Chapter 19, Problem 19.2QP

(a)

Interpretation Introduction

Interpretation: The given redox equation should be balanced by the half-reaction method.

Concept Introduction:

Half-reaction can either be oxidation or reduction reaction of a reactant component in a redox reaction.

Oxidation: Removal of electrons from a chemical species is known as oxidation.

Reduction: Addition of electrons to a chemical species is known as reduction.

Redox reaction: It is a reaction which involves both reduction and oxidation reactions.

To balance the equation, use the following steps:

The general steps involved balancing the redox reactions in both acidic and basic conditions:

• Write the oxidation number for all of the elements involved in reaction.
• Based on the change in the oxidation number separate the reaction into reduced and oxidized half-reactions.
• Balance the atoms other than Oxygen and Hydrogen atom.
• Balance the Oxygen atoms by adding water on the appropriate side.
• Balance the hydrogen atoms by adding equal number of ${\text{H}}^{+}$ on the opposite side.
• Only in the basic condition, the ${\text{H}}^{+}$ ions are needed to be balanced further. The ${\text{H}}^{+}$ ions are balanced by the addition of equal number of ${\text{OH}}^{\text{-}}$ ions on both the sides. If there are equal number of ${\text{H}}^{+}$ and ${\text{OH}}^{\text{-}}$ ions on any of the two sides, it will constitute the corresponding number of water molecules.
• Balance the charge by adding appropriate number of electrons at the appropriate side.
• Multiply the reduction and oxidation half-reactions, by the whole number such that the number of electrons released in the oxidation half-reaction should be equal to the number of electrons gained in the reduction half-reaction.
• Add both the oxidation and reduction half-reactions and cancel out the electrons on both the sides of the reaction.
• Write the final equation and count the number of atoms and check the charge to verify the balanced reaction.

Answer to Problem 19.2QP

The balanced redox reaction is ${\text{2OH}}^{\text{-}}+{\text{H}}_{\text{2}}\text{O}\text{+}{\text{Mn}}^{2+}\text{+}{\text{H}}_2{\text{O}}_{\text{2}}\to {\text{MnO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}+{\text{2H}}_{\text{2}}\text{O}$

Explanation of Solution

The given equation in basic condition is shown below with oxidation states:

${\text{Mn}}^{2+}\text{+}{\text{H}}_2{\text{O}}_{\text{2}}\to {\text{MnO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}$

The manganese atoms are being balanced on both sides.

Balancing oxygen:

Water molecule is needed to be added to balance the oxygen atom which is shown below:

${\text{H}}_{\text{2}}\text{O}\text{+}{\text{Mn}}^{2+}\text{+}{\text{H}}_2{\text{O}}_{\text{2}}\to {\text{MnO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}$

Balancing hydrogen:

Now the hydrogen atoms are needed to be balanced by the addition of protons which is shown below:

${\text{H}}_{\text{2}}\text{O}\text{+}{\text{Mn}}^{2+}\text{+}{\text{H}}_2{\text{O}}_{\text{2}}\to {\text{MnO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}+{\text{2H}}^{\text{+}}$

Balancing Protons:

The protons can be balanced by adding the equal number of hydroxide ions on both sides.

${\text{2OH}}^{\text{-}}+{\text{H}}_{\text{2}}\text{O}\text{+}{\text{Mn}}^{2+}\text{+}{\text{H}}_2{\text{O}}_{\text{2}}\to {\text{MnO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}+{\text{2H}}^{\text{+}}+{\text{2OH}}^{\text{-}}$

The equal number of protons and hydroxide ions on one particular side is converted into the equal number of water molecules.

${\text{2OH}}^{\text{-}}+{\text{H}}_{\text{2}}\text{O}\text{+}{\text{Mn}}^{2+}\text{+}{\text{H}}_2{\text{O}}_{\text{2}}\to {\text{MnO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}+{\text{2H}}_{\text{2}}\text{O}$

So this is the overall balanced reaction.

