Chapter 18, Problem 46QAP

Interpretation Introduction

(a)

Interpretation:

The given oxidation-reduction reaction should be balanced using the half-reaction method.

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 46QAP

$\text{2Al}\left(\text{s}\right)+6{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right)\to 2{\text{Al}}^{3+}\left(\text{a}\text{q}\right)+3{\text{H}}_2^{}\left(\text{g}\right)$.

Explanation of Solution

The given reaction is as follows:

$\text{Al}\left(\text{s}\right)+{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right)\to {\text{Al}}^{3+}\left(\text{a}\text{q}\right)+{\text{H}}_2^{}\left(\text{g}\right)$

First step is to separate the two half reaction,

$\text{Al}\left(\text{s}\right)\to {\text{Al}}^{3+}\left(\text{a}\text{q}\right)$....... (1)

And,

${\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right)\to {\text{H}}_2^{}\left(\text{g}\right)$...... (2)

Now to balance reaction (1), there is + 3 charge on right thus, 3 electrons are added to right,

$\text{Al}\left(\text{s}\right)\to {\text{Al}}^{3+}\left(\text{a}\text{q}\right)+3{\text{e}}^{-}$...... (3)

Similarly, reaction (2) can be balanced by adding hydrogen ion to the left or giving coefficient 2 to hydrogen ion on left thus,

$2{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right)\to {\text{H}}_2^{}\left(\text{g}\right)$

Now, to balance the charge, 2 electrons are added to left side of the reaction arrow thus,

$2{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\to {\text{H}}_2^{}\left(\text{g}\right)$..... (4)

Now, adding reaction (3) and (4) to get the net overall reaction,

$\begin{array}{l}2\left(\text{Al}\left(\text{s}\right)\to {\text{Al}}^{3+}\left(\text{a}\text{q}\right)+3{\text{e}}^{-}\right)\\ 3\left(2{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\to {\text{H}}_2^{}\left(\text{g}\right)\right)\\ \overline{\underset{¯}{\text{2Al}\left(\text{s}\right)+6{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right)\to 2{\text{Al}}^{3+}\left(\text{a}\text{q}\right)+3{\text{H}}_2^{}\left(\text{g}\right)}}\end{array}$

Thus, the balanced reaction by half reaction method will be:

$\text{2Al}\left(\text{s}\right)+6{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right)\to 2{\text{Al}}^{3+}\left(\text{a}\text{q}\right)+3{\text{H}}_2^{}\left(\text{g}\right)$.

Interpretation Introduction

(b)

Interpretation:

The given oxidation-reduction reaction should be balanced using the half-reaction method.

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 46QAP

${\text{3S}}^{2-}\left(\text{a}\text{q}\right){\text{+2NO}}_3^{-}\left(\text{a}\text{q}\right)+8{\text{H}}^{+}\left(\text{a}\text{q}\right)\to 3\text{S}\left(\text{s}\right)+2\text{NO}\left(\text{g}\right)+4{\text{H}}_{\text{2}}\text{O}$.

Explanation of Solution

The given reaction is as follows:

${\text{S}}^{2-}\left(\text{a}\text{q}\right)+{\text{NO}}_{\text{3}}{}^{-}\left(\text{a}\text{q}\right)\to \text{S}\left(\text{s}\right)+\text{NO}\left(\text{g}\right)$

First step is to separate the two half reaction,

${\text{S}}^{2-}\left(\text{a}\text{q}\right)\to \text{S}\left(\text{s}\right)$....... (1)

And,

${\text{NO}}_{\text{3}}{}^{-}\left(\text{a}\text{q}\right)\to \text{NO}\left(\text{g}\right)$...... (2)

First reaction 1 is balanced by adding 2 electrons to the right side of the reaction arrow:

${\text{S}}^{2-}\left(\text{a}\text{q}\right)\to \text{S}\left(\text{s}\right)+2{\text{e}}^{-}$...... (3)

Now, to balance reaction to 2 water molecules are added to right side of the reaction arrow:

