Chapter 21, Problem 21.12P

(a)

Interpretation Introduction

Interpretation:

The given skeleton redox reaction has to be balanced and also the oxidizing and reducing agents has to be identified.

Concept Introduction:

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

Oxidizing agent: The substance that is oxidized is called as a reducing agent.

Reducing agent: The substance that is reduced is called as an oxidizing agent.

Steps in balancing redox reactions:

1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction.
2. 2) Balance atoms other than $\text{O}$ and $\text{H}$.
3. 3) Balance $\text{O}$ by adding ${\text{H}}_{\text{2}}\text{O}$ as needed
4. 4) Balance $\text{H}$ by adding ${\text{H}}^{\text{+}}$ ion at the required side.
5. 5) Balance charges by adding, as needed number of electrons,
6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
9. 9) For the reaction in acidic medium, the presence of $H^{+}$ ion allowed.
10. 10) For the reaction in basic medium, the $H^{+}$ ion should be neutralized by the addition of required amount of $O{H}^{-}$ ions on both the sides.

11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Basic conditions to balance the half-reactions are as follows:

• Balance the oxidation number with electrons.
• Balance the elements except for O and H.
• Balance O with ${\text{H}}_{\text{2}}\text{O}$.
• Balance H with ${\text{H}}^{\text{+}}$ ion.
• Balance the net charge with the electrons.
• Add $\text{O}{\text{H}}^{\text{-}}$ on both sides to cancel all  ${\text{H}}^{\text{+}}$ ions.

Explanation of Solution

The reaction between $Cl{O}_3^{-}\left(aq\right)$ and $I_{}^{-}\left(aq\right)$ contains two half reactions. They are divided and grouped into reactants and products with similar atoms.

$\begin{array}{c}Cl{O}_3^{-}\left(aq\right)\to C{l}_{}^{-}\left(aq\right)\\ 2I^{-}\left(aq\right)\to I_2^{}\left(s\right)\end{array}$

For above half reactions the atoms other than oxygen and hydrogen are balanced first. Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using $H^{+}$. Finally, the charges are balanced using electrons.

$\begin{array}{c}Cl{O}_3^{-}\left(aq\right)\to C{l}_{}^{-}\left(aq\right)\\ 2I^{-}\left(aq\right)\to I_2^{}\left(s\right)\end{array}$

$\begin{array}{c}Cl{O}^{-}{}_3\left(aq\right)\to C{l}_{}^{-}\left(aq\right)+3H_{2}O\left(l\right)\\ 2I^{-}\left(aq\right)\to I_2^{}\left(a\right)Oarebalanced\\ \end{array}$

$\begin{array}{c}Cl{O}^{-}{}_3\left(aq\right)+6H^{+}\left(aq\right)\to C{l}_{}^{-}\left(aq\right)+3H_{2}O\left(l\right)\\ 2I^{-}\left(aq\right)\to I_2^{}\left(s\right)Harebalanced\\ \\ Cl{O}^{-}{}_3\left(aq\right)+6H^{+}\left(aq\right)+6e^{-}\to C{l}_{}^{-}\left(aq\right)+3H_{2}O\left(l\right)\\ 2I^{-}\left(aq\right)\to I_2^{}\left(s\right)+2e^{-}\text{charges}arebalanced\\ \end{array}$

Finally obtained above reactions are multiplied with integers which will balance the electrons lost with the electrons gained. Here 2 electrons are lost and 6 electrons are gained hence the oxidation reaction has to be multiplied with integer 3.

$\begin{array}{c}Cl{O}^{-}{}_3\left(aq\right)+6H^{+}\left(aq\right)+6e^{-}\to C{l}_{}^{-}\left(aq\right)+3H_{2}O\left(l\right)\\ 6I^{-}\left(aq\right)\to 3I_2^{}\left(s\right)+6e^{-}\text{charges}arebalanced\\ \end{array}$

The above half reactions are added as follows in which the substances appear on both reactants and products sides are gets cancelled and merge into single molecular equation.

