidation state of the species that are interacting in reaction. These reactions are helpful in the identification of a substance that is oxidized or reduced. Along with that these are also useful in predicting reaction species that act as an oxidizing agent/ reducing agent.

Answer to Problem 78E

The reaction, ${\text{Cr}}_2{\text{O}}_7{}^{2-}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)\to 2{\text{CrO}}_4{}^{2-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$ , is not the oxidation-reduction reaction.

Explanation of Solution

The given chemical equation is shown below.

  ${\text{Cr}}_2{\text{O}}_7{}^{2-}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)\to 2{\text{CrO}}_4{}^{2-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

The oxidation state of each chromium atom and oxygen atom in ${\text{Cr}}_2{\text{O}}_7{}^{2-}\left(aq\right)$ is $+6$ and $\text{}-2$ .

The oxidation state of oxygen and hydrogen atom in ${\text{OH}}^{-}\left(aq\right)$ is $\text{}-2$ and $+1$ .

The oxidation state of chromium atom and each oxygen atom in ${\text{CrO}}_4{}^{2-}\left(aq\right)$ is $+6$ and $-2$ .

The oxidation state of each hydrogen atom and oxygen atom in ${\text{H}}_2\text{O}\left(l\right)$ is $+1$ and $\text{}-2$ .

The oxidation state of all the elements present on reactant side are equal to the oxidation state of all the elements present on product side.

So, there is no change in oxidation state of any element.

Thus, the given reaction is not the oxidation-reduction reaction.

(d)

Interpretation Introduction

Interpretation: Whether the reaction, ${\text{O}}_3\left(g\right)+\text{NO}\left(g\right)\to {\text{O}}_2\left(g\right)+{\text{NO}}_2\left(g\right)$ , is oxidation-reduction reaction or not is to be identified. If the reaction is oxidation-reduction reaction then the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced are to be predicted.

Concept introduction: The oxidation-reduction reactions involve the changes with respect to the oxidation state of the species that are interacting in reaction. These reactions are helpful in the identification of a substance that is oxidized or reduced. Along with that these are also useful in predicting reaction species that act as an oxidizing agent/ reducing agent.

Answer to Problem 78E

The reaction, ${\text{O}}_3\left(g\right)+\text{NO}\left(g\right)\to {\text{O}}_2\left(g\right)+{\text{NO}}_2\left(g\right)$ , is an oxidation-reduction reaction. The oxidizing agent is ${\text{O}}_3\left(g\right)$ , reducing agent is $\text{NO}\left(g\right)$ , nitrogen is the substance that is being oxidized and oxygen is the substance that is being reduced.

Explanation of Solution

The given chemical equation is shown below.

  ${\text{O}}_3\left(g\right)+\text{NO}\left(g\right)\to {\text{O}}_2\left(g\right)+{\text{NO}}_2\left(g\right)$

The oxidation state of each atom of oxygen in ${\text{O}}_3\left(g\right)$ is $0$ .

The oxidation state of nitrogen atom and oxygen atom in $\text{NO}\left(g\right)$ is $+2$ and $\text{}-2$ .

The oxidation state of each oxygen atom in ${\text{O}}_2\left(g\right)$ is $0$ .

The oxidation state ofnitrogen atom and each oxygen atom in ${\text{NO}}_2\left(g\right)$ is $+4$ and $-2$ .

The oxidation state of nitrogen increases from $+2$ in $\text{NO}\left(g\right)$ to $+4$ in ${\text{NO}}_2\left(g\right)$ .

So, nitrogen is oxidized from $+2$ to $+4$ oxidation state.

Thus, $\text{NO}\left(g\right)$ is a reducing agent.

The oxidation state of oxygen decreases from $0$ in ${\text{O}}_3\left(g\right)$ to $-2$ in ${\text{NO}}_2\left(g\right)$ .

So, oxygen is reduced from $0$ to $-2$ oxidation state.

Thus, ${\text{O}}_3\left(g\right)$ is an oxidizing agent.

(e)

Interpretation Introduction

Interpretation: Whether the reaction, ${\text{2H}}_2{\text{O}}_2\left(l\right)\to 2{\text{H}}_2\text{O}\left(l\right)+{\text{O}}_2\left(g\right)$ , is oxidation-reduction reaction or not is to be identified. If the reaction is oxidation-reduction reaction then the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced are to be predicted.

Concept introduction: The oxidation-reduction reactions involve the changes with respect to the oxidation state of the species that are interacting in reaction. These reactions are helpful in the identification of a substance that is oxidized or reduced. Along with that these are also useful in predicting reaction species that act as an oxidizing agent/ reducing agent.

Answer to Problem 78E

The reaction, ${\text{2H}}_2{\text{O}}_2\left(l\right)\to 2{\text{H}}_2\text{O}\left(l\right)+{\text{O}}_2\left(g\right)$ , is oxidation-reduction reaction. The oxidizing agent and reducing agent is ${\text{H}}_2{\text{O}}_2\left(l\right)$ , oxygen is the substance that is being oxidized and reduced.

