Chapter 19, Problem 19.42QP

(a)

Interpretation Introduction

Interpretation:

Balanced equations for the given redox reactions, which are takes place in acidic and basic 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, which lose its electron to get oxidized.

In reduction half reaction, a chemical species, this gains electron from solution to get 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 written separately then 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.

• Basic medium redox reaction:

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

${\text{H}}^{\text{+}}$ Ions are eliminated by the addition of ${\text{OH}}^{\text{-}}$ ions in both sides.

In the reaction, Charges are balanced by the addition electrons.

If necessary, the two 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 19.42QP

The balanced equation for given redox reaction is,

${\text{18H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{ + 3Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}{\text{ + 6H}}^{\text{+}}\stackrel{}{\to }{\text{ 18O}}_{\text{2}}{\text{ + 2Cr}}_{\text{3}}{\text{ + 21H}}_{\text{2}}\text{O}$

Explanation of Solution

To write the equations for two half-reactions,

Oxidation half reaction:

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{O}}_{\text{2}}{\text{ + 2e}}^{\text{-}}$

Reduction half reaction,

${\text{3Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}{\text{ + 36e}}^{\text{-}}\stackrel{}{\to }{\text{2Cr}}_{\text{3}}$

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

Oxidation half reaction:

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{ O}}_{\text{2}}{\text{ + 2e}}^{\text{-}}{\text{ + 2H}}^{\text{+}}$

Reduction half reaction,

${\text{3Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}{\text{ + 36e}}^{\text{-}}{\text{ + 42H}}^{\text{+}}\stackrel{}{\to }{\text{ 2Cr}}_{\text{3}}{\text{ + 21H}}_{\text{2}}\text{O}$

Two half cell reactions are multiply the coefficients of all species by lowest common multipliers of half cell reactions, to make a equal number of electrons in both half reactions.

${\text{18H}}_{\text{2}}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{ 18O}}_{\text{2}}{\text{ + 36e}}^{\text{-}}{\text{ + 36H}}^{\text{+}}$

${\text{3Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}{\text{ + 36e}}^{\text{-}}{\text{ + 42H}}^{\text{+}}\stackrel{}{\to }{\text{ 2Cr}}_{\text{3}}{\text{ + 21H}}_{\text{2}}\text{O}$

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

${\text{18H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{ + 3Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}{\text{ + 36e}}^{\text{-}}{\text{ + 42H}}^{\text{+}}\stackrel{}{\to }{\text{18O}}_{\text{2}}{\text{ + 2Cr}}_{\text{3}}{\text{ + 36e}}^{\text{-}}{\text{ + 21H}}_{\text{2}}{\text{O + 36H}}^{\text{+}}$

Hence, the balanced equation is,

${\text{18H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{ + 3Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}{\text{ + 6H}}^{\text{+}}\stackrel{}{\to }{\text{ 18O}}_{\text{2}}{\text{ + 2Cr}}_{\text{3}}{\text{ + 21H}}_{\text{2}}\text{O}$

(b)

Interpretation Introduction

Interpretation:

Balanced equations for the given redox reactions, which are takes place in acidic and basic 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, which lose its electron to get oxidized.

In reduction half reaction, a chemical species, this gains electron from solution to get 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 written separately then 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.

• Basic medium redox reaction:

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

${\text{H}}^{\text{+}}$ Ions are eliminated by the addition of ${\text{OH}}^{\text{-}}$ ions in both sides.

In the reaction, Charges are balanced by the addition electrons.

