Chapter 18, Problem 30E

(a)

Interpretation Introduction

Interpretation:

The balanced equations for the given redox reactions in acid medium should be written.

Concept introduction:

Balanced equation:

If the chemical equation contains the equal number of atoms and charge in both reactant and product side is known as balanced equation.

Oxidation- Reduction Equations:

The chemical reaction, a chemical species, which is loosed its electron to gets oxidized and simultaneously another species gained this loosed electron to gets reduces is known as oxidation- reduction reaction or Redox reaction.

Half reactions:

In the redox reaction split into two half reactions one is oxidation and another one is reduction.

In oxidation half reaction, a chemical species loses its electron to gets oxidized.

In reduction half reaction, a chemical species, which is gained the electron from solution to gets reduced.

Balanced Oxidation- Reduction Equation:

In the oxidation- reduction contains the equal number of atoms and charge in both reactant and a product side is known as balanced oxidation- reduction equation.

• Acidic medium redox reaction:

  The Two half reactions are write separately then the balance the all elements except hydrogen and oxygen.

  The Oxygen is balanced by the addition of water and Hydrogen is balanced by addition of ${\text{H}}^{\text{+}}$ ion.

The charge is balanced by the addition electrons

If necessary, the two are reactions are multiplied by integer and add these balanced two half reactions and remove the spectator species to give a balanced equation of redox reaction.

Answer to Problem 30E

The balanced equations for the given redox reaction is,

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

Explanation of Solution

Given:

${\text{Cu }}_{\text{(s)}}{\text{ + NO}}_{\text{3}}{}^{\text{-}}{}_{\text{(aq) }}\stackrel{}{\to }{\text{Cu}}^{\text{2+}}{}_{\text{ (aq)}}{\text{ + NO}}_{\text{(g)}}$

To write the equations for two half-reactions,

Oxidation half reaction:

$\text{Cu}\stackrel{}{\to }{\text{Cu}}^{\text{2+}}$

Reduction half reaction,

${\text{NO}}_{\text{3}}{}^{\text{-}}\stackrel{}{\to }\text{NO}$

To balance the equations for two half-reactions,

The balance oxygen by adding ${\text{H}}_{\text{2}}\text{O}$ and balance hydrogen by adding ${\text{H}}^{\text{+}}$

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

Charge is balanced by addition of electrons

$\text{Cu}\stackrel{}{\to }{\text{Cu}}^{\text{2+}}{\text{+2e}}^{\text{-}}$

The two balanced half reactions are added and cancelling the electrons to give a balanced equation.

${\text{3Cu}}_{\text{(s)}}{\text{+8H}}^{\text{+}}{\text{+2NO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}{\text{+6e}}^{\text{-}}\stackrel{}{\to }{\text{2NO}}_{\text{(g)}}{\text{+4H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+3Cu}}^{\text{2+}}{\text{+6e}}^{\text{-}}$

Hence, the balanced equation is,

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

(b)

Interpretation Introduction

Interpretation:

The balanced equations for the given redox reactions in acid medium should be written.

Concept introduction:

Balanced equation:

If the chemical equation contains the equal number of atoms and charge in both reactant and product side is known as balanced equation.

Oxidation- Reduction Equations:

The chemical reaction, a chemical species, which is loosed its electron to gets oxidized and simultaneously another species gained this loosed electron to gets reduces is known as oxidation- reduction reaction or Redox reaction.

Half reactions:

In the redox reaction split into two half reactions one is oxidation and another one is reduction.

In oxidation half reaction, a chemical species loses its electron to gets oxidized.

In reduction half reaction, a chemical species, which is gained the electron from solution to gets reduced.

Balanced Oxidation- Reduction Equation:

In the oxidation- reduction contains the equal number of atoms and charge in both reactant and a product side is known as balanced oxidation- reduction equation.

• Acidic medium redox reaction:

  The Two half reactions are write separately then the balance the all elements except hydrogen and oxygen.

  The Oxygen is balanced by the addition of water and Hydrogen is balanced by addition of ${\text{H}}^{\text{+}}$ ion.

The charge is balanced by the addition electrons

If necessary, the two are reactions are multiplied by integer and add these balanced two half reactions and remove the spectator species to give a balanced equation of redox reaction.

Answer to Problem 30E

The balanced equations for the given redox reaction is,

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

Explanation of Solution

Given:

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{2-}}_{\text{(s)}}{\text{ + Cl}}^{\text{-}}{}_{\text{(aq) }}\stackrel{}{\to }{\text{Cr}}^{\text{3+}}{}_{\text{ (aq)}}{\text{ +Cl}}_{\text{2}}{}_{\text{(g)}}$

To balance the equations for two half-reactions,

To write the equations for two half-reactions,

Oxidation half reaction:

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

Reduction half reaction,

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

To balance the equations for two half-reactions,

The balance oxygen by adding ${\text{H}}_{\text{2}}\text{O}$ and balance hydrogen by adding ${\text{H}}^{\text{+}}$

Charge is balanced by addition of electrons

The two balanced half reactions are added and cancelling the electrons to give a balanced equation.