$\begin{array}{l}\underset{\_}{\begin{array}{c}Cl{O}^{-}{}_3\left(aq\right)+6H^{+}\left(aq\right)+\cancel{6e^{-}}\to C{l}_{}^{-}\left(aq\right)+3H_{2}O\left(l\right)\\ 6I^{-}\left(aq\right)\to 3I_2^{}\left(s\right)+\cancel{6e^{-}}\end{array}}\\ Cl{O}^{-}{}_3\left(aq\right)+6H^{+}\left(aq\right)+6I^{-}\left(aq\right)\to C{l}_{}^{-}\left(aq\right)+3H_{2}O\left(l\right)+3I_2^{}\left(s\right)\end{array}$

The substance that is reduced is called as an oxidizing agent. The substance that gets oxidized is termed as reducing agent.

From the balanced reaction it is clear that $Cl{O}_3^{-}$ is reduced to $C{l}_{}^{-}$ and $I_{}^{-}$ is oxidized to $I_2$.

Hence, the oxidizing agent is $Cl{O}_3^{-}$ and the reducing agent is $I_{}^{-}$.

(b)

Interpretation Introduction

Interpretation:

The given skeleton redox reaction has to be balanced and also the oxidizing and reducing agents has to be identified.

Concept Introduction:

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

Oxidizing agent: The substance that is oxidized is called as a reducing agent.

Reducing agent: The substance that is reduced is called as an oxidizing agent.

Steps in balancing redox reactions:

1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction.
2. 2) Balance atoms other than $\text{O}$ and $\text{H}$.
3. 3) Balance $\text{O}$ by adding ${\text{H}}_{\text{2}}\text{O}$ as needed
4. 4) Balance $\text{H}$ by adding ${\text{H}}^{\text{+}}$ ion at the required side.
5. 5) Balance charges by adding, as needed number of electrons,
6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
9. 9) For the reaction in acidic medium, the presence of $H^{+}$ ion allowed.
10. 10) For the reaction in basic medium, the $H^{+}$ ion should be neutralized by the addition of required amount of $O{H}^{-}$ ions on both the sides.

11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Basic conditions to balance the half-reactions are as follows:

• Balance the oxidation number with electrons.
• Balance the elements except for O and H.
• Balance O with ${\text{H}}_{\text{2}}\text{O}$.
• Balance H with ${\text{H}}^{\text{+}}$ ion.
• Balance the net charge with the electrons.
• Add $\text{O}{\text{H}}^{\text{-}}$ on both sides to cancel all ${\text{H}}^{\text{+}}$ ions.

Explanation of Solution

The reaction between $Mn{O}_4^{-}\left(aq\right)$ and $S{O}_3^{2-}\left(aq\right)$ contains two half reactions. They are divided and grouped into reactants and products with similar atoms.

$\begin{array}{c}Mn{O}_4^{-}\left(aq\right)\to Mn{O}_2^{}\left(s\right)\\ S{O}_3{}^{2-}\left(aq\right)\to S{O}_4{}^{2-}\left(aq\right)\end{array}$

For above half reactions the atoms other than oxygen and hydrogen are balanced first. Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using $H^{+}$. Finally, the charges are balanced using electrons.

$\begin{array}{c}Mn{O}_4^{-}\left(aq\right)\to Mn{O}_2^{}\left(s\right)\\ S{O}_3{}^{2-}\left(aq\right)\to S{O}_4{}^{2-}\left(aq\right)\end{array}$

$\begin{array}{c}Mn{O}_4^{-}\left(aq\right)\to Mn{O}_2^{}\left(s\right)+2H_{2}O\left(l\right)\\ S{O}_3{}^{2-}\left(aq\right)+H_{2}O\left(l\right)\to S{O}_4{}^{2-}\left(aq\right)Oarebalanced\end{array}$

$\begin{array}{c}Mn{O}_4^{-}\left(aq\right)+4H^{+}\left(aq\right)\to Mn{O}_2^{}\left(s\right)+2H_{2}O\left(l\right)\\ S{O}_3{}^{2-}\left(aq\right)+H_{2}O\left(l\right)\to S{O}_4{}^{2-}\left(aq\right)+2H^{+}\left(aq\right)Harebalanced\\ \\ Mn{O}_4^{-}\left(aq\right)+4H^{+}\left(aq\right)+3e^{-}\to Mn{O}_2^{}\left(s\right)+2H_{2}O\left(l\right)\\ S{O}_3{}^{2-}\left(aq\right)+H_{2}O\left(l\right)\to S{O}_4{}^{2-}\left(aq\right)+2H^{+}\left(aq\right)+2e^{-}\text{charges}arebalanced\\ \end{array}$

Finally obtained above reactions are multiplied with integers which will balance the electrons lost with the electrons gained. Here 2 electrons are lost and 3 electrons are gained hence the oxidation reaction has to be multiplied with integer 3 and reduction reaction has to be multiplied with 2 electrons.

