Chapter 21, Problem 21.15P

(a)

Interpretation Introduction

Interpretation:

The following skeleton reaction has to be balanced and the oxidizing and reducing agents has to be identified.

  ${\text{BH}}_{\text{4}}^{\text{-}}{}_{\text{(aq)}}{\text{+ ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{+ Cl}}^{\text{-}}{}_{\text{(aq)}}\text{[basic]}$

Concept Introduction:

Oxidizing agent:

A compound or group that causes the oxidation in another compound or group is said to be an oxidizing agent.  Most common oxidizing agents are oxygen, hydrogen peroxide and halogens.

Reducing agent:

A compound or group that causes the reduction in another compound or group is said to be a reducing agent.

Oxidation:

A reaction in which a compound gains electrons is said to be oxidation.

Reduction:

A reaction in which a compound loses electrons is said to be reduction.

Explanation of Solution

The given reaction is

  ${\text{BH}}_{\text{4}}^{\text{-}}{}_{\text{(aq)}}{\text{+ ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{+ Cl}}^{\text{-}}{}_{\text{(aq)}}\text{[basic]}$

The balancing of the reaction is as follows:

Step-1:

The given reaction has to be divided into half-reactions.

  ${\text{BH}}_{\text{4}}^{\text{-}}{}_{\text{(aq)}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}\to {\text{ Cl}}^{\text{-}}{}_{\text{(aq)}}$

Step-2:

Balance the half-reactions.

To balance the reactions, atoms other than oxygen and hydrogen are not needed.

Balance $\text{O}$ atoms with ${\text{H}}_{\text{2}}\text{O}$:

  $\begin{array}{l}{\text{BH}}_{\text{4}}^{\text{-}}{}_{\text{(aq)}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}\to {\text{ Cl}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{3H}}_{\text{2}}{\text{O+BH}}_{\text{4}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{\text{ClO}}_{\text{3}}^{\text{-}}\to {\text{ Cl}}^{\text{-}}{\text{+3H}}_{\text{2}}\text{O}\end{array}$

Balance $\text{H}$ atoms with ${\text{H}}^{\text{+}}$:

  $\begin{array}{l}{\text{BH}}_{\text{4}}^{\text{-}}{}_{\text{(aq)}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}\to {\text{ Cl}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{3H}}_{\text{2}}{\text{O+BH}}_{\text{4}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{\text{ClO}}_{\text{3}}^{\text{-}}\to {\text{ Cl}}^{\text{-}}{\text{+3H}}_{\text{2}}\text{O}\\ {\text{3H}}_{\text{2}}{\text{O+BH}}_{\text{4}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{\text{+8H}}^{\text{+}}{\text{6H}}^{\text{+}}+{\text{ClO}}_{\text{3}}^{\text{-}}\to {\text{ Cl}}^{\text{-}}{\text{+3H}}_{\text{2}}\text{O}\end{array}$

Addition of electrons:

  $\begin{array}{l}{\text{BH}}_{\text{4}}^{\text{-}}{}_{\text{(aq)}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{ClO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}\to {\text{ Cl}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{3H}}_{\text{2}}{\text{O+BH}}_{\text{4}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{\text{ClO}}_{\text{3}}^{\text{-}}\to {\text{ Cl}}^{\text{-}}{\text{+3H}}_{\text{2}}\text{O}\\ {\text{3H}}_{\text{2}}{\text{O+BH}}_{\text{4}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{\text{+8H}}^{\text{+}}{\text{6H}}^{\text{+}}+{\text{ClO}}_{\text{3}}^{\text{-}}\to {\text{ Cl}}^{\text{-}}{\text{+3H}}_{\text{2}}\text{O}\\ {\text{3H}}_{\text{2}}{\text{O+BH}}_{\text{4}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{\text{+8H}}^{\text{+}}{\text{+8e}}^{\text{-}}{\text{6e}}^{\text{-}}{\text{+6H}}^{\text{+}}+{\text{ClO}}_{\text{3}}^{\text{-}}\to {\text{ Cl}}^{\text{-}}{\text{+3H}}_{\text{2}}\text{O}\\ \text{(oxidation)}\text{(reduction)}\end{array}$

Step-3:

Each half reaction is multiplied with some integer to make electrons lost equal to electrons gained.

