Chapter 17, Problem 30E

(a)

Interpretation Introduction

Interpretation:

The five oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

Answer to Problem 30E

The balanced equations are as follows,

${\text{2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3Cu}}_{\left(\text{s}\right)}\to {\text{3Cu}}^{\text{+2}}{}_{\left(\text{aq}\right)}{\text{+2NO}}_{\left(\text{g}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Explanation of Solution

The balanced equation is defined as follows,

${\text{2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3Cu}}_{\left(\text{s}\right)}\to {\text{3Cu}}^{\text{+2}}{}_{\left(\text{aq}\right)}{\text{+2NO}}_{\left(\text{g}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The reduction half cell reaction is,

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

The change in the oxidation number of nitrogen is from $+5$ to $+2$.

The oxidation half reaction is,

${\text{Cu}}_{\left(\text{s}\right)}\to {\text{Cu}}^{\text{+2}}{}_{\left(\text{aq}\right)}$

The change in oxidation number of copper is from zero to $+2$.

As the atoms other than hydrogen and oxygen are already balanced, so directly balance the oxygen atom in the reduction half reaction by adding water to right hand side,

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

Balance the hydrogen atoms in the reduction half reaction by adding ${\text{H}}^{+}$ ions to the left hand side,

${\text{NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+4H}}^{\text{+}}{}_{\left(\text{aq}\right)}\to {\text{NO}}_{\left(\text{g}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the charge by adding appropriate number of electrons to the left hand side,

${\text{NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+4H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3e}}^{\text{-}}\to {\text{NO}}_{\left(\text{g}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ (1)

As the atoms other than hydrogen and oxygen are already balanced in the oxidation half reaction and there are also no hydrogen or oxygen atom in the reaction. Therefore directly balance the charge by adding electrons to the right hand side.

${\text{Cu}}_{\left(\text{s}\right)}\to {\text{Cu}}^{\text{+2}}{}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}$ (2)

Multiply equation (1) by $2$ and equation (2) by $3$ and then add the oxidation and reduction half reaction to get the final equation,

$\begin{array}{c}{\text{2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3e}}^{\text{-}}\to {\text{2NO}}_{\left(\text{g}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{3Cu}}_{\left(\text{s}\right)}\to {\text{3Cu}}^{\text{+2}}{}_{\left(\text{aq}\right)}{\text{+6e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides. The final equation is,

${\text{2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3Cu}}_{\left(\text{s}\right)}\to {\text{3Cu}}^{\text{+2}}{}_{\left(\text{aq}\right)}{\text{+2NO}}_{\left(\text{g}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

(b)

Interpretation Introduction

Interpretation:

The five oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

Answer to Problem 30E

The balanced equations are as follows,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+14H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+6Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{2Cr}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}{\text{+3Cl}}_{\text{2}}{}_{\left(\text{g}\right)}$

Explanation of Solution

The balanced equation is defined as follows,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+14H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+6Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{2Cr}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}{\text{+3Cl}}_{\text{2}}{}_{\left(\text{g}\right)}$

The reduction half cell reaction is,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{Cr}}^{\text{3+}}{}_{\left(\text{aq}\right)}$

The change in the oxidation number of chromium is from $+6$ to $+3$.

The oxidation half reaction is,

${\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Cl}}_{\text{2}}{}_{\left(\text{g}\right)}$

The change in oxidation number of chlorine is from $-1$ to zero.

Balance all the atoms except hydrogen and oxygen in the reduction half reaction,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{2Cr}}^{\text{3+}}{}_{\left(\text{aq}\right)}$

Balance the oxygen in the reduction half reaction by adding water to right hand side,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{2Cr}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the hydrogen atoms in the reduction half reaction by adding ${\text{H}}^{+}$ ions to the left hand side,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+14H}}^{\text{+}}{}_{\left(\text{aq}\right)}\to {\text{2Cr}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the charge by adding appropriate number of electrons to the left hand side,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+14H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+6e}}^{\text{-}}\to {\text{2Cr}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ (3)

Balance all the atoms except hydrogen and oxygen in the oxidation half,

${\text{2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Cl}}_{\text{2}}{}_{\left(\text{g}\right)}$