(b)

Interpretation Introduction

Interpretation: The given redox equation should be balanced by the half-reaction method.

Concept Introduction:

Half-reaction can either be oxidation or reduction reaction of a reactant component in a redox reaction.

Oxidation: Removal of electrons from a chemical species is known as oxidation.

Reduction: Addition of electrons to a chemical species is known as reduction.

Redox reaction: It is a reaction which involves both reduction and oxidation reactions.

To balance the equation, use the following steps:

The general steps involved balancing the redox reactions in both acidic and basic conditions:

• Write the oxidation number for all of the elements involved in reaction.
• Based on the change in the oxidation number separate the reaction into reduced and oxidized half-reactions.
• Balance the atoms other than Oxygen and Hydrogen atom.
• Balance the Oxygen atoms by adding water on the appropriate side.
• Balance the hydrogen atoms by adding equal number of ${\text{H}}^{+}$ on the opposite side.
• Only in the basic condition, the ${\text{H}}^{+}$ ions are needed to be balanced further. The ${\text{H}}^{+}$ ions are balanced by the addition of equal number of ${\text{OH}}^{\text{-}}$ ions on both the sides. If there are equal number of ${\text{H}}^{+}$ and ${\text{OH}}^{\text{-}}$ ions on any of the two sides, it will constitute the corresponding number of water molecules.
• Balance the charge by adding appropriate number of electrons at the appropriate side.
• Multiply the reduction and oxidation half-reactions, by the whole number such that the number of electrons released in the oxidation half-reaction should be equal to the number of electrons gained in the reduction half-reaction.
• Add both the oxidation and reduction half-reactions and cancel out the electrons on both the sides of the reaction.
• Write the final equation and count the number of atoms and check the charge to verify the balanced reaction.

Answer to Problem 19.2QP

The balanced redox reaction is ${\text{e}}^{\text{-}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{Bi}{\left(\text{OH}\right)}_{\text{3}}\text{+}{\text{SnO}}_{\text{2}}^{\text{2-}}\stackrel{}{\to }{\text{SnO}}_{\text{3}}^{\text{2-}}\text{+}\text{Bi}\text{+}{\text{2H}}_{\text{2}}\text{O}\text{+}{\text{OH}}^{-}$

Explanation of Solution

The given equation in basic condition is shown below with oxidation states:

$\text{Bi}{\left(\text{OH}\right)}_{\text{3}}\text{+}{\text{SnO}}_{\text{2}}^{\text{2-}}\stackrel{}{\to }{\text{SnO}}_{\text{3}}^{\text{2-}}\text{+}\text{Bi}$

Both bismuth and tin atoms are being balanced on both sides.

Balancing oxygen:

Water molecule is needed to be added to balance the oxygen atom which is shown below:

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

Balancing hydrogen:

Now the hydrogen atoms are needed to be balanced by the addition of protons which is shown below:

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

Balancing Protons:

The protons can be balanced by adding the equal number of hydroxide ions on both sides.

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

The equal number of protons and hydroxide ions on one particular side is converted into the equal number of water molecules.

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

Balancing charge:

The charge has to be balanced by adding the required number of electrons to the more positively charged side.

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

So this is the overall balanced reaction.

(c)

Interpretation Introduction

Interpretation: The given redox equation should be balanced by the half-reaction method.

Concept Introduction:

Half-reaction can either be oxidation or reduction reaction of a reactant component in a redox reaction.

Oxidation: Removal of electrons from a chemical species is known as oxidation.

Reduction: Addition of electrons to a chemical species is known as reduction.

Redox reaction: It is a reaction which involves both reduction and oxidation reactions.

