Chapter 19, Problem 19.1QP

(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 19.1QP

The balanced redox reaction is ${\text{e}}^{-}\text{+}{\text{2H}}^{+}+{\text{H}}_2{\text{O}}_{\text{2}}\text{+}{\text{Fe}}^{\text{2+}}\to {\text{Fe}}^{\text{3+}}\text{+}{\text{H}}_2\text{O}\text{+}{\text{H}}_2\text{O}$

Explanation of Solution

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

${\text{H}}_2{\text{O}}_{\text{2}}\text{+}{\text{Fe}}^{\text{2+}}\to {\text{Fe}}^{\text{3+}}\text{+}{\text{H}}_2\text{O}$

• The hydrogen atoms and the $\text{Fe}$ ions are balanced on both the sides.
• Only the oxygen atoms are needed to be balanced.

Balancing oxygen:

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

${\text{H}}_2{\text{O}}_{\text{2}}\text{+}{\text{Fe}}^{\text{2+}}\to {\text{Fe}}^{\text{3+}}\text{+}{\text{H}}_2\text{O}\text{+}{\text{H}}_2\text{O}$

Balancing hydrogen:

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

${\text{2H}}^{+}+{\text{H}}_2{\text{O}}_{\text{2}}\text{+}{\text{Fe}}^{\text{2+}}\to {\text{Fe}}^{\text{3+}}\text{+}{\text{H}}_2\text{O}\text{+}{\text{H}}_2\text{O}$

Balancing charge:

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

${\text{e}}^{-}\text{+}{\text{2H}}^{+}+{\text{H}}_2{\text{O}}_{\text{2}}\text{+}{\text{Fe}}^{\text{2+}}\to {\text{Fe}}^{\text{3+}}\text{+}{\text{H}}_2\text{O}\text{+}{\text{H}}_2\text{O}$

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 19.1QP

The balanced redox reaction is ${\text{e}}^{-}\text{+}{\text{3H}}^{+}\text{+}\text{Cu}\text{+}{\text{HNO}}_{\text{3}}\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\text{+}\text{NO}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}{\text{H}}_2\text{O}$

Explanation of Solution

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

$\text{Cu}\text{+}{\text{HNO}}_{\text{3}}\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\text{+}\text{NO}\text{+}{\text{H}}_{\text{2}}\text{O}$

• The nitrogen atoms are balanced on both sides.
•  The oxygen and hydrogen atoms are needed to be balanced.

Balancing oxygen:

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

$\text{Cu}\text{+}{\text{HNO}}_{\text{3}}\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\text{+}\text{NO}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}{\text{H}}_2\text{O}$

Balancing hydrogen:

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

${\text{3H}}^{+}\text{+}\text{Cu}\text{+}{\text{HNO}}_{\text{3}}\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\text{+}\text{NO}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}{\text{H}}_2\text{O}$

Balancing charge:

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

${\text{e}}^{-}\text{+}{\text{3H}}^{+}\text{+}\text{Cu}\text{+}{\text{HNO}}_{\text{3}}\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\text{+}\text{NO}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}{\text{H}}_2\text{O}$

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 19.1QP

The balanced redox reaction is ${\text{2H}}_{\text{2}}\text{O}\text{+}{\text{CN}}^{\text{-}}\text{+}{\text{MnO}}_4^{-}\to {\text{CNO}}^{-}\text{+}{\text{MnO}}_2+{\text{H}}_{\text{2}}\text{O}\text{+}{\text{2OH}}^{-}$

Explanation of Solution

The given equation in basic condition is given below:

${\text{CN}}^{\text{-}}\text{+}{\text{MnO}}_4^{-}\to {\text{CNO}}^{-}\text{+}{\text{MnO}}_2$

• The carbon, nitrogen and manganese are being balanced on both sides.

Balancing Oxygen:

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

${\text{CN}}^{\text{-}}\text{+}{\text{MnO}}_4^{-}\to {\text{CNO}}^{-}\text{+}{\text{MnO}}_2+{\text{H}}_{\text{2}}\text{O}$

Balancing Hydrogen:

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

${\text{2H}}^{+}\text{+}{\text{CN}}^{\text{-}}\text{+}{\text{MnO}}_4^{-}\to {\text{CNO}}^{-}\text{+}{\text{MnO}}_2+{\text{H}}_{\text{2}}\text{O}$

Balancing Protons:

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

${\text{2OH}}^{-}+{\text{2H}}^{+}\text{+}{\text{CN}}^{\text{-}}\text{+}{\text{MnO}}_4^{-}\to {\text{CNO}}^{-}\text{+}{\text{MnO}}_2+{\text{H}}_{\text{2}}\text{O}\text{+}{\text{2OH}}^{-}$

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

${\text{2H}}_{\text{2}}\text{O}\text{+}{\text{CN}}^{\text{-}}\text{+}{\text{MnO}}_4^{-}\to {\text{CNO}}^{-}\text{+}{\text{MnO}}_2+{\text{H}}_{\text{2}}\text{O}\text{+}{\text{2OH}}^{-}$

This is the overall balanced reaction.