Explanation of Solution

The given chemical equation is shown below.

  ${\text{2H}}_2{\text{O}}_2\left(l\right)\to 2{\text{H}}_2\text{O}\left(l\right)+{\text{O}}_2\left(g\right)$

The oxidation state of each atom of oxygen in ${\text{H}}_2{\text{O}}_2\left(l\right)$ is $-1$ .

The oxidation state of each hydrogen atom in ${\text{H}}_2{\text{O}}_2\left(l\right)$ is $+1$ .

The oxidation state of each hydrogen atom and oxygen atom in ${\text{H}}_2\text{O}\left(l\right)$ is $+1$ and $\text{}-2$ .

The oxidation state of each oxygen atom in ${\text{O}}_2\left(g\right)$ is $0$ .

The oxidation state of oxygen increases from $-1$ in ${\text{H}}_2{\text{O}}_2\left(l\right)$ to $0$ in ${\text{O}}_2\left(g\right)$ .

So, oxygen is oxidized from $-1$ to $0$ oxidation state.

Thus, ${\text{H}}_2{\text{O}}_2\left(l\right)$ is a reducing agent.

The oxidation state of oxygen decreases from $-1$ in ${\text{H}}_2{\text{O}}_2\left(l\right)$ to $-2$ in ${\text{H}}_2\text{O}\left(l\right)$ .

So, oxygen is reduced from $-1$ to $-2$ oxidation state.

Thus, ${\text{H}}_2{\text{O}}_2\left(l\right)$ is an oxidizing agent.

(f)

Interpretation Introduction

Interpretation: Whether the reaction, $\text{2CuCl}\left(aq\right)\to {\text{CuCl}}_2\left(aq\right)+\text{Cu}\left(s\right)$ , is oxidation-reduction reaction or not is to be identified. If the reaction is oxidation-reduction reaction then the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced are to be predicted.

Concept introduction: The oxidation-reduction reactions involve the changes with respect to the oxidation state of the species that are interacting in reaction. These reactions are helpful in the identification of a substance that is oxidized or reduced. Along with that these are also useful in predicting reaction species that act as an oxidizing agent/ reducing agent.

Answer to Problem 78E

The reaction, $\text{2CuCl}\left(aq\right)\to {\text{CuCl}}_2\left(aq\right)+\text{Cu}\left(s\right)$ , is oxidation-reduction reaction. The oxidizing agent and reducing agent is $\text{CuCl}\left(aq\right)$ , copper is the substance that is being oxidized and reduced.

Explanation of Solution

The given chemical equation is shown below.

  $\text{2CuCl}\left(aq\right)\to {\text{CuCl}}_2\left(aq\right)+\text{Cu}\left(s\right)$

The oxidation state of copper atom in $\text{CuCl}\left(aq\right)$ is $+1$ .

The oxidation state of chlorine atom in $\text{CuCl}\left(aq\right)$ is $-1$ .

The oxidation state of copper atom and each chlorine atom in ${\text{CuCl}}_2\left(aq\right)$ is $+2$ and $\text{}-1$ .

The oxidation state of copper atom in $\text{Cu}\left(s\right)$ is $0$ .

The oxidation state of copper increases from $+1$ in $\text{CuCl}\left(aq\right)$ to $+2$ in ${\text{CuCl}}_2\left(aq\right)$ .

So, copper is oxidized from $+1$ to $+2$ oxidation state.

Thus, $\text{CuCl}\left(aq\right)$ is a reducing agent.

The oxidation state of copper decreases from $+1$ in $\text{CuCl}\left(aq\right)$ to $0$ in $\text{Cu}\left(s\right)$ .

So, copper is reduced from $+1$ to $0$ oxidation state.

Thus, $\text{CuCl}\left(aq\right)$ is an oxidizing agent.

(g)

Interpretation Introduction

Interpretation: Whether the reaction, $\text{HCl}\left(g\right)+{\text{NH}}_3\left(g\right)\to {\text{NH}}_4\text{Cl}\left(s\right)$ , is oxidation-reduction reaction or not is to be identified. If the reaction is oxidation-reduction reaction then the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced are to be predicted.

Concept introduction: The oxidation-reduction reactions involve the changes with respect to the oxidation state of the species that are interacting in reaction. These reactions are helpful in the identification of a substance that is oxidized or reduced. Along with that these are also useful in predicting reaction species that act as an oxidizing agent/ reducing agent.

Answer to Problem 78E

The reaction, $\text{HCl}\left(g\right)+{\text{NH}}_3\left(g\right)\to {\text{NH}}_4\text{Cl}\left(s\right)$ , is not the oxidation-reduction reaction.

Explanation of Solution

The given chemical equation is shown below.

  $\text{HCl}\left(g\right)+{\text{NH}}_3\left(g\right)\to {\text{NH}}_4\text{Cl}\left(s\right)$

The oxidation state of hydrogen atom and chlorine atom in $\text{HCl}\left(g\right)$ is $+1$ and $\text{}-1$ .