To balance the equation, use the following steps:

The general steps involved balancing the redox reactions in both acidic and basic conditions:

• Write the oxidation number for all of the elements involved in reaction.
• Based on the change in the oxidation number separate the reaction into reduced and oxidized half-reactions.
• Balance the atoms other than Oxygen and Hydrogen atom.
• Balance the Oxygen atoms by adding water on the appropriate side.
• Balance the hydrogen atoms by adding equal number of ${\text{H}}^{+}$ on the opposite side.
• Only in the basic condition, the ${\text{H}}^{+}$ ions are needed to be balanced further. The ${\text{H}}^{+}$ ions are balanced by the addition of equal number of ${\text{OH}}^{\text{-}}$ ions on both the sides. If there are equal number of ${\text{H}}^{+}$ and ${\text{OH}}^{\text{-}}$ ions on any of the two sides, it will constitute the corresponding number of water molecules.
• Balance the charge by adding appropriate number of electrons at the appropriate side.
• Multiply the reduction and oxidation half-reactions, by the whole number such that the number of electrons released in the oxidation half-reaction should be equal to the number of electrons gained in the reduction half-reaction.
• Add both the oxidation and reduction half-reactions and cancel out the electrons on both the sides of the reaction.
• Write the final equation and count the number of atoms and check the charge to verify the balanced reaction.

Answer to Problem 19.2QP

The balanced redox reaction is $6{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}+28{\text{H}}^{\text{+}}+2{\text{Cr}}_{\text{2}}\text{O}{}_{\text{7}}^{\text{2-}}\stackrel{}{\to }4{\text{Cr}}^{\text{3+}}+{\text{14H}}_2\text{O}\text{+}{\text{12CO}}_{\text{2}}$

Explanation of Solution

The given equation in acidic condition is given below:

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}\text{+}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\stackrel{}{\to }{\text{Cr}}^{\text{3+}}\text{+}{\text{CO}}_{\text{2}}$

The half reactions are:

Oxidation half-reaction is:

${\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\stackrel{}{\to }{\text{CO}}_{\text{2}}$

Balancing carbon:

${\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\stackrel{}{\to }2{\text{CO}}_{\text{2}}$

Balancing charge:

${\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\stackrel{}{\to }2{\text{CO}}_{\text{2}}+{\text{2e}}^{\text{-}}$

Reduction half-reaction is:

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

Balancing chromium:

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

Balancing oxygen:

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}\stackrel{}{\to }{\text{2Cr}}^{\text{3+}}\text{+}{\text{7H}}_2\text{O}$

Balancing hydrogen:

${\text{14H}}^{\text{+}}+{\text{Cr}}_{\text{2}}\text{O}{}_{\text{7}}^{\text{2-}}\stackrel{}{\to }{\text{2Cr}}^{\text{3+}}+{\text{7H}}_2\text{O}$

Balancing charge:

$6{\text{e}}^{\text{-}}+{\text{14H}}^{\text{+}}\text{+}{\text{Cr}}_{\text{2}}\text{O}{}_{\text{7}}^{\text{2-}}\stackrel{}{\to }{\text{2Cr}}^{\text{3+}}+{\text{7H}}_2\text{O}$

Determining the overall balanced equation:

The balanced reduction half-reaction is $6{\text{e}}^{\text{-}}+{\text{14H}}^{\text{+}}\text{+}{\text{Cr}}_{\text{2}}\text{O}{}_{\text{7}}^{\text{2-}}\stackrel{}{\to }{\text{2Cr}}^{\text{3+}}+{\text{7H}}_2\text{O}$

The balanced oxidation half-reaction is ${\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\stackrel{}{\to }2{\text{CO}}_{\text{2}}+{\text{2e}}^{\text{-}}$

$\begin{array}{l}2\left(6{\text{e}}^{\text{-}}+{\text{14H}}^{\text{+}}\text{+}{\text{Cr}}_{\text{2}}\text{O}{}_{\text{7}}^{\text{2-}}\stackrel{}{\to }{\text{2Cr}}^{\text{3+}}+{\text{7H}}_2\text{O}\right)\\ 6\left({\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\stackrel{}{\to }2{\text{CO}}_{\text{2}}+{\text{2e}}^{\text{-}}\right)\\ ---------------------------------\end{array}$

