Chapter F, Problem K.17E

(a)

Interpretation Introduction

Interpretation:

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

    ${\text{Cl}}_{\text{2}}\text{(g)}+{\text{H}}_{\text{2}}\text{O}(\text{l})\to \text{HClO}(\text{aq})+\text{HCl}(\text{aq})$

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.17E

Balanced chemical equation is ${\text{Cl}}_{\text{2}}\text{(g)}+{\text{H}}_{\text{2}}\text{O}(\text{l})\to \text{HClO}(\text{aq})+\text{HCl}(\text{aq})$.  Oxidizing agent and reducing agent is ${\text{Cl}}_{\text{2}}$.

Explanation of Solution

The given reaction is written as follows;

    ${\text{Cl}}_{\text{2}}\text{(g)}+{\text{H}}_{\text{2}}\text{O}(\text{l})\to \text{HClO}(\text{aq})+\text{HCl}(\text{aq})$

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 chlorine is increased from $0$ to $+1$.  Therefore, oxidation has taken place.  ${\text{Cl}}_{\text{2}}$ is oxidized.  Therefore, ${\text{Cl}}_{\text{2}}$ is reducing agent.

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

(b)

Interpretation Introduction

Interpretation:

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

    ${\text{NaClO}}_{\text{3}}\text{(aq)}+{\text{SO}}_{\text{2}}(\text{g})+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)\to {\text{NaHSO}}_{\text{4}}(\text{aq})+{\text{ClO}}_{\text{2}}(\text{g})$

Concept Introduction:

Refer part (a).

Answer to Problem K.17E

Balanced chemical equation is;

    ${\text{2NaClO}}_{\text{3}}\text{(aq)}+{\text{SO}}_{\text{2}}(\text{g})+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)\to 2{\text{NaHSO}}_{\text{4}}(\text{aq})+2{\text{ClO}}_{\text{2}}(\text{g})$

Oxidizing agent is ${\text{NaClO}}_{\text{3}}$ and reducing agent is ${\text{SO}}_{\text{2}}$.

Explanation of Solution

The given reaction is written as follows;

    ${\text{NaClO}}_{\text{3}}\text{(aq)}+{\text{SO}}_{\text{2}}(\text{g})+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)\to {\text{NaHSO}}_{\text{4}}(\text{aq})+{\text{ClO}}_{\text{2}}(\text{g})$

Balancing Hydrogen atoms:  In the left side of the equation there are two hydrogen atoms while on the product side only one hydrogen atom is present.  Adding coefficient $2$ before ${\text{NaHSO}}_{\text{4}}$ balances the hydrogen atoms on both sides of the equation.  The chemical equation obtained is given as follows;

    ${\text{NaClO}}_{\text{3}}\text{(aq)}+{\text{SO}}_{\text{2}}(\text{g})+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)\to 2{\text{NaHSO}}_{\text{4}}(\text{aq})+{\text{ClO}}_{\text{2}}(\text{g})$

Balancing Sodium atoms:  In the left side of the equation there is one sodium atom while on the product side there are two sodium atoms.  Adding coefficient $2$ before ${\text{NaClO}}_{\text{3}}$ balances the sodium atoms on both sides of the equation.  The chemical equation obtained is given as follows;

    ${\text{2NaClO}}_{\text{3}}\text{(aq)}+{\text{SO}}_{\text{2}}(\text{g})+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)\to 2{\text{NaHSO}}_{\text{4}}(\text{aq})+{\text{ClO}}_{\text{2}}(\text{g})$

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

    ${\text{2NaClO}}_{\text{3}}\text{(aq)}+{\text{SO}}_{\text{2}}(\text{g})+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)\to 2{\text{NaHSO}}_{\text{4}}(\text{aq})+2{\text{ClO}}_{\text{2}}(\text{g})$

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 sulfur is increased from $+4$ to $+6$.  Therefore, oxidation has taken place.  ${\text{SO}}_{\text{2}}$ is oxidized.  Therefore, ${\text{SO}}_{\text{2}}$ is reducing agent.

The oxidation state of chlorine decreases from $+5$ to $+4$ as shown in the equation.  Therefore, ${\text{NaClO}}_{\text{3}}$ is reduced.  Therefore, the oxidizing agent is ${\text{NaClO}}_{\text{3}}$.

(c)

Interpretation Introduction

Interpretation:

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

    $\text{CuI(aq)}\to \text{Cu}(\text{s})+{\text{I}}_{\text{2}}\text{(s)}$

Concept Introduction:

Refer part (a).

Answer to Problem K.17E

Balanced chemical equation is $\text{2CuI(aq)}\to 2\text{Cu}(\text{s})+{\text{I}}_{\text{2}}\text{(s)}$.  Oxidizing agent and reducing agent are $\text{CuI}$.

Explanation of Solution

The given reaction is written as follows;

    $\text{2CuI(aq)}\to 2\text{Cu}(\text{s})+{\text{I}}_{\text{2}}\text{(s)}$

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

    $\text{2CuI(aq)}\to \text{Cu}(\text{s})+{\text{I}}_{\text{2}}\text{(s)}$

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

    $\text{2CuI(aq)}\to 2\text{Cu}(\text{s})+{\text{I}}_{\text{2}}\text{(s)}$

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 iodine is increased from $-1$ to $0$.  Therefore, oxidation has taken place.  $\text{CuI}$ is oxidized.  Therefore, $\text{CuI}$ is reducing agent.

The oxidation state of copper decreases from $+1$ to $0$ as shown in the equation.  Therefore, $\text{CuI}$ is reduced.  Therefore, the oxidizing agent is $\text{CuI}$.