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

Now, to balance hydrogen atoms, 4 hydrogen ions can be added to left side of the reaction arrow:

${\text{NO}}_{\text{3}}{}^{-}\left(\text{a}\text{q}\right)+4{\text{H}}^{+}\to \text{NO}\left(\text{g}\right)+{\text{2H}}_{\text{2}}\text{O}$

To balance the charge, 3 electrons must be added to left,

${\text{NO}}_{\text{3}}{}^{-}\left(\text{a}\text{q}\right)+4{\text{H}}^{+}+3{\text{e}}^{-}\to \text{NO}\left(\text{g}\right)+{\text{2H}}_{\text{2}}\text{O}$..... (4)

Now, adding reaction (3) and (4) to get the net overall reaction,

$\begin{array}{l}3\times \left({\text{S}}^{2-}\left(\text{a}\text{q}\right)\to \text{S}\left(\text{s}\right)+2{\text{e}}^{-}\right)\\ \text{2}\times \left({\text{NO}}_{\text{3}}{}^{-}\left(\text{a}\text{q}\right)+4{\text{H}}^{+}+3{\text{e}}^{-}\to \text{NO}\left(\text{g}\right)+{\text{2H}}_{\text{2}}\text{O}\right)\\ \overline{\underset{¯}{{\text{3S}}^{2-}\left(\text{a}\text{q}\right){\text{+2NO}}_3^{-}\left(\text{a}\text{q}\right)+8{\text{H}}^{+}\left(\text{a}\text{q}\right)\to 3\text{S}\left(\text{s}\right)+2\text{NO}\left(\text{g}\right)+4{\text{H}}_{\text{2}}\text{O}}}\end{array}$

Thus, the balanced reaction by half reaction method will be:

${\text{3S}}^{2-}\left(\text{a}\text{q}\right){\text{+2NO}}_3^{-}\left(\text{a}\text{q}\right)+8{\text{H}}^{+}\left(\text{a}\text{q}\right)\to 3\text{S}\left(\text{s}\right)+2\text{NO}\left(\text{g}\right)+4{\text{H}}_{\text{2}}\text{O}$.

Interpretation Introduction

(c)

Interpretation:

The given oxidation-reduction reaction should be balanced using the half-reaction method.

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 46QAP

${\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+5Cl}}_{\text{2}}\left(\text{a}\text{q}\right)\to 2{\text{IO}}_3^{-}\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)+10\text{HCl}\left(\text{g}\right)$.

Explanation of Solution

The given reaction is as follows:

${\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)+{\text{Cl}}_{\text{2}}\left(\text{a}\text{q}\right)\to {\text{IO}}_3^{-}\left(\text{a}\text{q}\right)+\text{HCl}\left(\text{g}\right)$

First step is to separate the two half reaction,

${\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)\to {\text{IO}}_3^{-}\left(\text{a}\text{q}\right)$....... (1)

And,

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

First reaction 1 is balanced by giving coefficient 2 to ${\text{IO}}_3^{-}\left(\text{a}\text{q}\right)$ thus,

${\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)\to 2{\text{IO}}_3^{-}\left(\text{a}\text{q}\right)$

Next step is to balance oxygen atom by adding 6 water molecule on left thus,

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

Now, hydrogen atoms are balanced adding 12 hydrogen ions to the right:

${\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to 2{\text{IO}}_3^{-}\left(\text{a}\text{q}\right)+12{\text{H}}^{+}\left(\text{a}\text{q}\right)$

Last step is to balance the charge, thus, 10 electrons are added to right side of the reaction arrow.