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

${\text{6Cl}}^{\text{-}}{}_{\text{(aq)}}\underset{}{\to }{\text{3Cl}}_{\text{2(g)}}\text{+6e-}$

${\text{14H}}^{\text{+}}{\text{+Cr}}_{\text{2}}{\text{O}}_{\text{7}}{{}^{\text{2-}}}_{\text{(s)}}{\text{ +6e}}^{\text{-}}{\text{+6Cl}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{2Cr}}^{\text{3+}}{}_{\text{ (aq)}}{\text{+7H}}_{\text{2}}{\text{O+3Cl}}_{\text{2(g)}}$

Hence, the balanced equation is,

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

(c)

Interpretation Introduction

Interpretation:

The balanced equations for the given redox reactions in acid medium should be written.

Concept introduction:

Balanced equation:

If the chemical equation contains the equal number of atoms and charge in both reactant and product side is known as balanced equation.

Oxidation- Reduction Equations:

The chemical reaction, a chemical species, which is loosed its electron to gets oxidized and simultaneously another species gained this loosed electron to gets reduces is known as oxidation- reduction reaction or Redox reaction.

Half reactions:

In the redox reaction split into two half reactions one is oxidation and another one is reduction.

In oxidation half reaction, a chemical species loses its electron to gets oxidized.

In reduction half reaction, a chemical species, which is gained the electron from solution to gets reduced.

Balanced Oxidation- Reduction Equation:

In the oxidation- reduction contains the equal number of atoms and charge in both reactant and a product side is known as balanced oxidation- reduction equation.

• Acidic medium redox reaction:

  The Two half reactions are write separately then the balance the all elements except hydrogen and oxygen.

  The Oxygen is balanced by the addition of water and Hydrogen is balanced by addition of ${\text{H}}^{\text{+}}$ ion.

The charge is balanced by the addition electrons

If necessary, the two are reactions are multiplied by integer and add these balanced two half reactions and remove the spectator species to give a balanced equation of redox reaction.

Answer to Problem 30E

The balanced equations for the given redox reaction is,

${\text{Pb}}_{\text{(s)}}{\text{+PbO}}_{\text{2}}{}_{\text{(s)}}{\text{+2H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{2PbSO4}}_{\text{(s)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Explanation of Solution

Given:

${\text{Pb}}_{\text{(s)}}{\text{+PbO}}_{\text{2(s)}}{\text{+H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{PbSO}}_{\text{4}}{}_{\text{(s)}}$

To write the equations for two half-reactions,

Oxidation half reaction:

${\text{Pb}}_{\text{(s)}}\stackrel{}{\to }{\text{PbSO}}_{\text{4(s)}}{\text{+2e}}^{\text{-}}$

Reduction half reaction:

${\text{PbO}}_{\text{2}}{}_{\text{(s)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{PbSO}}_{\text{4(s)}}$

To balance the equations for two half-reactions,

The balance oxygen by adding ${\text{H}}_{\text{2}}\text{O}$ and balance hydrogen by adding ${\text{H}}^{\text{+}}$

Charge is balanced by addition of electrons

${\text{Pb}}_{\text{(s)}}{\text{+H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{PbSO}}_{\text{4(aq)}}{\text{+2e}}^{\text{-}}{\text{+2H}}^{\text{+}}$

${\text{PbO}}_{\text{2}}{}_{\text{(s)}}{\text{+H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}{\text{+2e}}^{\text{-}}{\text{+2H}}^{\text{+}}\stackrel{}{\to }{\text{PbSO}}_{\text{4(aq)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

The two balanced half reactions are added and cancelling the electrons to give a balanced equation.

${\text{Pb}}_{\text{(s)}}{\text{+PbO}}_{\text{2}}{}_{\text{(s)}}{\text{+2H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}{\text{+2H}}^{\text{+}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{2PbSO4}}_{\text{(s)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+2e}}^{\text{-}}{\text{+2H}}^{\text{+}}{}_{\text{(aq)}}$

Hence, the balanced equation is,

${\text{Pb}}_{\text{(s)}}{\text{+PbO}}_{\text{2}}{}_{\text{(s)}}{\text{+2H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{2PbSO4}}_{\text{(s)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(d)

Interpretation Introduction

Interpretation:

The balanced equations for the given redox reactions in acid medium should be written.

Concept introduction:

Balanced equation:

If the chemical equation contains the equal number of atoms and charge in both reactant and product side is known as balanced equation.

