Chapter F, Problem K.18E

(a)

Interpretation Introduction

Interpretation:

Given chemical equation has to be balanced and also the oxidizing agent and reducing agent has to be identified.

    $\text{CO(g)}+{\text{H}}_{\text{2}}\text{O}(\text{g})\to {\text{CO}}_{\text{2}}(\text{g})+{\text{H}}_{\text{2}}(\text{g})$

Concept Introduction:

In redox reactions, oxidizing agent is the one that gets reduced by causing oxidation.  These agents can be ions, elements, or even compounds.  In oxidation, the oxidation number increases due to loss of electrons.

In redox reactions, reducing agent is the one that gets oxidized by causing reduction.  These agents can be ions, elements, or even compounds.  In reduction, the oxidation number decreases due to gain of electrons.

Answer to Problem K.18E

Balanced chemical equation is $\text{CO(g)}+{\text{H}}_{\text{2}}\text{O}(\text{g})\to {\text{CO}}_{\text{2}}(\text{g})+{\text{H}}_{\text{2}}(\text{g})$.  Oxidizing agent is ${\text{H}}_{\text{2}}\text{O}$ and reducing agent is $\text{CO}$.

Explanation of Solution

The given reaction is written as follows;

    $\text{CO(g)}+{\text{H}}_{\text{2}}\text{O}(\text{g})\to {\text{CO}}_{\text{2}}(\text{g})+{\text{H}}_{\text{2}}(\text{g})$

The above chemical equation has the same number of atoms of elements equal on both sides.  Hence, this itself is a balanced equation.

Oxidation number of the atoms present in the above equation is indicated as follows;

From the above equation, it is found that the oxidation state of carbon is increased from $+2$ to $+4$.  Therefore, oxidation has taken place.  $\text{CO}$ is oxidized.  Therefore, $\text{CO}$ is reducing agent.

The oxidation state of hydrogen decreases from $+1$ to $0$ as shown in the equation.  Therefore, ${\text{H}}_{\text{2}}\text{O}$ is reduced.  Therefore, the oxidizing agent is ${\text{H}}_{\text{2}}\text{O}$.

(b)

Interpretation Introduction

Interpretation:

Given chemical equation has to be balanced and also the oxidizing agent and reducing agent has to be identified.

    ${\text{ClO}}_2\text{(g)}+{\text{O}}_{\text{3}}(\text{g})\to {\text{Cl}}_{\text{2}}{\text{O}}_{\text{6}}(\text{l})+{\text{O}}_{\text{2}}(\text{g})$

Concept Introduction:

Refer part (a).

Answer to Problem K.18E

Balanced chemical equation is;

    ${\text{2ClO}}_2\text{(g)}+2{\text{O}}_{\text{3}}(\text{g})\to {\text{Cl}}_{\text{2}}{\text{O}}_{\text{6}}(\text{l})+2{\text{O}}_{\text{2}}(\text{g})$

Oxidizing agent is ${\text{O}}_{\text{3}}$ and reducing agent is ${\text{ClO}}_{\text{2}}$.

Explanation of Solution

The given reaction is written as follows;

    ${\text{ClO}}_2\text{(g)}+{\text{O}}_{\text{3}}(\text{g})\to {\text{Cl}}_{\text{2}}{\text{O}}_{\text{6}}(\text{l})+{\text{O}}_{\text{2}}(\text{g})$

Balancing Chlorine atoms:  In the left side of the equation there is one chlorine atom while on the product side there are two chlorine atoms.  Adding coefficient $2$ before ${\text{ClO}}_{\text{2}}$ balances the chlorine atoms on both sides of the equation.  The chemical equation obtained is given as follows;

    ${\text{2ClO}}_2\text{(g)}+{\text{O}}_{\text{3}}(\text{g})\to {\text{Cl}}_{\text{2}}{\text{O}}_{\text{6}}(\text{l})+{\text{O}}_{\text{2}}(\text{g})$

Balancing Oxygen atoms:  In the left side of the equation there are seven oxygen atoms while on the product side there are eight oxygen atoms.  Adding coefficient $2$ before ${\text{O}}_{\text{3}}$ and ${\text{O}}_{\text{2}}$ balances the oxygen atoms on both sides of the equation.  The balanced chemical equation obtained is given as follows;

    ${\text{2ClO}}_2\text{(g)}+2{\text{O}}_{\text{3}}(\text{g})\to {\text{Cl}}_{\text{2}}{\text{O}}_{\text{6}}(\text{l})+{\text{2O}}_{\text{2}}(\text{g})$

Oxidation number of the atoms present in the above equation is indicated as follows;

From the above equation, it is found that the oxidation state of chlorine is increased from $+4$ to $+6$.  Therefore, oxidation has taken place.  ${\text{ClO}}_{\text{2}}$ is oxidized.  Therefore, ${\text{ClO}}_{\text{2}}$ is reducing agent.

The oxidation state of oxygen decreases from $0$ to $-2$ as shown in the equation.  Therefore, ${\text{O}}_{\text{3}}$ is reduced.  Therefore, the oxidizing agent is ${\text{O}}_{\text{3}}$.

(c)

Interpretation Introduction

Interpretation:

Given chemical equation has to be balanced and also the oxidizing agent and reducing agent has to be identified.

    ${\text{Cl}}_{\text{2}}{\text{(g) + F}}_{\text{2}}\text{(g)}\to {\text{ClF}}_{\text{3}}\text{(g)}$

Concept Introduction:

Refer part (a).

Answer to Problem K.18E

Balanced chemical equation is ${\text{Cl}}_{\text{2}}{\text{(g) + 3F}}_{\text{2}}\text{(g)}\to 2{\text{ClF}}_{\text{3}}\text{(g)}$.  Oxidizing agent is ${\text{F}}_{\text{2}}$ and reducing agent are ${\text{Cl}}_{\text{2}}$.

Explanation of Solution

The given reaction is written as follows;

    ${\text{Cl}}_{\text{2}}{\text{(g) + F}}_{\text{2}}\text{(g)}\to {\text{ClF}}_{\text{3}}\text{(g)}$

Balancing chlorine atom:  In the reactant side, there are two chlorine atoms while on the product side, there is one chlorine atom.  Adding coefficient $2$ before ${\text{ClF}}_{\text{3}}$ balances the chlorine atom on both sides of the equation.  The chemical equation obtained is written as follows;

    ${\text{Cl}}_{\text{2}}{\text{(g) + F}}_{\text{2}}\text{(g)}\to 2{\text{ClF}}_{\text{3}}\text{(g)}$

Balancing fluorine atom:  In the reactant side, there are two fluorine atoms while on the product side, there are six fluorine atoms.  Adding coefficient $3$ before ${\text{F}}_{\text{2}}$ balances the fluorine atom on both sides of the equation.  The balanced chemical equation obtained is written as follows;

    ${\text{Cl}}_{\text{2}}{\text{(g) + 3F}}_{\text{2}}\text{(g)}\to 2{\text{ClF}}_{\text{3}}\text{(g)}$

Oxidation number of the atoms present in the above equation is indicated as follows;

From the above equation, it is found that the oxidation state of chlorine is increased from $0$ to $+3$.  Therefore, oxidation has taken place.  ${\text{Cl}}_{\text{2}}$ is oxidized.  Therefore, ${\text{Cl}}_{\text{2}}$ is reducing agent.

The oxidation state of fluorine decreases from $0$ to $-1$ as shown in the equation.  Therefore, ${\text{F}}_{\text{2}}$ is reduced.  Therefore, the oxidizing agent is ${\text{F}}_{\text{2}}$.