  $\begin{array}{l}{\text{6(3H}}_{\text{2}}{\text{O+BH}}_{\text{4}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{\text{+8H}}^{\text{+}}{\text{+8e}}^{\text{-}})8({\text{6e}}^{\text{-}}{\text{+6H}}^{\text{+}}+{\text{ClO}}_{\text{3}}^{\text{-}}\to {\text{ Cl}}^{\text{-}}{\text{+3H}}_{\text{2}}\text{O)}\\ {\text{18H}}_{\text{2}}{\text{O+6BH}}_{\text{4}}^{\text{-}}\to 6{\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{\text{+48H}}^{\text{+}}{\text{+48e}}^{\text{-}}{\text{48e}}^{\text{-}}{\text{+48H}}^{\text{+}}+8{\text{ClO}}_{\text{3}}^{\text{-}}\to {\text{ 8Cl}}^{\text{-}}{\text{+24H}}_{\text{2}}\text{O}\\ \end{array}$

Step-4:

Half-reactions are added and cancel the species appearing on both sides.

  $\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{18H}}_{\text{2}}{\text{O + 6BH}}_{\text{4}}^{\text{-}}\to 6{\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}\text{+}\cancel{{\text{48H}}^{\text{+}}}\text{+}\cancel{{\text{48e}}^{\text{-}}}\\ \cancel{{\text{48e}}^{\text{-}}}\text{+}\cancel{{\text{48H}}^{\text{+}}}+8{\text{ClO}}_{\text{3}}^{\text{-}}\to {\text{ 8Cl}}^{\text{-}}{\text{+ 24H}}_{\text{2}}\text{O}\end{array}}\\ {\text{6BH}}_{\text{4}}^{\text{-}}+8{\text{ClO}}_{\text{3}}^{\text{-}}\to 6{\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{\text{+8Cl}}^{\text{-}}{\text{+ 6H}}_{\text{2}}\text{O}\end{array}$

To get the final balanced equation, the above equation is divided by 2.  The final balanced equation is as follows:

  ${\text{3BH}}_{\text{4}}^{\text{-}}+4{\text{ClO}}_{\text{3}}^{\text{-}}\to 3{\text{H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}{\text{+4Cl}}^{\text{-}}{\text{+ 3H}}_{\text{2}}\text{O}$

In the above equation, the oxidizing agent is ${\text{ClO}}_{\text{3}}^{\text{-}}$ and the reducing agent is ${\text{BH}}_{\text{4}}^{\text{-}}$.

(b)

Interpretation Introduction

Interpretation:

The following skeleton reaction has to be balanced and the oxidizing and reducing agents has to be identified.

  ${\text{CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}{\text{+ N}}_{\text{2}}{\text{O}}_{\text{(g)}}\to {\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}{\text{+NO}}_{\text{(g)}}\text{[acidic]}$

Concept Introduction:

Oxidizing agent:

A compound or group that causes the oxidation in another compound or group is said to be an oxidizing agent.  Most common oxidizing agents are oxygen, hydrogen peroxide and halogens.

Reducing agent:

A compound or group that causes the reduction in another compound or group is said to be a reducing agent.

Oxidation:

A reaction in which a compound gains electrons is said to be oxidation.

Reduction:

A reaction in which a compound loses electrons is said to be reduction.

Explanation of Solution

The given reaction is

  ${\text{CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}{\text{+ N}}_{\text{2}}{\text{O}}_{\text{(g)}}\to {\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}{\text{+NO}}_{\text{(g)}}\text{[acidic]}$

The balancing of the reaction is as follows:

Step-1:

The given reaction has to be divided into half-reactions.

