Chapter 18, Problem 89AP

. Balance each of the following oxidation-reduction reactions, which take place in acidic solution.

a. ${\text{MnO}}_{\text{4}}{}^{-}\left(aq\right)+{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)\to {\text{Mn}}^{\text{2+}}\left(aq\right)+{\text{O}}_{\text{2}}\left(g\right)$   b. ${\text{BrO}}_{\text{3}}{}^{-}\left(aq\right)+{\text{Cu}}^{+}\left(aq\right)\to {\text{Br}}^{-}\left(aq\right)+{\text{Cu}}^{2+}\left(aq\right)$   c. ${\text{HNO}}_{\text{2}}\left(aq\right)+\text{I}\left(aq\right)\to \text{NO}\left(g\right)+{\text{I}}_{\text{2}}\left(aq\right)$

Interpretation Introduction

(a)

Interpretation:

The given half reaction should be balanced in acidic medium.

Concept Introduction:

The oxidation-reduction reaction is also known as a redox reaction. In this reaction, one reactant is oxidized and other is reduced. In balancing an oxidation-reduction reaction, they must be first divided into two half reactions: one is oxidation reaction and other is reduction reaction.

The balancing of redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half-reactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.

The following rules must be followed in balancing redox reaction by half equation method:

1. Initially, redox reaction is separated into two half equations; oxidation and reduction.
2. Atoms other than hydrogen and oxygen are balanced first in the unbalanced half equations.
3. Oxygen atoms are balanced by addition of water on either side of the reaction.
4. Hydrogen ion/s is added to balance the hydrogen atom.
5. Electrons are added to balance the charge.
6. Half reactions are added to get the net total equation.
7. The further addition of hydroxide ion takes place on both side of the reaction, if the solution is basic in nature to neutralise the hydrogen ion present in the solution.

Answer to Problem 89AP

${\text{2MnO}}_4^{-}\left(\text{a}\text{q}\right)+6{\text{H}}^{+}\left(\text{a}\text{q}\right)+5{\text{H}}_{\text{2}}\text{O}\left(\text{a}\text{q}\right)\to 2{\text{Mn}}^{2+}\left(\text{a}\text{q}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+5O}}_{\text{2}}\left(\text{g}\right)$.

Explanation of Solution

The given reaction is as follows:

${\text{MnO}}_4^{-}\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{a}\text{q}\right)\to {\text{Mn}}^{2+}\left(\text{a}\text{q}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)$

Separate the two half reactions as follows:

${\text{MnO}}_4^{-}\left(\text{a}\text{q}\right)\to {\text{Mn}}^{2+}\left(\text{a}\text{q}\right)$....... (1)

And,

${\text{H}}_{\text{2}}\text{O}\left(\text{a}\text{q}\right)\to {\text{O}}_{\text{2}}\left(\text{g}\right)$...... (2)

In reaction (1), add 4 water molecules to the right to balance the oxygen atoms,

${\text{MnO}}_4^{-}\left(\text{a}\text{q}\right)\to {\text{Mn}}^{2+}\left(\text{a}\text{q}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Add 8 hydrogen ions to the left to balance the number of hydrogen atoms thus,

${\text{MnO}}_4^{-}\left(\text{a}\text{q}\right)+8{\text{H}}^{+}\left(\text{a}\text{q}\right)\to {\text{Mn}}^{2+}\left(\text{a}\text{q}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Last step is to balance the charge, add 5 electrons to the left to balance the charge.

${\text{MnO}}_4^{-}\left(\text{a}\text{q}\right)+8{\text{H}}^{+}\left(\text{a}\text{q}\right)+5{\text{e}}^{-}\to {\text{Mn}}^{2+}\left(\text{a}\text{q}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$...... (3)

In reaction (2), add 2 hydrogen ions to the right to balance the hydrogen atoms.