If necessary, the two 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 19.42QP

The balanced equation for given redox reaction is,

${\text{3CN}}^{\text{-}}{\text{ + 2MnO}}_{\text{4}}{}^{\text{-}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{3CNO}}^{\text{-}}{\text{ + 2MnO}}_{\text{2}}{\text{ + 2OH}}^{\text{-}}$

Explanation of Solution

To write the equations for two half-reactions,

Oxidation half reaction:

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

Reduction half reaction,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}{\text{ + 6e}}^{\text{-}}\stackrel{}{\to }{\text{2Cr}}^{\text{3+}}$

To balance the equations for two half-reactions,

The balance of oxygen and hydrogen by adding water and ${\text{OH}}^{\text{-}}$

Charge is balanced by addition of electrons

Oxidation half reaction:

${\text{CN}}^{\text{-}}{\text{ + 2OH}}^{\text{-}}\stackrel{}{\to }{\text{CNO}}^{\text{-}}{\text{ + 2e}}^{\text{-}}{\text{ + H}}_{\text{2}}\text{O}$

Reduction half reaction,

${\text{MnO}}_{\text{4}}{}^{\text{-}}{\text{ + 3e}}^{\text{-}}{\text{ + 2H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{MnO}}_{\text{2}}{\text{ + 4OH}}^{\text{-}}$

Two half cell reactions are multiply the coefficients of all species by lowest common multipliers of half cell reactions, to make a equal number of electrons in both half reactions.

${\text{3CN}}^{\text{-}}{\text{ + 6OH}}^{\text{-}}\stackrel{}{\to }{\text{3CNO}}^{\text{-}}{\text{ + 6e}}^{\text{-}}{\text{ + 3H}}_{\text{2}}\text{O}$

${\text{2MnO}}_{\text{4}}{}^{\text{-}}{\text{ + 6e}}^{\text{-}}{\text{ + 4H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{ 2MnO}}_{\text{2}}{\text{ + 8OH}}^{\text{-}}$

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

${\text{3CN}}^{\text{-}}{\text{ + 2MnO}}_{\text{4}}{}^{\text{-}}{\text{ + 6OH}}^{\text{-}}{\text{ + 6e}}^{\text{-}}{\text{ + 4H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{ 3CNO}}^{\text{-}}{\text{ + 2MnO}}_{\text{2}}{\text{ + 6e}}^{\text{-}}{\text{ + 8OH}}^{\text{-}}{\text{ + 3H}}_{\text{2}}\text{O}$

Hence, the balanced equation is,

${\text{3CN}}^{\text{-}}{\text{ + 2MnO}}_{\text{4}}{}^{\text{-}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{3CNO}}^{\text{-}}{\text{ + 2MnO}}_{\text{2}}{\text{ + 2OH}}^{\text{-}}$

(c)

Interpretation Introduction

Interpretation:

Balanced equations for the given redox reactions, which are takes place in acidic and basic 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, which lose its electron to get oxidized.

In reduction half reaction, a chemical species, this gains electron from solution to get 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 written separately then 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.

• Basic medium redox reaction:

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

${\text{H}}^{\text{+}}$ Ions are eliminated by the addition of ${\text{OH}}^{\text{-}}$ ions in both sides.

In the reaction, Charges are balanced by the addition electrons.

If necessary, the two 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 19.42QP

The balanced equation for given redox reaction is,

$\text{2Cr}{\left(\text{OH}\right)}_{\text{4}}{}^{\text{-}}{\text{ + 3OCl}}^{\text{-}}{\text{ + 2OH}}^{\text{-}}\stackrel{}{\to }{\text{ 2CrO}}_{\text{4}}{}^{\text{2-}}{\text{ + 3Cl}}^{\text{-}}{\text{ + 5H}}_{\text{2}}\text{O}$

Explanation of Solution

To write the equations for two half-reactions,

Oxidation half reaction:

${\text{Cr(OH)}}_{\text{4}}{}^{\text{-}}\stackrel{}{\to }{\text{ CrO}}_{\text{4}}{}^{\text{2-}}{\text{ + 3e}}^{\text{-}}$

Reduction half reaction,

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

To balance the equations for two half-reactions,

The balance of oxygen and hydrogen by adding water and ${\text{OH}}^{\text{-}}$

Charge is balanced by addition of electrons

Oxidation half reaction:

${\text{Cr(OH)}}_{\text{4}}{}^{\text{-}}{\text{ + 4OH}}^{\text{-}}\stackrel{}{\to }{\text{CrO}}_{\text{4}}{}^{\text{2-}}{\text{ + 3e}}^{\text{-}}{\text{ + 4H}}_{\text{2}}\text{O}$

Reduction half reaction,

${\text{OCl}}^{\text{-}}{\text{ + 2e}}^{\text{-}}{\text{ + H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{ Cl}}^{\text{-}}{\text{ + 2OH}}^{\text{-}}$

Two half cell reactions are multiply the coefficients of all species by lowest common multipliers of half cell reactions, to make a equal number of electrons in both half reactions.

${\text{2Cr(OH)}}_{\text{4}}{}^{\text{-}}{\text{ + 8OH}}^{\text{-}}\stackrel{}{\to }{\text{ 2CrO}}_{\text{4}}{}^{\text{2-}}{\text{ + 6e}}^{\text{-}}{\text{ + 8H}}_{\text{2}}\text{O}$

${\text{3OCl}}^{\text{-}}{\text{ + 6e}}^{\text{-}}{\text{ + 3H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{ 3Cl}}^{\text{-}}{\text{ + 6OH}}^{\text{-}}$

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

${\text{2Cr(OH)}}_{\text{4}}{}^{\text{-}}{\text{ + 3OCl}}^{\text{-}}{\text{ + 8OH}}^{\text{-}}{\text{ + 6e}}^{\text{-}}{\text{ + 3H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{2CrO}}_{\text{4}}{}^{\text{2-}}{\text{ + 3Cl}}^{\text{-}}{\text{ + 6e}}^{\text{-}}{\text{ + 6OH}}^{\text{-}}{\text{ + 8H}}_{\text{2}}\text{O}$

Hence, the balanced equation is,

$\text{2Cr}{\left(\text{OH}\right)}_{\text{4}}{}^{\text{-}}{\text{ + 3OCl}}^{\text{-}}{\text{ + 2OH}}^{\text{-}}\stackrel{}{\to }{\text{ 2CrO}}_{\text{4}}{}^{\text{2-}}{\text{ + 3Cl}}^{\text{-}}{\text{ + 5H}}_{\text{2}}\text{O}$

(d)

Interpretation Introduction

Interpretation:

Balanced equations for the given redox reactions, which are takes place in acidic and basic 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, which lose its electron to get oxidized.

In reduction half reaction, a chemical species, this gains electron from solution to get 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 written separately then 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.

• Basic medium redox reaction:

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

${\text{H}}^{\text{+}}$ Ions are eliminated by the addition of ${\text{OH}}^{\text{-}}$ ions in both sides.

In the reaction, Charges are balanced by the addition electrons.

If necessary, the two 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 19.42QP

The balanced equation for given redox reaction is,

${\text{SO}}_{\text{2}}{\text{ + Br}}_{\text{2}}{\text{ + 2H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{SO}}_{\text{4}}{}^{\text{2-}}{\text{ + 2Br}}^{\text{-}}{\text{ + 4H}}^{\text{+}}$

Explanation of Solution

To write the equations for two half-reactions,

Oxidation half reaction:

${\text{SO}}_{\text{2}}\stackrel{}{\to }{\text{SO}}_{\text{4}}{}^{\text{2-}}{\text{ + 2e}}^{\text{-}}$

Reduction half reaction,

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

To balance the equations for two half-reactions,

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

Charge is balanced by addition of electrons

Oxidation half reaction:

${\text{SO}}_{\text{2}}{\text{ + 2H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{ SO}}_{\text{4}}{}^{\text{2-}}{\text{ + 2e}}^{\text{-}}{\text{ + 4H}}^{\text{+}}$