${\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to 2{\text{IO}}_3^{-}\left(\text{a}\text{q}\right)+12{\text{H}}^{+}\left(\text{a}\text{q}\right)+10{\text{e}}^{-}$..... (3)

Now to balance reaction (2), chlorine atom is balanced first by giving coefficient 2 to $\text{HCl}\left(\text{g}\right)$ as follows:

${\text{Cl}}_{\text{2}}\left(\text{a}\text{q}\right)\to 2\text{HCl}\left(\text{g}\right)$

To balance the hydrogen atom, 2 hydrogen ions are added to left thus,

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

Last step is to balance the charge, thus, two electrons are added to left side of the reaction arrow:

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

Now, adding reaction (3) and (4) to get the net overall reaction,

$\begin{array}{l}{\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to 2{\text{IO}}_3^{-}\left(\text{a}\text{q}\right)+12{\text{H}}^{+}\left(\text{a}\text{q}\right)+10{\text{e}}^{-}\\ \text{ 5}\times \left({\text{Cl}}_{\text{2}}\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\to 2\text{HCl}\left(\text{g}\right)\right)\\ \overline{\underset{¯}{{\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+5Cl}}_{\text{2}}\left(\text{a}\text{q}\right)\to 2{\text{IO}}_3^{-}\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)+10\text{HCl}\left(\text{g}\right)}}\end{array}$

Thus, the balanced reaction by half reaction method will be:

${\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+5Cl}}_{\text{2}}\left(\text{a}\text{q}\right)\to 2{\text{IO}}_3^{-}\left(\text{a}\text{q}\right)+2{\text{H}}^{+}\left(\text{a}\text{q}\right)+10\text{HCl}\left(\text{g}\right)$.

Interpretation Introduction

(d)

Interpretation:

The given oxidation-reduction reaction should be balanced using the half-reaction method.

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 46QAP

${\text{AsO}}_4^{-}\left(\text{a}\text{q}\right)+2{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right){\text{+S}}_{}^{2-}\left(\text{a}\text{q}\right)\to {\text{AsO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+\text{S}\left(\text{s}\right)$.

Explanation of Solution

The given reaction is as follows:

${\text{AsO}}_4^{-}\left(\text{a}\text{q}\right)+{\text{S}}_{}^{2-}\left(\text{a}\text{q}\right)\to {\text{AsO}}_3^{-}\left(\text{a}\text{q}\right)+\text{S}\left(\text{s}\right)$

First step is to separate the two half reaction,

${\text{AsO}}_4^{-}\left(\text{a}\text{q}\right)\to {\text{AsO}}_3^{-}\left(\text{a}\text{q}\right)$....... (1)

And,

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

In reaction 1, oxygen atom can be balanced by adding 1 water molecule to the right:

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

Now, hydrogen atom can be balanced by adding 2 hydrogen ions to the left thus,

${\text{AsO}}_4^{-}\left(\text{a}\text{q}\right)+2{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right)\to {\text{AsO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Last step is to balance the charge, since, there is + 1 charge on left and -1 charge on right thus, two electrons must be added to left thus,

${\text{AsO}}_4^{-}\left(\text{a}\text{q}\right)+2{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\to {\text{AsO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$...... (3) Now to balance reaction (2), charge can be balanced by adding 2 electrons to the right,

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

Now, adding reaction (3) and (4) to get the net overall reaction,

$\begin{array}{l}{\text{AsO}}_4^{-}\left(\text{a}\text{q}\right)+2{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\to {\text{AsO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ {\text{S}}_{}^{2-}\left(\text{a}\text{q}\right)\to \text{S}\left(\text{s}\right)+2{\text{e}}^{-}\\ \overline{\underset{¯}{{\text{AsO}}_4^{-}\left(\text{a}\text{q}\right)+2{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right){\text{+S}}_{}^{2-}\left(\text{a}\text{q}\right)\to {\text{AsO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+\text{S}\left(\text{s}\right)}}\end{array}$

Thus, the balanced reaction by half reaction method will be:

${\text{AsO}}_4^{-}\left(\text{a}\text{q}\right)+2{\mathrm{H<mo>+</mo>}}^{}\left(\text{a}\text{q}\right){\text{+S}}_{}^{2-}\left(\text{a}\text{q}\right)\to {\text{AsO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+\text{S}\left(\text{s}\right)$.