${\text{H}}_{\text{2}}\text{O}\left(\text{a}\text{q}\right)\to {\text{O}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)$

Balance the charge by adding 2 electrons to the right:

${\text{H}}_{\text{2}}\text{O}\left(\text{a}\text{q}\right)\to {\text{O}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}$...... (4)

To get the net balanced reaction, add reaction (3) and (4)

$\begin{array}{l}2\times \left({\text{MnO}}_4^{-}\left(\text{a}\text{q}\right)+8{\text{H}}^{+}\left(\text{a}\text{q}\right)+5{\text{e}}^{-}\to {\text{Mn}}^{2+}\left(\text{a}\text{q}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\right)\\ 5\times \left({\text{H}}_{\text{2}}\text{O}\left(\text{a}\text{q}\right)\to {\text{O}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\right)\\ \overline{\underset{¯}{{\text{2MnO}}_4^{-}\left(\text{a}\text{q}\right)+6{\text{H}}^{+}\left(\text{a}\text{q}\right)+5{\text{H}}_{\text{2}}\text{O}\left(\text{a}\text{q}\right)\to 2{\text{Mn}}^{2+}\left(\text{a}\text{q}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+5O}}_{\text{2}}\left(\text{g}\right)}}\end{array}$

The balanced oxidation-reduction reaction is as follows:

${\text{2MnO}}_4^{-}\left(\text{a}\text{q}\right)+6{\text{H}}^{+}\left(\text{a}\text{q}\right)+5{\text{H}}_{\text{2}}\text{O}\left(\text{a}\text{q}\right)\to 2{\text{Mn}}^{2+}\left(\text{a}\text{q}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+5O}}_{\text{2}}\left(\text{g}\right)$.

Interpretation Introduction

(b)

Interpretation:

The given half reaction should be balanced in acidic medium.

Concept Introduction:

The oxidation-reduction reaction is also known as a redox reaction. In this reaction, one reactant is oxidized and other is reduced. In balancing an oxidation-reduction reaction, they must be first divided into two half reactions: one is oxidation reaction and other is reduction reaction.

The balancing of redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half-reactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.

The following rules must be followed in balancing redox reaction by half equation method:

1. Initially, redox reaction is separated into two half equations; oxidation and reduction.
2. Atoms other than hydrogen and oxygen are balanced first in the unbalanced half equations.
3. Oxygen atoms are balanced by addition of water on either side of the reaction.
4. Hydrogen ion/s is added to balance the hydrogen atom.
5. Electrons are added to balance the charge.
6. Half reactions are added to get the net total equation.
7. The further addition of hydroxide ion takes place on both side of the reaction, if the solution is basic in nature to neutralise the hydrogen ion present in the solution.

Answer to Problem 89AP

${\text{BrO}}_3^{-}\left(\text{a}\text{q}\right)+6{\text{H}}^{+}\left(\text{a}\text{q}\right)+6{\text{Cu}}^{+}\left(\text{a}\text{q}\right)\to {\text{Br}}^{-}\left(\text{a}\text{q}\right)+{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)+6{\text{Cu}}^{2+}\left(\text{a}\text{q}\right)$.

Explanation of Solution

${\text{BrO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{Cu}}^{+}\left(\text{a}\text{q}\right)\to {\text{Br}}^{-}\left(\text{a}\text{q}\right)+{\text{Cu}}^{2+}\left(\text{a}\text{q}\right)$

The above reaction can be separated into two half reactions as follows:

${\text{BrO}}_3^{-}\left(\text{a}\text{q}\right)\to {\text{Br}}^{-}\left(\text{a}\text{q}\right)$...... (1)

And,

${\text{Cu}}^{+}\left(\text{a}\text{q}\right)\to {\text{Cu}}^{2+}\left(\text{a}\text{q}\right)$...... (2)

To reaction (1) add 3 water molecule to the right to balance the number of oxygen atoms,

${\text{BrO}}_3^{-}\left(\text{a}\text{q}\right)\to {\text{Br}}^{-}\left(\text{a}\text{q}\right)+{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)$

To balance the hydrogen atom, add 6 hydrogen ions to the left

${\text{BrO}}_3^{-}\left(\text{a}\text{q}\right)+6{\text{H}}^{+}\left(\text{a}\text{q}\right)\to {\text{Br}}^{-}\left(\text{a}\text{q}\right)+{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Last step is to balance the charge, add 6 electrons to the left to balance the charge.