$\begin{array}{c}2Mn{O}_4^{-}\left(aq\right)+8H^{+}\left(aq\right)+6e^{-}\to 2Mn{O}_2^{}\left(s\right)+4H_{2}O\left(l\right)\\ 3S{O}_3{}^{2-}\left(aq\right)+3H_{2}O\left(l\right)\to 3S{O}_4{}^{2-}\left(aq\right)+6H^{+}\left(aq\right)+6e^{-}\end{array}$

The above half reactions are added as follows in which the substances appear on both reactants and products sides are gets cancelled and merge into single equation.

$\begin{array}{l}\underset{\_}{\begin{array}{c}2Mn{O}_4^{-}\left(aq\right)+2\cancel{8H^{+}}\left(aq\right)+\cancel{6e^{-}}\to 2Mn{O}_2^{}\left(s\right)+\cancel{4}{H}_{2}O\left(l\right)\\ 3S{O}_3{}^{2-}\left(aq\right)+\cancel{3H_{2}O\left(l\right)}\to 3S{O}_4{}^{2-}\left(aq\right)+\cancel{6H^{+}}\left(aq\right)+\cancel{6e^{-}}\end{array}}\\ 2Mn{O}_4^{-}\left(aq\right)+2H^{+}\left(aq\right)+3S{O}_3{}^{2-}\left(aq\right)\to 2Mn{O}_2^{}\left(s\right)+H_{2}O\left(l\right)+3S{O}_4{}^{2-}\left(aq\right)\end{array}$

The above reaction is balanced one in acidic solution. In basic solution the reaction is obtained by adding $O{H}^{-}$ on both sides in order to neutralize $H^{+}$.

$\begin{array}{l}2Mn{O}_4^{-}\left(aq\right)+2H^{+}\left(aq\right)+3S{O}_3{}^{2-}\left(aq\right)+2O{H}^{-}\to 2Mn{O}_2^{}\left(s\right)+H_{2}O\left(l\right)+3S{O}_4{}^{2-}\left(aq\right)+2O{H}^{-}\\ 2Mn{O}_4^{-}\left(aq\right)+1\cancel{2}\cancel{H_{2}O\left(l\right)}+3S{O}_3{}^{2-}\left(aq\right)\to 2Mn{O}_2^{}\left(s\right)+\cancel{H_{2}O\left(l\right)}+3S{O}_4{}^{2-}\left(aq\right)+2O{H}^{-}\\ \underset{\_}{\underset{\_}{Balancedreaction:}}\\ 2Mn{O}_4^{-}\left(aq\right)+H_{2}O\left(l\right)+3S{O}_3{}^{2-}\left(aq\right)\to 2Mn{O}_2^{}\left(s\right)+3S{O}_4{}^{2-}\left(aq\right)+2O{H}^{-}\left(aq\right)\end{array}$

The substance that is reduced is called as an oxidizing agent. The substance that gets oxidized is termed as reducing agent.

From the balanced reaction it is clear that $Mn{O}_4^{-}$ is reduced to $Mn{O}_2$ and $S{O}_3^{2-}$ is oxidized to $S{O}_4^{2-}$.

Hence, the oxidizing agent is $Mn{O}_4^{-}$ and the reducing agent is $S{O}_3^{2-}$.

(c)

Interpretation Introduction

Interpretation:

The given skeleton redox reaction has to be balanced and also the oxidizing and reducing agents has to be identified.

Concept Introduction:

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called as oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or the gain of an electron in a species during a redox reaction is called as reduction.

Oxidizing agent: The substance that is oxidized is called as a reducing agent.

Reducing agent: The substance that is reduced is called as an oxidizing agent.

Steps in balancing redox reactions:

1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction.
2. 2) Balance atoms other than $\text{O}$ and $\text{H}$.
3. 3) Balance $\text{O}$ by adding ${\text{H}}_{\text{2}}\text{O}$ as needed
4. 4) Balance $\text{H}$ by adding ${\text{H}}^{\text{+}}$ ion at the required side.
5. 5) Balance charges by adding, as needed number of electrons,
6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
9. 9) For the reaction in acidic medium, the presence of $H^{+}$ ion allowed.
10. 10) For the reaction in basic medium, the $H^{+}$ ion should be neutralized by the addition of required amount of $O{H}^{-}$ ions on both the sides.