Reduction half reaction,

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

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

${\text{SO}}_{\text{2}}{\text{ + Br}}_{\text{2}}{\text{ + 2H}}_{\text{2}}{\text{O + 2e}}^{\text{-}}\stackrel{}{\to }{\text{SO}}_{\text{4}}{}^{\text{2-}}{\text{ + 2Br}}^{\text{-}}{\text{ + 2e}}^{\text{-}}{\text{ + 4H}}^{\text{+}}$

Hence, the balanced equation is,

${\text{SO}}_{\text{2}}{\text{ + Br}}_{\text{2}}{\text{ + 2H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{SO}}_{\text{4}}{}^{\text{2-}}{\text{ + 2Br}}^{\text{-}}{\text{ + 4H}}^{\text{+}}$

(e)

Interpretation Introduction

Interpretation:

Balanced equations for the given redox reactions, which are takes place in acidic and basic 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, which lose its electron to get oxidized.

In reduction half reaction, a chemical species, this gains electron from solution to get 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 written separately then 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.

• Basic medium redox reaction:

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

${\text{H}}^{\text{+}}$ Ions are eliminated by the addition of ${\text{OH}}^{\text{-}}$ ions in both sides.

In the reaction, Charges are balanced by the addition electrons.

If necessary, the two 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 19.42QP

The balanced equation for given redox reaction is,

${\text{24CuS + 16NO}}_{\text{3}}{}^{\text{-}}{\text{ + 64H}}^{\text{+}}\stackrel{}{\to }{\text{3S}}_{\text{8}}{\text{ + 16NO + 24Cu}}^{\text{2+}}{\text{ + 32H}}_{\text{2}}\text{O}$

Explanation of Solution

To write the equations for two half-reactions,

Oxidation half reaction:

$\text{8CuS }\stackrel{}{\to }{\text{ S}}_{\text{8}}{\text{ + 16e}}^{\text{-}}$

Reduction half reaction,

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

To balance the equations for two half-reactions,

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

Charge is balanced by addition of electrons

Oxidation half reaction:

$\text{8CuS }\stackrel{}{\to }{\text{ S}}_{\text{8}}{\text{ + 8Cu}}^{\text{2+}}{\text{ + 16e}}^{\text{-}}$

Reduction half reaction,

${\text{NO}}_{\text{3}}{}^{\text{-}}{\text{ + 3e}}^{\text{-}}{\text{ + 4H}}^{\text{+}}\stackrel{}{\to }{\text{ NO + 2H}}_{\text{2}}\text{O}$

Two half cell reactions are multiply the coefficients of all species by lowest common multipliers of half cell reactions, to make a equal number of electrons in both half reactions.

$\text{24CuS}\stackrel{}{\to }{\text{ 3S}}_{\text{8}}{\text{ + 24Cu}}^{\text{2+}}{\text{ + 48e}}^{\text{-}}$

${\text{16NO}}_{\text{3}}{}^{\text{-}}{\text{ + 48e}}^{\text{-}}{\text{ + 64H}}^{\text{+}}\stackrel{}{\to }{\text{ 16NO+ 32H}}_{\text{2}}\text{O}$

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

${\text{24CuS+ 16NO}}_{\text{3}}{}^{\text{-}}{\text{ + 48e}}^{\text{-}}{\text{ + 64H}}^{\text{+}}\stackrel{}{\to }{\text{ 3S}}_{\text{8}}{\text{ + 16NO+ 24Cu}}^{\text{2+}}{\text{ + 32H}}_{\text{2}}{\text{O + 48e}}^{\text{-}}$

Hence, the balanced equation is,

${\text{24CuS + 16NO}}_{\text{3}}{}^{\text{-}}{\text{ + 64H}}^{\text{+}}\stackrel{}{\to }{\text{3S}}_{\text{8}}{\text{ + 16NO + 24Cu}}^{\text{2+}}{\text{ + 32H}}_{\text{2}}\text{O}$