${\text{BrO}}_3^{-}\left(\text{a}\text{q}\right)+6{\text{H}}^{+}\left(\text{a}\text{q}\right)+6{\text{e}}^{-}\to {\text{Br}}^{-}\left(\text{a}\text{q}\right)+{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)$...... (3)

In reaction (2), charge can be balanced by adding 1 electron to the right thus,

${\text{Cu}}^{+}\left(\text{a}\text{q}\right)\to {\text{Cu}}^{2+}\left(\text{a}\text{q}\right)+{\text{e}}^{-}$...... (4)

To get the net balanced reaction, add reaction (3) and (4)

$\begin{array}{l}{\text{BrO}}_3^{-}\left(\text{a}\text{q}\right)+6{\text{H}}^{+}\left(\text{a}\text{q}\right)+6{\text{e}}^{-}\to {\text{Br}}^{-}\left(\text{a}\text{q}\right)+{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ 6\times \left({\text{Cu}}^{+}\left(\text{a}\text{q}\right)\to {\text{Cu}}^{2+}\left(\text{a}\text{q}\right)+{\text{e}}^{-}\right)\\ \overline{\underset{¯}{{\text{BrO}}_3^{-}\left(\text{a}\text{q}\right)+6{\text{H}}^{+}\left(\text{a}\text{q}\right)+6{\text{Cu}}^{+}\left(\text{a}\text{q}\right)\to {\text{Br}}^{-}\left(\text{a}\text{q}\right)+{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)+6{\text{Cu}}^{2+}\left(\text{a}\text{q}\right)}}\end{array}$

The balanced oxidation-reduction reaction is as follows:

${\text{BrO}}_3^{-}\left(\text{a}\text{q}\right)+6{\text{H}}^{+}\left(\text{a}\text{q}\right)+6{\text{Cu}}^{+}\left(\text{a}\text{q}\right)\to {\text{Br}}^{-}\left(\text{a}\text{q}\right)+{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)+6{\text{Cu}}^{2+}\left(\text{a}\text{q}\right)$.

Interpretation Introduction

(c)

Interpretation:

The given half reaction should be balanced in acidic medium.

Concept Introduction:

The oxidation-reduction reaction is also known as a redox reaction. In this reaction, one reactant is oxidized and other is reduced. In balancing an oxidation-reduction reaction, they must be first divided into two half reactions: one is oxidation reaction and other is reduction reaction.

The balancing of redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half-reactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.

The following rules must be followed in balancing redox reaction by half equation method:

1. Initially, redox reaction is separated into two half equations; oxidation and reduction.
2. Atoms other than hydrogen and oxygen are balanced first in the unbalanced half equations.
3. Oxygen atoms are balanced by addition of water on either side of the reaction.
4. Hydrogen ion/s is added to balance the hydrogen atom.
5. Electrons are added to balance the charge.
6. Half reactions are added to get the net total equation.
7. The further addition of hydroxide ion takes place on both side of the reaction, if the solution is basic in nature to neutralise the hydrogen ion present in the solution.

Answer to Problem 89AP

${\text{2HNO}}_{\text{2}}\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right){\text{+2I}}^{-}\left(\text{a}\text{q}\right)\to 2\text{NO}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)$.

Explanation of Solution

The given reaction is as follows:

${\text{HNO}}_{\text{2}}\left(\text{a}\text{q}\right)+{\text{I}}^{-}\left(\text{a}\text{q}\right)\to \text{NO}\left(\text{g}\right)+{\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)$

The two half reactions will be:

${\text{HNO}}_{\text{2}}\left(\text{a}\text{q}\right)\to \text{NO}\left(\text{g}\right)$...... (1)

And,

${\text{I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)$...... (2)

In reaction (1) add 1 water molecule to the right to balance number of oxygen atoms:

${\text{HNO}}_{\text{2}}\left(\text{a}\text{q}\right)\to \text{NO}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Now, add one hydrogen ion to balance the hydrogen atom.

${\text{HNO}}_{\text{2}}\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)\to \text{NO}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Charge can be balanced by adding 1 electron to the left,

${\text{HNO}}_{\text{2}}\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{e}}^{-}\to \text{NO}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$...... (3)

In reaction (2), give coefficient 2 to ${\text{I}}^{-}\left(\text{a}\text{q}\right)$ to balance the number of iodine atom.

${\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)$

Now, add two electrons to the right to balance the negative charge.

${\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}$...... (4)

To get the net balanced reaction, add reaction (3) and (4)

$\begin{array}{l}2\times \left({\text{HNO}}_{\text{2}}\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{e}}^{-}\to \text{NO}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\right)\\ {\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\\ \overline{\underset{¯}{{\text{2HNO}}_{\text{2}}\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right){\text{+2I}}^{-}\left(\text{a}\text{q}\right)\to 2\text{NO}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)}}\end{array}$

The balanced oxidation-reduction reaction is as follows:

${\text{2HNO}}_{\text{2}}\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right){\text{+2I}}^{-}\left(\text{a}\text{q}\right)\to 2\text{NO}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)$.