Chapter 18, Problem 44QAP

Interpretation Introduction

(a)

Interpretation:

The given half reaction should be balanced

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 44QAP

${\text{O}}_2\left(\text{g}\right)+4{\text{H}}^{+}\left(\text{a}\text{q}\right)+4{\text{e}}^{-}\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$.

Explanation of Solution

The given reaction is as follows:

${\text{O}}_2\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Oxygen atom is balanced by adding one water molecule on right side of the reaction arrow:

${\text{O}}_2\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Now, to balance hydrogen atoms, 4 hydrogen ions on left side of the reaction arrow.

${\text{O}}_2\left(\text{g}\right)+4{\text{H}}^{+}\left(\text{a}\text{q}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The last step is to balance the charge, to do so, 4 electrons can be added to left side of the reaction arrow as follows:

${\text{O}}_2\left(\text{g}\right)+4{\text{H}}^{+}\left(\text{a}\text{q}\right)+4{\text{e}}^{-}\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The above reaction is the balanced half reaction.

Interpretation Introduction

(b)

Interpretation:

The given half reaction should be balanced

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 44QAP

${\text{SO}}_4^{2-}\left(\text{a}\text{q}\right)+4{\text{H}}^{+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\to {\text{H}}_{\text{2}}{\text{SO}}_3\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$.

Explanation of Solution

The given reaction is as follows:

${\text{SO}}_4^{2-}\left(\text{a}\text{q}\right)\to {\text{H}}_{\text{2}}{\text{SO}}_3\left(\text{a}\text{q}\right)$

According to the rule, atom other than hydrogen and oxygen is balanced first. Here, sulfur atom is already balanced. Next step is to balance the oxygen and hydrogen atoms. To balance the oxygen atom, one water molecule can be added to right side of the reaction arrow thus,

${\text{SO}}_4^{2-}\left(\text{a}\text{q}\right)\to {\text{H}}_{\text{2}}{\text{SO}}_3\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Now, to balance hydrogen atom, add 4 hydrogen ions to left side as follows:

${\text{SO}}_4^{2-}\left(\text{a}\text{q}\right)+4{\text{H}}^{+}\left(\text{a}\text{q}\right)\to {\text{H}}_{\text{2}}{\text{SO}}_3\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The net charge on left side will be + 2, to balance the charge add 2 electrons to the left thus,

${\text{SO}}_4^{2-}\left(\text{a}\text{q}\right)+4{\text{H}}^{+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\to {\text{H}}_{\text{2}}{\text{SO}}_3\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The above reaction is the balanced half reaction.

Interpretation Introduction

(c)

Interpretation:

The given half reaction should be balanced

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 44QAP

${\text{H}}_{\text{2}}{\text{O}}_2\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$.

Explanation of Solution

The given reaction is as follows:

${\text{H}}_{\text{2}}{\text{O}}_2\left(\text{a}\text{q}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

First 1 water molecule can be added to right side to balance the oxygen atom.

${\text{H}}_{\text{2}}{\text{O}}_2\left(\text{a}\text{q}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Next step is to balance the hydrogen atom, to do so, 2 hydrogen ions must be added to left side thus,

${\text{H}}_{\text{2}}{\text{O}}_2\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Last step is to balance the charge, there is + 2 charge on left side, to balance the charge 2 electrons must be added to left side of the reaction arrow thus,

${\text{H}}_{\text{2}}{\text{O}}_2\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The above reaction is the balanced half reaction.

Interpretation Introduction

(d)

Interpretation:

The given half reaction should be balanced

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 44QAP

${\text{NO}}_2{}^{-}\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{NO}}_3{}^{-}\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}$.

Explanation of Solution

The given reaction is as follows:

${\text{NO}}_2{}^{-}\left(\text{a}\text{q}\right)\to {\text{NO}}_3{}^{-}\left(\text{a}\text{q}\right)$

According to the rule, atom other than hydrogen and oxygen is balanced first. Here, nitrogen atom is already balanced. The next step is to balance the oxygen atom. To balance the oxygen atom, one water molecule should be added to left side of the reaction arrow.

Thus,

${\text{NO}}_2{}^{-}\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{NO}}_3{}^{-}\left(\text{a}\text{q}\right)$

Now, to balance hydrogen atom, hydrogen ion is added to right side of the reaction. Since, there are 2 hydrogen atoms on the left side, 2 hydrogen ions should be added.

${\text{NO}}_2{}^{-}\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{NO}}_3{}^{-}\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)$

Next step is to balance the charge, there is + 1 charge on right side and -1 on the left, to balance the charge 2 electrons must be added to right side of the reaction arrow thus,

${\text{NO}}_2{}^{-}\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{NO}}_3{}^{-}\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}$

The above reaction is the balanced half reaction.