As no hydrogen or oxygen atom is present in the above reaction. So, directly balance the charge by adding electrons at the appropriate side,

${\text{2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Cl}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+2e}}^{\text{-}}$ (4)

Multiply equation (4) by $3$ and then add the oxidation and reduction half reaction to get the final equation,

$\begin{array}{c}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+14H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+6e}}^{\text{-}}\to {\text{2Cr}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{6Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{3Cl}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+6e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides to get the final equation as,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+14H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+6Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{2Cr}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)\text{+}}{\text{3Cl}}_{\text{2}}{}_{\left(\text{g}\right)}$

(c)

Interpretation Introduction

Interpretation:

The five oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

Answer to Problem 30E

The balanced equations are as follows,

${\text{PbO}}_{\text{2}}{}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+Pb}}_{\left(\text{s}\right)}\to {\text{2PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Explanation of Solution

The balanced equation is defined as follows,

${\text{PbO}}_{\text{2}}{}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+Pb}}_{\left(\text{s}\right)}\to {\text{2PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The reduction half cell reaction is,

${\text{PbO}}_{\text{2}}{}_{\left(\text{s}\right)}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\to {\text{PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}$

The change in the oxidation number of lead is from $+4$ to $+2$.

The oxidation half reaction is,

${\text{Pb}}_{\left(\text{s}\right)}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\to {\text{PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}$

The change in oxidation number of bromine is from zero to $+2$.

All the elements except hydrogen and oxygen are already balanced so directly balance the oxygen atoms in the reduction half reaction by adding water molecules to the right hand side,

${\text{PbO}}_{\text{2}}{}_{\left(\text{s}\right)}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\to {\text{PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the hydrogen atoms by adding ${\text{H}}^{+}$ to left hand side,

${\text{PbO}}_{\text{2}}{}_{\left(\text{s}\right)}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+3H}}^{\text{+}}{}_{\left(\text{aq}\right)}\to {\text{PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the charge by adding appropriate number of electrons to the right hand side,

${\text{PbO}}_{\text{2}}{}_{\left(\text{s}\right)}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+3H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}\to {\text{PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ (5)

All the atoms except hydrogen and oxygen in oxidation half reaction are already balanced and oxygen is also balanced. So, directly balance hydrogen atoms by adding ${\text{H}}^{+}$ ions,

${\text{Pb}}_{\left(\text{s}\right)}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\to {\text{PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+H}}^{\text{+}}{}_{\left(\text{aq}\right)}$

Balance the charge by adding electrons at the appropriate side,

${\text{Pb}}_{\left(\text{s}\right)}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\to {\text{PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}$ (6)

Add the oxidation and reduction half reaction to get the final equation,

$\begin{array}{c}{\text{PbO}}_{\text{2}}{}_{\left(\text{s}\right)}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+3H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}\to {\text{PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{Pb}}_{\left(\text{s}\right)}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\to {\text{PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides. The final equation is,

$\begin{array}{l}{\text{PbO}}_{\text{2}}{}_{\left(\text{s}\right)}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+3H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+Pb}}_{\left(\text{s}\right)}{\text{+HSO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\to \\ {\text{PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}{\text{+PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+H}}^{\text{+}}{}_{\left(\text{aq}\right)}\end{array}$

The final equation is,

${\text{PbO}}_{\text{2}}{}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+Pb}}_{\left(\text{s}\right)}\to {\text{2PbSO}}_{\text{4}}{}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

(d)

Interpretation Introduction

Interpretation:

The five oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

Answer to Problem 30E

The balanced equations are as follows,

${\text{2MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+10Br}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+16H}}^{\text{+}}{}_{\left(\text{aq}\right)}\to {\text{2Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+5Br}}_{\text{2}}{}_{\left(\text{l}\right)}{\text{+8H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Explanation of Solution

The balanced equation is defined as follows,

$\begin{array}{l}{\text{2Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+5NaBiO}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+14H}}^{\text{+}}{}_{\left(\text{aq}\right)}\to \\ {\text{2MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+5Bi}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+5Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\end{array}$

The reduction half cell reaction is,

${\text{NaBiO}}_{\text{3}}{}_{\left(\text{s}\right)}\to {\text{Bi}}^{\text{3+}}{}_{\left(\text{aq}\right)}$

The change in the oxidation number of bismuth is from $+5$ to $+3$.