  ${\text{CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to {\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{N}}_{\text{2}}\text{O}\to \text{NO}$

Step-2:

Balance the half-reactions.

To balance the reactions, atoms other than oxygen and hydrogen are not needed.

Balance $\text{O}$ atoms with ${\text{H}}_{\text{2}}\text{O}$:

  $\begin{array}{l}{\text{CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to {\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{N}}_{\text{2}}\text{O}\to 2\text{NO}\\ {\text{CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to {\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{+4H}}_{\text{2}}\text{O}{\text{H}}_{\text{2}}{\text{O+N}}_{\text{2}}\text{O}\to 2\text{NO}\end{array}$

Balance $\text{H}$ atoms with ${\text{H}}^{\text{+}}$:

  $\begin{array}{l}{\text{CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to {\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{N}}_{\text{2}}\text{O}\to 2\text{NO}\\ {\text{CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to {\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{+4H}}_{\text{2}}\text{O}{\text{H}}_{\text{2}}{\text{O+N}}_{\text{2}}\text{O}\to 2\text{NO}\\ {\text{8H}}^{+}{\text{+CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to {\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{+4H}}_{\text{2}}\text{O}{\text{H}}_{\text{2}}{\text{O+N}}_{\text{2}}\text{O}\to 2\text{NO}+{\text{2H}}^{\text{+}}\end{array}$

Addition of electrons:

  $\begin{array}{l}{\text{CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to {\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{N}}_{\text{2}}\text{O}\to 2\text{NO}\\ {\text{CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to {\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{+4H}}_{\text{2}}\text{O}{\text{H}}_{\text{2}}{\text{O+N}}_{\text{2}}\text{O}\to 2\text{NO}\\ {\text{8H}}^{+}{\text{+CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to {\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{+4H}}_{\text{2}}\text{O}{\text{H}}_{\text{2}}{\text{O+N}}_{\text{2}}\text{O}\to 2\text{NO}+{\text{2H}}^{\text{+}}\\ {\text{3e}}^{-}{\text{+8H}}^{+}{\text{+CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to {\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{+4H}}_{\text{2}}\text{O}{\text{H}}_{\text{2}}{\text{O+N}}_{\text{2}}\text{O}\to 2\text{NO}+{\text{2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\\ \text{(reduction)}\text{(oxidation)}\end{array}$

Step-3:

Each half reaction is multiplied with some integer to make electrons lost equal to electrons gained.

  $\begin{array}{l}{\text{2(3e}}^{-}{\text{+8H}}^{+}{\text{+CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to {\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{+4H}}_{\text{2}}\text{O})3({\text{H}}_{\text{2}}{\text{O+N}}_{\text{2}}\text{O}\to 2\text{NO}+{\text{2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\text{)}\\ {\text{6e}}^{-}{\text{+16H}}^{+}{\text{+2CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to 2{\text{Cr}}^{3+}{}_{\text{(aq)}}{\text{+8H}}_{\text{2}}\text{O}3{\text{H}}_{\text{2}}{\text{O+3N}}_{\text{2}}\text{O}\to 6\text{NO}+{\text{6H}}^{\text{+}}{\text{+6e}}^{\text{-}}\\ \end{array}$

Step-4:

Half-reactions are added and cancel the species appearing on both sides.

