Chapter 19, Problem 19.101QP

Balance the following skeleton equations. The reactions occur in acidic or basic aqueous solution, as indicated.

1. a ${\text{MnO}}_4{}^{-}+{\text{S}}^{2-}\stackrel{}{\to }{\text{MnO}}_2+{\text{S}}_8(\text{basic})$
2. b ${\text{IO}}_3{}^{-}+{\text{HSO}}_3{}^{-}\stackrel{}{\to }{\text{I}}^{-}+{\text{SO}}_4{}^{2-}\text{(acidic)}$
3. c $\begin{array}{l}{\text{Fe(OH)}}_2+{\text{CrO}}_4{}^{2-}\stackrel{}{\to }\\ {\text{Fe(OH)}}_3+{\text{Ce(OH)}}_4{}^{-}\text{(basic)}\end{array}$
4. d ${\text{Cl}}_2\stackrel{}{\to }{\text{Cl}}^{-}+{\text{ClO}}^{-}\text{(basic)}$

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:

In chemical reaction, a chemical species, which loses its electron to gets oxidized and simultaneously another species gains this lost electron gets reduced 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, which gains 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 written separately then the balance 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 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 19.101QP

The balanced equation for given redox reaction is,

${\text{24S}}^{\text{2-}}{\text{ + 16MnO}}_{\text{4}}{}^{\text{-}}{\text{ + 32H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{3S}}_{\text{8}}{\text{ + 16MnO}}_{\text{2}}{\text{ + 64OH}}^{\text{-}}$

Explanation of Solution

To write the equations for two half-reactions,

Oxidation half reaction:

${\text{S}}^{\text{2-}}\stackrel{}{\to }{\text{ S}}_{\text{8}}$

Reduction half reaction,

${\text{MnO}}_{\text{4}}{}^{\text{-}}\stackrel{}{\to }{\text{ MnO}}_{\text{2}}$

To balance the equations for two half-reactions,

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

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

Charge is balanced by addition of electrons

Oxidation half reaction:

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

Reduction half reaction,

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

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

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

${\text{16MnO}}_{\text{4}}{}^{\text{-}}{\text{ + 64H}}_{\text{2}}{\text{O + 48e}}^{\text{-}}\stackrel{}{\to }{\text{16MnO}}_{\text{2}}{\text{ + 32H}}_{\text{2}}{\text{O + 64OH}}^{\text{-}}$

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

${\text{24S}}^{\text{2-}}{\text{ + 16MnO}}_{\text{4}}{}^{\text{-}}{\text{ + 64H}}_{\text{2}}{\text{O + 48e}}^{\text{-}}\stackrel{}{\to }{\text{ 3S}}_{\text{8}}{\text{ + 16MnO}}_{\text{2}}{\text{ + 48e}}^{\text{-}}{\text{ + 32H}}_{\text{2}}{\text{O + 64OH}}^{\text{-}}$

Hence, the balanced equation is,

${\text{24S}}^{\text{2-}}{\text{ + 16MnO}}_{\text{4}}{}^{\text{-}}{\text{ + 32H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{3S}}_{\text{8}}{\text{ + 16MnO}}_{\text{2}}{\text{ + 64OH}}^{\text{-}}$

Conclusion

Balanced equations for the given redox reactions, which takes place in acidic and basic medium were written.

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:

In chemical reaction, a chemical species, which loses its electron to gets oxidized and simultaneously another species gains this lost electron gets reduced 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, which gains 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 written separately then the balance 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 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 19.101QP

The balanced equation for given redox reaction is,

${\text{3HSO}}_{\text{3}}{}^{\text{-}}{\text{ + IO}}_{\text{3}}{}^{\text{-}}\stackrel{}{\to }{\text{3SO}}_{\text{4}}{}^{\text{2-}}{\text{ + I}}^{\text{-}}{\text{ + 3H}}^{\text{+}}$

Explanation of Solution

To write the equations for two half-reactions,

Oxidation half reaction:

${\text{HSO}}_{\text{3}}{}^{\text{-}}\stackrel{}{\to }{\text{SO}}_{\text{4}}{}^{\text{2-}}$

Reduction half reaction,

${\text{IO}}_{\text{3}}{}^{\text{-}}\stackrel{}{\to }{\text{ I}}^{\text{-}}$

To balance the equations for two half-reactions,

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

Charge is balanced by addition of electrons

Oxidation half reaction:

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

Reduction half reaction,

${\text{IO}}_{\text{3}}{}^{\text{-}}{\text{ + 6H}}^{\text{+}}{\text{ + 6e}}^{\text{-}}\stackrel{}{\to }{\text{I}}^{\text{-}}{\text{ + 3H}}_{\text{2}}\text{O}$

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

${\text{3HSO}}_{\text{3}}{}^{\text{-}}{\text{ + 3H}}_{\text{2}}\text{O }\stackrel{}{\to }{\text{3SO}}_{\text{4}}{}^{\text{2-}}{\text{ + 9H}}^{\text{+}}{\text{ + 6e}}^{\text{-}}$

${\text{IO}}_{\text{3}}{}^{\text{-}}{\text{ + 6H}}^{\text{+}}{\text{ + 6e}}^{\text{-}}\stackrel{}{\to }{\text{I}}^{\text{-}}{\text{ + 3H}}_{\text{2}}\text{O}$

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

${\text{3HSO}}_{\text{3}}{}^{\text{-}}{\text{ + IO}}_{\text{3}}{}^{\text{-}}{\text{ + 3H}}_{\text{2}}{\text{O + 6H}}^{\text{+}}{\text{ + 6e}}^{\text{-}}\stackrel{}{\to }{\text{3SO}}_{\text{4}}{}^{\text{2-}}{\text{ + I}}^{\text{-}}{\text{ + 9H}}^{\text{+}}{\text{ + 3H}}_{\text{2}}{\text{O + 6e}}^{\text{-}}$

Hence, the balanced equation is,

${\text{3HSO}}_{\text{3}}{}^{\text{-}}{\text{ + IO}}_{\text{3}}{}^{\text{-}}\stackrel{}{\to }{\text{3SO}}_{\text{4}}{}^{\text{2-}}{\text{ + I}}^{\text{-}}{\text{ + 3H}}^{\text{+}}$

Conclusion

Balanced equations for the given redox reactions, which takes place in acidic and basic medium were written.