The oxidation half reaction is,

${\text{Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}\to {\text{MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

The change in oxidation number of manganese is from $+2$ to $+7$.

Balance all the atoms except hydrogen and oxygen in the reduction half cell,

${\text{NaBiO}}_{\text{3}}{}_{\left(\text{s}\right)}\to {\text{Bi}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+Na}}^{\text{+}}{}_{\left(\text{aq}\right)}$

Balance the oxygen in the reduction half reaction by adding water to right hand side,

${\text{NaBiO}}_{\text{3}}{}_{\left(\text{s}\right)}\to {\text{Bi}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the hydrogen atoms in the reduction half reaction by adding ${\text{H}}^{+}$ ions to the left hand side,

${\text{NaBiO}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+6H}}^{\text{+}}{}_{{}_{\left(\text{aq}\right)}}\to {\text{Bi}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the charge by adding appropriate number of electrons to the left hand side,

${\text{NaBiO}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+6H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}\to {\text{Bi}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ (7)

As all the atoms except hydrogen are already balanced in the oxidation half reaction. So, directly balance the oxygen atoms by adding water molecule to the left hand side,

${\text{Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

Balance the hydrogen atoms by adding ${\text{H}}^{+}$ to the right hand side,

${\text{Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}$

Balance the charge by adding electrons at the appropriate side,

${\text{Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+5e}}^{\text{-}}$ (8)

Multiply equation (7) by $5$ and equation (8) by $2$ to cancel out the electrons and then add the oxidation and reduction half reaction to get the final equation,

$\begin{array}{l}{\text{5NaBiO}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+30H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+10e}}^{\text{-}}\to {\text{5Bi}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+5Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+15H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{2Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+8H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to {\text{2MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+16H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+10e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides. The final equation is,

$\begin{array}{l}{\text{2Mn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+5NaBiO}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+14H}}^{\text{+}}{}_{\left(\text{aq}\right)}\to \\ {\text{2MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+5Bi}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+5Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+7H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The five oxidation-reduction reactions are given. The balancing of all the reactions in acidic media using half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

Explanation of Solution

The balanced equation is defined as follows,

${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+4Zn}}_{\left(\text{s}\right)}\to {\text{AsH}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+4Zn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The reduction half cell reaction is,

${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}\to {\text{AsH}}_{\text{3}}{}_{\left(\text{s}\right)}$

The change in the oxidation number of arsenic is from $+5$ to $-3$.

The oxidation half reaction is,

${\text{Zn}}_{\left(\text{s}\right)}\to {\text{Zn}}^{\text{2+}}{}_{\left(\text{aq}\right)}$

The change in oxidation number of zinc is from zero to $+2$.

All the atoms except hydrogen and oxygen in the reduction half cell are already balanced. So, directly balance the oxygen by adding water to right hand side,

${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}\to {\text{AsH}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the hydrogen atoms in the reduction half reaction by adding ${\text{H}}^{+}$ ions to the left hand side,

${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}\to {\text{AsH}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the charge by adding appropriate number of electrons to the left hand side,

${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+8e}}^{\text{-}}\to {\text{AsH}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ (9)

As all the atoms except hydrogen are already balanced in the oxidation half reaction and there are no hydrogen or oxygen atom. So, directly balance the charge by adding electrons to the right hand side,

${\text{Zn}}_{\left(\text{s}\right)}\to {\text{Zn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}$ (10)

Multiply equation (10) by $4$ to cancel out the electrons and then add the oxidation and reduction half reaction to get the final equation,

$\begin{array}{c}{\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+8e}}^{\text{-}}\to {\text{AsH}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{4Zn}}_{\left(\text{s}\right)}\to {\text{4Zn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+8e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides. The final equation is,

${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+4Zn}}_{\left(\text{s}\right)}\to {\text{AsH}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+4Zn}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$