Oxidation- Reduction Equations:

The chemical reaction, a chemical species, which is loosed its electron to gets oxidized and simultaneously another species gained this loosed electron to gets reduces is known as oxidation- reduction reaction or Redox reaction.

Half reactions:

In the redox reaction split into two half reactions one is oxidation and another one is reduction.

In oxidation half reaction, a chemical species loses its electron to gets oxidized.

In reduction half reaction, a chemical species, which is gained the electron from solution to gets reduced.

Balanced Oxidation- Reduction Equation:

In the oxidation- reduction contains the equal number of atoms and charge in both reactant and a product side is known as balanced oxidation- reduction equation.

• Acidic medium redox reaction:

  The Two half reactions are write separately then the balance the all elements except hydrogen and oxygen.

  The Oxygen is balanced by the addition of water and Hydrogen is balanced by addition of ${\text{H}}^{\text{+}}$ ion.

The charge is balanced by the addition electrons

If necessary, the two are reactions are multiplied by integer and add these balanced two half reactions and remove the spectator species to give a balanced equation of redox reaction.

Answer to Problem 30E

The balanced equations for the given redox reaction is,

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

Explanation of Solution

Given:

${\text{Mn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+NaBiO}}_{\text{3}}{}_{\text{(s)}}\stackrel{}{\to }{\text{Bi}}^{\text{3+}}{}_{\text{(aq)}}{\text{+MnO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}$

To write the equations for two half-reactions,

Oxidation half reaction:

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

Reduction half reaction:

${\text{NaBIO}}_{\text{3}}\stackrel{}{\to }{\text{Bi}}^{\text{3+}}{}_{\text{(aq)}}$

To balance the equations for two half-reactions,

The balance oxygen by adding ${\text{H}}_{\text{2}}\text{O}$ and balance hydrogen by adding ${\text{H}}^{\text{+}}$

Charge is balanced by addition of electrons

${\text{2Mn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+14H}}^{\text{+}}\stackrel{}{\to }{\text{2MnO}}^{\text{4-}}{}_{\text{(aq)}}{\text{+5e}}^{\text{-}}$

${\text{5NaBiO}}_{\text{3(s)}}{\text{+5e}}^{\text{-}}\stackrel{}{\to }{\text{5Bi}}^{\text{3+}}{}_{\text{(aq)}}{\text{+5Na}}^{\text{+}}{}_{\text{(aq)}}\text{+7}{\text{.5H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

The two balanced half reactions are added and cancelling the electrons to give a balanced equation.

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

Hence, the balanced equation is,

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

(e)

Interpretation Introduction

Interpretation:

The balanced equations for the given redox reactions in acid medium should be written.

Concept introduction:

Balanced equation:

If the chemical equation contains the equal number of atoms and charge in both reactant and product side is known as balanced equation.

Oxidation- Reduction Equations:

The chemical reaction, a chemical species, which is loosed its electron to gets oxidized and simultaneously another species gained this loosed electron to gets reduces is known as oxidation- reduction reaction or Redox reaction.

Half reactions:

In the redox reaction split into two half reactions one is oxidation and another one is reduction.

In oxidation half reaction, a chemical species loses its electron to gets oxidized.

In reduction half reaction, a chemical species, which is gained the electron from solution to gets reduced.

Balanced Oxidation- Reduction Equation:

In the oxidation- reduction contains the equal number of atoms and charge in both reactant and a product side is known as balanced oxidation- reduction equation.

• Acidic medium redox reaction:

  The Two half reactions are write separately then the balance the all elements except hydrogen and oxygen.

  The Oxygen is balanced by the addition of water and Hydrogen is balanced by addition of ${\text{H}}^{\text{+}}$ ion.

The charge is balanced by the addition electrons

If necessary, the two are reactions are multiplied by integer and add these balanced two half reactions and remove the spectator species to give a balanced equation of redox reaction.

Answer to Problem 30E

The balanced equations for the given redox reaction is,

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

Explanation of Solution

Given:

${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4(aq)}}{\text{+Zn}}_{\text{(s)}}\stackrel{}{\to }{\text{AsH}}_{\text{3(g)}}{\text{+Zn}}_{\text{(aq)}}$

To write the equations for two half-reactions,

Oxidation half reaction:

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

Reduction half reaction:

${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}{\text{+8e}}^{\text{-}}\stackrel{}{\to }{\text{AsH}}_{\text{3}}$

To balance the equations for two half-reactions,

The balance oxygen by adding ${\text{H}}_{\text{2}}\text{O}$ and balance hydrogen by adding ${\text{H}}^{\text{+}}$

Charge is balanced by addition of electrons

$\text{Zn}\underset{}{\to }{\text{Zn}}^{\text{2+}}{\text{+2e}}^{\text{-}}$

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

The two balanced half reactions are added and cancelling the electrons to give a balanced equation.

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

Hence, the balanced equation is,

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