  $\begin{array}{l}\underset{\_}{\begin{array}{l}\cancel{{\text{6e}}^{-}}\text{+}\cancel{{\text{16H}}^{+}}{\text{+2CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}\to 2{\text{Cr}}^{3+}{}_{\text{(aq)}}\text{+}\cancel{{\text{8H}}_{\text{2}}\text{O}}\\ \cancel{3{\text{H}}_{\text{2}}\text{O}}{\text{+3N}}_{\text{2}}\text{O}\to 6\text{NO}+\cancel{{\text{6H}}^{\text{+}}}\text{+}\cancel{{\text{6e}}^{\text{-}}}\end{array}}\\ {\text{2CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}{\text{+3N}}_{\text{2}}{\text{O+10H}}^{+}\to 2{\text{Cr}}^{3+}{}_{\text{(aq)}}+6{\text{NO+5H}}_{\text{2}}\text{O}\end{array}$

The final balanced equation is as follows:

  ${\text{2CrO}}_{\text{4}}^{\text{2-}}{}_{\text{(aq)}}{\text{+3N}}_{\text{2}}{\text{O+10H}}^{+}\to 2{\text{Cr}}^{3+}{}_{\text{(aq)}}+6{\text{NO+5H}}_{\text{2}}\text{O}$

In the above equation, the oxidizing agent is ${\text{CrO}}_4^{\text{2-}}$ and the reducing agent is ${\text{N}}_{\text{2}}\text{O}$.

(c)

Interpretation Introduction

Interpretation:

The following skeleton reaction has to be balanced and the oxidizing and reducing agents has to be identified.

  ${\text{Br}}_{2(l)}\to {\text{BrO}}_3^{-}{}_{\text{(aq)}}{\text{+Br}}^{-}{}_{(aq)}\text{[basic]}$

Concept Introduction:

Oxidizing agent:

A compound or group that causes the oxidation in another compound or group is said to be an oxidizing agent.  Most common oxidizing agents are oxygen, hydrogen peroxide and halogens.

Reducing agent:

A compound or group that causes the reduction in another compound or group is said to be a reducing agent.

Oxidation:

A reaction in which a compound gains electrons is said to be oxidation.

Reduction:

A reaction in which a compound loses electrons is said to be reduction.

Explanation of Solution

The given reaction is

  ${\text{Br}}_{2(l)}\to {\text{BrO}}_3^{-}{}_{\text{(aq)}}{\text{+Br}}^{-}{}_{(aq)}\text{[basic]}$

The balancing of the reaction is as follows:

Step-1:

The given reaction has to be divided into half-reactions.

  ${\text{Br}}_2\to {\text{BrO}}_{\text{3}}^{\text{-}}{\text{Br}}_{\text{2}}\to {\text{Br}}^{\text{-}}$

Step-2:

Balance the half-reactions.

To balance the reactions, atoms other than oxygen and hydrogen are not needed.

Balance $\text{O}$ atoms with ${\text{H}}_{\text{2}}\text{O}$:

  $\begin{array}{l}{\text{Br}}_2\to {\text{BrO}}_{\text{3}}^{\text{-}}{\text{Br}}_{\text{2}}\to {\text{Br}}^{\text{-}}\\ {\text{Br}}_2\to 2{\text{BrO}}_{\text{3}}^{\text{-}}{\text{Br}}_{\text{2}}\to 2{\text{Br}}^{\text{-}}\\ {\text{6H}}_{\text{2}}\text{O}+{\text{Br}}_2\to 2{\text{BrO}}_{\text{3}}^{\text{-}}{\text{Br}}_{\text{2}}\to 2{\text{Br}}^{\text{-}}\end{array}$

Balance $\text{H}$ atoms with ${\text{H}}^{\text{+}}$:

  $\begin{array}{l}{\text{Br}}_2\to {\text{BrO}}_{\text{3}}^{\text{-}}{\text{Br}}_{\text{2}}\to {\text{Br}}^{\text{-}}\\ {\text{Br}}_2\to 2{\text{BrO}}_{\text{3}}^{\text{-}}{\text{Br}}_{\text{2}}\to 2{\text{Br}}^{\text{-}}\\ {\text{6H}}_{\text{2}}\text{O}+{\text{Br}}_2\to 2{\text{BrO}}_{\text{3}}^{\text{-}}{\text{Br}}_{\text{2}}\to 2{\text{Br}}^{\text{-}}\\ {\text{6H}}_{\text{2}}\text{O}+{\text{Br}}_2\to 2{\text{BrO}}_{\text{3}}^{\text{-}}+{\text{12H}}^{\text{+}}{\text{Br}}_{\text{2}}\to 2{\text{Br}}^{\text{-}}\end{array}$