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:

In chemical reaction, a chemical species, which loses its electron to gets oxidized and simultaneously another species gains this lost electron gets reduced 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, which gains 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 written separately then the balance 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 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 19.101QP

The balanced equation for given redox reaction is,

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

Explanation of Solution

To write the equations for two half-reactions,

Oxidation half reaction:

$\text{Fe}{\left(\text{OH}\right)}_{\text{2}}\stackrel{}{\to }\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$

Reduction half reaction,

${\text{CrO}}_{\text{4}}{}^{\text{2-}}\stackrel{}{\to }\text{Cr}{\left(\text{OH}\right)}_{\text{4}}{}^{\text{-}}$

To balance the equations for two half-reactions,

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

Charge is balanced by addition of electrons

Oxidation half reaction:

$\text{Fe}{\left(\text{OH}\right)}_{\text{2}}{\text{ + H}}_{\text{2}}{\text{O + OH}}^{\text{-}}\stackrel{}{\to }\text{Fe}{\left(\text{OH}\right)}_{\text{3}}{\text{ + H}}_{\text{2}}{\text{O + e}}^{\text{-}}$

Reduction half reaction,

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

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

$\text{3Fe}{\left(\text{OH}\right)}_{\text{2}}{\text{ + 3H}}_{\text{2}}{\text{O + 3OH}}^{\text{-}}\stackrel{}{\to }\text{3Fe}{\left(\text{OH}\right)}_{\text{3}}{\text{ + 3H}}_{\text{2}}{\text{O + 3e}}^{\text{-}}$

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

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

$\text{3Fe}{\left(\text{OH}\right)}_{\text{2}}{\text{ + CrO}}_{\text{4}}{}^{\text{2-}}{\text{ + 7H}}_{\text{2}}{\text{O + 3OH}}^{\text{-}}{\text{ + 3e}}^{\text{-}}\stackrel{}{\to }\text{3Fe}{\left(\text{OH}\right)}_{\text{3}}\text{ + Cr}{\left(\text{OH}\right)}_{\text{4}}{}^{\text{-}}{\text{ + 3H}}_{\text{2}}{\text{O + 4OH}}^{\text{-}}{\text{ + 3e}}^{\text{-}}$

Hence, the balanced equation is,

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

Conclusion

Balanced equations for the given redox reactions, which takes place in acidic and basic medium were written.

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:

In chemical reaction, a chemical species, which loses its electron to gets oxidized and simultaneously another species gains this lost electron gets reduced 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, which gains 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 written separately then the balance 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 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 19.101QP

The balanced equation for given redox reaction is,

${\text{40H}}_{\text{2}}{\text{S + 16MnO}}_{\text{4}}{}^{\text{-}}{\text{ + 48H}}^{\text{+}}\stackrel{}{\to }{\text{ 5S}}_{\text{8}}{\text{ + 16Mn}}^{\text{2+}}{\text{ + 64H}}_{\text{2}}\text{O}$

Explanation of Solution

To write the equations for two half-reactions,

Oxidation half reaction:

${\text{H}}_{\text{2}}\text{S }\stackrel{}{\to }{\text{ S}}_{\text{8}}$

Reduction half reaction,

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

To balance the equations for two half-reactions,

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

Charge is balanced by addition of electrons

Oxidation half reaction:

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

Reduction half reaction,

${\text{MnO}}_{\text{4}}{}^{\text{-}}{\text{ + 8H}}^{\text{+}}{\text{ + 5e}}^{\text{-}}\stackrel{}{\to }{\text{ Mn}}^{\text{2+}}{\text{ + 4H}}_{\text{2}}\text{O}$

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

${\text{40H}}_{\text{2}}\text{S }\stackrel{}{\to }{\text{5S}}_{\text{8}}{\text{ + 80H}}^{\text{+}}{\text{ + 80e}}^{\text{-}}$

${\text{16MnO}}_{\text{4}}{}^{\text{-}}{\text{ + 128H}}^{\text{+}}{\text{ + 80e}}^{\text{-}}\stackrel{}{\to }{\text{16Mn}}^{\text{2+}}{\text{ + 64H}}_{\text{2}}\text{O}$

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

${\text{40H}}_{\text{2}}{\text{S + 16MnO}}_{\text{4}}{}^{\text{-}}{\text{ + 128H}}^{\text{+}}{\text{ + 80e}}^{\text{-}}\stackrel{}{\to }{\text{5S}}_{\text{8}}{\text{ + 16Mn}}^{\text{2+}}{\text{ + 80H}}^{\text{+}}{\text{ + 64H}}_{\text{2}}{\text{O + 80e}}^{\text{-}}$

Hence, the balanced equation is,

${\text{40H}}_{\text{2}}{\text{S + 16MnO}}_{\text{4}}{}^{\text{-}}{\text{ + 48H}}^{\text{+}}\stackrel{}{\to }{\text{ 5S}}_{\text{8}}{\text{ + 16Mn}}^{\text{2+}}{\text{ + 64H}}_{\text{2}}\text{O}$

Conclusion

Balanced equations for the given redox reactions, which takes place in acidic and basic medium were written.