(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 19.1QP

The balanced redox reaction is ${\text{6OH}}^{\text{-}}+{\text{3H}}_2\text{O}\text{+}{\text{Br}}_{\text{2}}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}\text{+}{\text{Br}}^{\text{-}}+{\text{6H}}_2\text{O}\text{+}4{\text{e}}^{-}$

Explanation of Solution

The given equation in basic condition is given below:

${\text{Br}}_{\text{2}}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}\text{+}{\text{Br}}^{\text{-}}$

• The bromine atoms are being balanced on both the sides.

Balancing Oxygen:

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

${\text{3H}}_2\text{O}\text{+}{\text{Br}}_{\text{2}}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}\text{+}{\text{Br}}^{\text{-}}$

Balancing Hydrogen:

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

${\text{3H}}_2\text{O}\text{+}{\text{Br}}_{\text{2}}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}\text{+}{\text{Br}}^{\text{-}}+{\text{6H}}^{\text{+}}$

Balancing Protons:

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

${\text{6OH}}^{\text{-}}+{\text{3H}}_2\text{O}\text{+}{\text{Br}}_{\text{2}}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}\text{+}{\text{Br}}^{\text{-}}+{\text{6H}}^{\text{+}}+{\text{6OH}}^{\text{-}}$

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

${\text{6OH}}^{\text{-}}+{\text{3H}}_2\text{O}\text{+}{\text{Br}}_{\text{2}}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}\text{+}{\text{Br}}^{\text{-}}+{\text{6H}}_2\text{O}$

Balancing the charge:

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

${\text{6OH}}^{\text{-}}+{\text{3H}}_2\text{O}\text{+}{\text{Br}}_{\text{2}}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}\text{+}{\text{Br}}^{\text{-}}+{\text{6H}}_2\text{O}\text{+}4{\text{e}}^{-}$

This is the overall balanced reaction.

(e)

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 19.1QP

The balanced redox reaction is $2{\text{I}}_2+{\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\stackrel{}{\to }4{\text{I}}^{-}+{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}$

Explanation of Solution

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

${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}{\text{+I}}_2\to {\text{I}}^{\text{-}}+{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}$

The half reactions are:

Oxidation half-reaction:

${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\stackrel{}{\to }{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}$

Balancing sulfur atom:

${\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\stackrel{}{\to }{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}$

So the oxygen atoms are balanced.

Balancing charge:

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

${\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\stackrel{}{\to }{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}+{\text{2e}}^{\text{-}}$

Reduction half-reaction:

${\text{I}}_2\to {\text{I}}^{-}$

Balancing iodine atom:

${\text{I}}_2\to 2{\text{I}}^{-}$

Balancing charge:

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

${\text{e}}^{-}\text{+}{\text{I}}_2\to 2{\text{I}}^{-}$

Determining the overall balanced equation:

The balanced reduction half-reaction is ${\text{e}}^{-}\text{+}{\text{I}}_2\to 2{\text{I}}^{-}$

The balanced oxidation half-reaction is ${\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\stackrel{}{\to }{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}+{\text{2e}}^{\text{-}}$

$\begin{array}{l}2\left({\text{e}}^{-}\text{+}{\text{I}}_2\to 2{\text{I}}^{-}\right)\\ \left({\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\stackrel{}{\to }{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}+{\text{2e}}^{\text{-}}\right)\\ ------------------------\end{array}$

$\begin{array}{l}{\cancel{\text{2}}\cancel{\text{e}}\text{-}}^{}+2{\text{I}}_2+{\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\stackrel{}{\to }4{\text{I}}^{-}+{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}+{\cancel{\text{2}}\cancel{\text{e}}\text{-}}^{}\\ ------------------------\end{array}$

$\begin{array}{l}2{\text{I}}_2+{\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\stackrel{}{\to }4{\text{I}}^{-}+{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}\\ \text{This}\text{is}\text{the}\text{overall}\text{balanced}\text{equation in the acidic condition}\text{.}\end{array}$