Addition of electrons:

  $\begin{array}{l}{\text{Br}}_2\to {\text{BrO}}_{\text{3}}^{\text{-}}{\text{Br}}_{\text{2}}\to {\text{Br}}^{\text{-}}\\ {\text{Br}}_2\to 2{\text{BrO}}_{\text{3}}^{\text{-}}{\text{Br}}_{\text{2}}\to 2{\text{Br}}^{\text{-}}\\ {\text{6H}}_{\text{2}}\text{O}+{\text{Br}}_2\to 2{\text{BrO}}_{\text{3}}^{\text{-}}{\text{Br}}_{\text{2}}\to 2{\text{Br}}^{\text{-}}\\ {\text{6H}}_{\text{2}}\text{O}+{\text{Br}}_2\to 2{\text{BrO}}_{\text{3}}^{\text{-}}+{\text{12H}}^{\text{+}}{\text{Br}}_{\text{2}}\to 2{\text{Br}}^{\text{-}}\\ {\text{6H}}_{\text{2}}\text{O}+{\text{Br}}_2\to 2{\text{BrO}}_{\text{3}}^{\text{-}}+{\text{12H}}^{\text{+}}{\text{+10e}}^{\text{-}}{\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to 2{\text{Br}}^{\text{-}}\end{array}$

Step-3:

Each half reaction is multiplied with some integer to make electrons lost equal to electrons gained.

  $\begin{array}{l}1\left({\text{6H}}_{\text{2}}\text{O}+{\text{Br}}_2\to 2{\text{BrO}}_{\text{3}}^{\text{-}}+{\text{12H}}^{\text{+}}{\text{+10e}}^{\text{-}}\right)5\left({\text{Br}}_{\text{2}}{\text{+2e}}^{\text{-}}\to 2{\text{Br}}^{\text{-}}\right)\\ {\text{6H}}_{\text{2}}\text{O}+{\text{Br}}_2\to 2{\text{BrO}}_{\text{3}}^{\text{-}}+{\text{12H}}^{\text{+}}{\text{+10e}}^{\text{-}}5{\text{Br}}_{\text{2}}{\text{+10e}}^{\text{-}}\to 10{\text{Br}}^{\text{-}}\end{array}$

Step-4:

Half-reactions are added and cancel the species appearing on both sides.

  $\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{6H}}_{\text{2}}\text{O}+{\text{Br}}_2\to 2{\text{BrO}}_{\text{3}}^{\text{-}}+{\text{12H}}^{\text{+}}\text{+}\cancel{{\text{10e}}^{\text{-}}}\\ 5{\text{Br}}_{\text{2}}\text{+}\cancel{{\text{10e}}^{\text{-}}}\to 10{\text{Br}}^{\text{-}}\end{array}}\\ {\text{6H}}_{\text{2}}\text{O}+6{\text{Br}}_{\text{2}}\to 2{\text{BrO}}_{\text{3}}^{\text{-}}+10{\text{Br}}^{\text{-}}+{\text{12H}}^{\text{+}}\end{array}$

To get the final balanced equation, the above equation is divided by 2.  The final balanced equation is as follows:

  ${\text{3H}}_{\text{2}}\text{O}+3{\text{Br}}_{\text{2}}\to {\text{BrO}}_{\text{3}}^{\text{-}}+5{\text{Br}}^{\text{-}}+{\text{6H}}^{\text{+}}$

In the above equation, the oxidizing agent and the reducing agent is ${\text{Br}}_{\text{2}}$.