The oxidation state of nitrogen and each hydrogen atom in ${\text{NH}}_3\left(g\right)$ is $\text{}-3$ and $+1$ .

The oxidation state of nitrogen atom and each hydrogen atom in ${\text{NH}}_4\text{Cl}\left(s\right)$ is $-3$ and $+1$ .

The oxidation state of chlorine atom in ${\text{NH}}_4\text{Cl}\left(s\right)$ is $-1$ .

The oxidation state of all the elements present on reactant side are equal to the oxidation state of all the elements present on product side.

So, there is no change in oxidation state of any element

Thus, the given reaction is not the oxidation-reduction reaction.

(h)

Interpretation Introduction

Interpretation: Whether the reaction, ${\text{SiCl}}_4\left(l\right)+2{\text{H}}_2\text{O}\left(l\right)\to 4\text{HCl}\left(aq\right)+{\text{SiO}}_2\left(s\right)$ , is oxidation-reduction reaction or not is to be identified. If the reaction is oxidation-reduction reaction then the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced are to be predicted.

Concept introduction: The oxidation-reduction reactions involve the changes with respect to the oxidation state of the species that are interacting in reaction. These reactions are helpful in the identification of a substance that is oxidized or reduced. Along with that these are also useful in predicting reaction species that act as an oxidizing agent/ reducing agent.

Answer to Problem 78E

The reaction, ${\text{SiCl}}_4\left(l\right)+2{\text{H}}_2\text{O}\left(l\right)\to 4\text{HCl}\left(aq\right)+{\text{SiO}}_2\left(s\right)$ , is not the oxidation-reduction reaction.

Explanation of Solution

The given chemical equation is shown below.

  ${\text{SiCl}}_4\left(l\right)+2{\text{H}}_2\text{O}\left(l\right)\to 4\text{HCl}\left(aq\right)+{\text{SiO}}_2\left(s\right)$

The oxidation state of silicon atom and each chlorine atom in ${\text{SiCl}}_4\left(l\right)$ is $+4$ and $\text{}-1$ .

The oxidation state of chlorine and hydrogen atom in $\text{HCl}\left(aq\right)$ is $\text{}-1$ and $+1$ .

The oxidation state of silicon atom and each oxygen atom in ${\text{SiO}}_2\left(s\right)$ is $+4$ and $-2$ .

The oxidation state of all the elements present on reactant side are equal to the oxidation state of all the elements present on product side.

So, there is no change in oxidation state of any element

Thus, the given reaction is not the oxidation-reduction reaction.

(i)

Interpretation Introduction

Interpretation: Whether the reaction, ${\text{SiCl}}_4\left(l\right)+2\text{Mg}\left(s\right)\to 2{\text{MgCl}}_2\left(s\right)+\text{Si}\left(s\right)$ , is oxidation-reduction reaction or not is to be identified. If the reaction is oxidation-reduction reaction then the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced are to be predicted.

Concept introduction: The oxidation-reduction reactions involve the changes with respect to the oxidation state of the species that are interacting in reaction. These reactions are helpful in the identification of a substance that is oxidized or reduced. Along with that these are also useful in predicting reaction species that act as an oxidizing agent/ reducing agent.

Answer to Problem 78E

The reaction, ${\text{SiCl}}_4\left(l\right)+2\text{Mg}\left(s\right)\to 2{\text{MgCl}}_2\left(s\right)+\text{Si}\left(s\right)$ , is oxidation-reduction reaction. The oxidizing agent is ${\text{SiCl}}_4\left(l\right)$ , reducing agent is $\text{Mg}\left(s\right)$ , magnesium is the substance that being oxidized and silicon is the substance that being reduced.

Explanation of Solution

The given chemical equation is shown below.

  ${\text{SiCl}}_4\left(l\right)+2\text{Mg}\left(s\right)\to 2{\text{MgCl}}_2\left(s\right)+\text{Si}\left(s\right)$

The oxidation state of magnesium in $\text{Mg}\left(s\right)$ is $0$ .

The oxidation state of magnesium atom and each chlorine atom in ${\text{MgCl}}_2\left(s\right)$ is $+2$ and $\text{}-1$ .

The oxidation state of silicon atom and each chlorine atom in ${\text{SiCl}}_4\left(l\right)$ is $+4$ and $\text{}-1$ .

The oxidation state of silicon in $\text{Si}\left(s\right)$ is $0$ .

The oxidation state of magnesium increases from $0$ in $\text{Mg}\left(s\right)$ to $+2$ in ${\text{MgCl}}_2\left(s\right)$ .

So, magnesium is oxidized from $0$ to $+2$ oxidation state.

Thus, $\text{Mg}\left(s\right)$ is a reducing agent.

The oxidation state of silicon decreases from $+4$ in ${\text{SiCl}}_4\left(l\right)$ to $0$ in $\text{Si}\left(s\right)$ .

So, silicon is reduced from $+4$ to $0$ oxidation state.

Thus, ${\text{SiCl}}_4\left(l\right)$ is an oxidizing agent.