11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Basic conditions to balance the half-reactions are as follows:

• Balance the oxidation number with electrons.
• Balance the elements except for O and H.
• Balance O with ${\text{H}}_{\text{2}}\text{O}$.
• Balance H with ${\text{H}}^{\text{+}}$ ion.
• Balance the net charge with the electrons.
• Add $\text{O}{\text{H}}^{\text{-}}$ on both sides to cancel all  ${\text{H}}^{\text{+}}$ ions.

Explanation of Solution

The reaction between $Mn{O}_4^{-}$ and $H_2{O}_2\left(aq\right)$ contains two half reactions. They are divided and grouped into reactants and products with similar atoms.

$\begin{array}{c}Mn{O}_4^{-}\left(aq\right)\to M{n}^{2+}\left(aq\right)\\ H_2{O}_2\left(aq\right)\to O_2\left(g\right)\end{array}$

For above half reactions the atoms other than oxygen and hydrogen are balanced first. Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using $H^{+}$. Finally, the charges are balanced using electrons.

$\begin{array}{c}Mn{O}_4^{-}\left(aq\right)\to M{n}^{2+}\left(aq\right)\\ H_2{O}_2\left(aq\right)\to O_2\left(g\right)\end{array}$

$\begin{array}{c}Mn{O}_4^{-}\left(aq\right)\to M{n}^{2+}\left(aq\right)+4H_{2}O\left(l\right)\\ H_2{O}_2\left(aq\right)\to O_2\left(g\right)Oarebalanced\end{array}$

$\begin{array}{c}Mn{O}_4^{-}\left(aq\right)+8H^{+}\left(aq\right)\to M{n}^{2+}\left(aq\right)+4H_{2}O\left(l\right)\\ H_2{O}_2\left(aq\right)\to O_2\left(g\right)+2H^{+}\left(aq\right)Harebalanced\\ \\ Mn{O}_4^{-}\left(aq\right)+8H^{+}\left(aq\right)+5e^{-}\to M{n}^{2+}\left(aq\right)+4H_{2}O\left(l\right)\\ H_2{O}_2\left(aq\right)\to O_2\left(g\right)+2H^{+}\left(aq\right)+2e^{-}\text{charges}arebalanced\\ \end{array}$

Finally obtained above reactions are multiplied with integers which will balance the electrons lost with the electrons gained. Here 2 electrons are lost and 5 electrons are gained hence the oxidation reaction has to be multiplied with integer 5 and the reduction reaction is multiplied with 2.

$\begin{array}{c}2Mn{O}_4^{-}\left(aq\right)+16H^{+}\left(aq\right)+10e^{-}\to 2M{n}^{2+}\left(aq\right)+8H_{2}O\left(l\right)\\ 5H_2{O}_2\left(aq\right)\to 5O_2\left(g\right)+10H^{+}\left(aq\right)+10e^{-}\end{array}$

The above half reactions are added as follows in which the substances appear on both reactants and products sides are gets cancelled and merge into single molecular equation.

$\begin{array}{l}\underset{\_}{\begin{array}{c}2Mn{O}_4^{-}\left(aq\right)+6\cancel{16H^{+}\left(aq\right)}+1\cancel{0e^{-}}\to 2M{n}^{2+}\left(aq\right)+8H_{2}O\left(l\right)\\ 5H_2{O}_2\left(aq\right)\to 5O_2\left(g\right)+\cancel{10H^{+}\left(aq\right)}+\cancel{10e^{-}}\end{array}}\\ 2Mn{O}_4^{-}\left(aq\right)+6H^{+}\left(aq\right)+5H_2{O}_2\left(aq\right)\to 2M{n}^{2+}\left(aq\right)+8H_{2}O\left(l\right)+5O_2\left(g\right)\end{array}$

The substance that is reduced is called as an oxidizing agent. The substance that gets oxidized is termed as reducing agent.

From the balanced reaction it is clear that $Mn{O}_4^{-}$ is reduced to $M{n}_{}^{2+}$ and $H_2{O}_2$ is oxidized to $H_{2}O$.

Hence, the oxidizing agent is $Mn{O}_4^{-}$ and the reducing agent is $H_2{O}_2$.