$\begin{array}{l}6{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}+1\bcancel{2}{\bcancel{\text{e}}}^{-}+28{\text{H}}^{\text{+}}+2{\text{Cr}}_{\text{2}}\text{O}{}_{\text{7}}^{\text{2-}}\stackrel{}{\to }4{\text{Cr}}^{\text{3+}}+{\text{14H}}_2\text{O}\text{+}{\text{12CO}}_{\text{2}}+1\bcancel{2}{\bcancel{\text{e}}}^{-}\\ ---------------------------------\end{array}$

$\begin{array}{l}6{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}+28{\text{H}}^{\text{+}}+2{\text{Cr}}_{\text{2}}\text{O}{}_{\text{7}}^{\text{2-}}\stackrel{}{\to }4{\text{Cr}}^{\text{3+}}+{\text{14H}}_2\text{O}\text{+}{\text{12CO}}_{\text{2}}\\ \text{This}\text{is}\text{the}\text{overall}\text{balanced}\text{equation in the acidic condition}\text{.}\end{array}$

(d)

Interpretation Introduction

Interpretation: The given redox equation should be balanced by the half-reaction method.

Concept Introduction:

Half-reaction can either be oxidation or reduction reaction of a reactant component in a redox reaction.

Oxidation: Removal of electrons from a chemical species is known as oxidation.

Reduction: Addition of electrons to a chemical species is known as reduction.

Redox reaction: It is a reaction which involves both reduction and oxidation reactions.

To balance the equation, use the following steps:

The general steps involved balancing the redox reactions in both acidic and basic conditions:

• Write the oxidation number for all of the elements involved in reaction.
• Based on the change in the oxidation number separate the reaction into reduced and oxidized half-reactions.
• Balance the atoms other than Oxygen and Hydrogen atom.
• Balance the Oxygen atoms by adding water on the appropriate side.
• Balance the hydrogen atoms by adding equal number of ${\text{H}}^{+}$ on the opposite side.
• Only in the basic condition, the ${\text{H}}^{+}$ ions are needed to be balanced further. The ${\text{H}}^{+}$ ions are balanced by the addition of equal number of ${\text{OH}}^{\text{-}}$ ions on both the sides. If there are equal number of ${\text{H}}^{+}$ and ${\text{OH}}^{\text{-}}$ ions on any of the two sides, it will constitute the corresponding number of water molecules.
• Balance the charge by adding appropriate number of electrons at the appropriate side.
• Multiply the reduction and oxidation half-reactions, by the whole number such that the number of electrons released in the oxidation half-reaction should be equal to the number of electrons gained in the reduction half-reaction.
• Add both the oxidation and reduction half-reactions and cancel out the electrons on both the sides of the reaction.
• Write the final equation and count the number of atoms and check the charge to verify the balanced reaction.

Answer to Problem 19.2QP

The balanced redox reaction is ${\text{2H}}^{+}\text{+}{\text{ClO}}_{\text{3}}^{\text{-}}\text{+}{\text{Cl}}^{\text{-}}\stackrel{}{\to }{\text{Cl}}_{\text{2}}\text{+}{\text{ClO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

The given equation in acidic condition is given below:

${\text{ClO}}_{\text{3}}^{\text{-}}\text{+}{\text{Cl}}^{\text{-}}\stackrel{}{\to }{\text{Cl}}_{\text{2}}\text{+}{\text{ClO}}_{\text{2}}$

The chlorine atoms are being balanced.

Balancing Oxygen:

Water molecule is needed to be added to balance the oxygen atom which is shown below:

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

Balancing Hydrogen:

Now the hydrogen atoms are needed to be balanced by the addition of protons which is shown below:

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

This is the overall balanced reaction.