Chapter 18, Problem 47QAP

. Iodide ion, I- , is one of the most easily oxidized species. Balance each of the following oxidation-reduction reactions, which take place in acidic solution, by using the “half-reaction” method.

a. ${\text{IO}}_3{}^{-}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{aq}\right)$   b. ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\to {\text{Cr}}^{3+}\left(\text{aq}\right)+{\text{I}}_{\text{2}}\left(\text{aq}\right)$   c. ${\text{Cu}}^{2+}\left(\text{aq}\right)+{\text{I}}^{-}\left(\text{aq}\right)\to \text{CuI}\left(\text{s}\right)+{\text{I}}_{\text{2}}\left(\text{aq}\right)$

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 47QAP

${\text{2IO}}_3^{-}\left(\text{a}\text{q}\right)+12{\text{H}}^{+}\left(\text{a}\text{q}\right){\text{+10I}}^{-}\left(\text{a}\text{q}\right)\to 6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+6I}}_2\left(\text{a}\text{q}\right)$.

Explanation of Solution

The given reaction is as follows:

${\text{IO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)$

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

${\text{IO}}_3^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)$...... (1)

${\text{I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)$...... (2)

Here, ${\text{IO}}_3^{-}\left(\text{a}\text{q}\right)$ gets reduced to ${\text{I}}_2\left(\text{a}\text{q}\right)$ and ${\text{I}}^{-}\left(\text{a}\text{q}\right)$ gets oxidized to ${\text{I}}_2\left(\text{a}\text{q}\right)$.

To balance the reaction (1), give coefficient 2 to ${\text{IO}}_3^{-}\left(\text{a}\text{q}\right)$ to balance the iodine atom.

${\text{2IO}}_3^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)$

Now, oxygen atom can be balanced by adding 6 water molecules to right thus,

${\text{2IO}}_3^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Next step is to balance hydrogen atoms by adding 12 hydrogen ions to the left:

${\text{2IO}}_3^{-}\left(\text{a}\text{q}\right)+12{\text{H}}^{+}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The last step is to balance the charge thus, 4 electrons are added to left thus,

${\text{2IO}}_3^{-}\left(\text{a}\text{q}\right)+12{\text{H}}^{+}\left(\text{a}\text{q}\right)+10{\text{e}}^{-}\to {\text{I}}_2\left(\text{a}\text{q}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$...... (3)

Now, balance reaction (2) by given coefficient 2 to ${\text{I}}^{-}\left(\text{a}\text{q}\right)$ thus,

${\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)$

Now, charge can be balanced by adding 2 electrons to the right thus,

${\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)+2{\text{e}}^{-}$...... (4)

To get the net reaction, add reaction (3) and (4) as follows:

$\begin{array}{l}{\text{2IO}}_3^{-}\left(\text{a}\text{q}\right)+12{\text{H}}^{+}\left(\text{a}\text{q}\right)+10{\text{e}}^{-}\to {\text{I}}_2\left(\text{a}\text{q}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ 5\times \left({\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\right)\\ \overline{\underset{¯}{{\text{2IO}}_3^{-}\left(\text{a}\text{q}\right)+12{\text{H}}^{+}\left(\text{a}\text{q}\right){\text{+10I}}^{-}\left(\text{a}\text{q}\right)\to 6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+6I}}_2\left(\text{a}\text{q}\right)}}\end{array}$

Thus, the balanced chemical reaction is as follows:

${\text{2IO}}_3^{-}\left(\text{a}\text{q}\right)+12{\text{H}}^{+}\left(\text{a}\text{q}\right){\text{+10I}}^{-}\left(\text{a}\text{q}\right)\to 6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+6I}}_2\left(\text{a}\text{q}\right)$.

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 47QAP

${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{a}\text{q}\right)+14{\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{6I}}^{-}\left(\text{a}\text{q}\right)\to 2{\text{Cr}}^{3+}\left(\text{a}\text{q}\right)+7{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{I}}_2\left(\text{a}\text{q}\right)$.

Explanation of Solution

The given reaction is as follows:

${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{a}\text{q}\right)+{\text{I}}^{-}\left(\text{a}\text{q}\right)\to {\text{Cr}}^{3+}\left(\text{a}\text{q}\right){\text{+I}}_2\left(\text{a}\text{q}\right)$

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

${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{a}\text{q}\right)\to {\text{Cr}}^{3+}\left(\text{a}\text{q}\right)$...... (1)

And,

${\text{I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)$...... (2)

Reaction (1) can be balanced by giving coefficient 2 to ${\text{Cr}}^{3+}\left(\text{a}\text{q}\right)$ thus,

${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{a}\text{q}\right)\to 2{\text{Cr}}^{3+}\left(\text{a}\text{q}\right)$

Add 7 water molecules to the right to balance oxygen atoms thus,

${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{a}\text{q}\right)\to 2{\text{Cr}}^{3+}\left(\text{a}\text{q}\right)+7{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Hydrogen atom can be balanced by adding 14 hydrogen ions to the left

${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{a}\text{q}\right)+14{\text{H}}^{+}\left(\text{a}\text{q}\right)\to 2{\text{Cr}}^{3+}\left(\text{a}\text{q}\right)+7{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Last step is to balance the charge, the net charge on left side is + 12 and on right side is + 6 thus, 6 electrons are added to left to balance the charge thus,

${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{a}\text{q}\right)+14{\text{H}}^{+}\left(\text{a}\text{q}\right)+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}\left(\text{a}\text{q}\right)+7{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$...... (3)

In reaction (2), iodine atom is balanced by giving coefficient 2 to ${\text{I}}^{-}\left(\text{a}\text{q}\right)$ thus,

${\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)$

Next step is to balance the charge; two electrons are added to the right to balance the two negative charges on left side:

${\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)+2{\text{e}}^{-}$...... (4)

To get the net reaction, add reaction (3) and (4) as follows:

$\begin{array}{l}{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{a}\text{q}\right)+14{\text{H}}^{+}\left(\text{a}\text{q}\right)+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}\left(\text{a}\text{q}\right)+7{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ 3\times \left({\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\right)\\ \overline{\underset{¯}{{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{a}\text{q}\right)+14{\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{6I}}^{-}\left(\text{a}\text{q}\right)\to 2{\text{Cr}}^{3+}\left(\text{a}\text{q}\right)+7{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{I}}_2\left(\text{a}\text{q}\right)}}\end{array}$

Thus, the balanced chemical reaction is as follows:

${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\left(\text{a}\text{q}\right)+14{\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{6I}}^{-}\left(\text{a}\text{q}\right)\to 2{\text{Cr}}^{3+}\left(\text{a}\text{q}\right)+7{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{I}}_2\left(\text{a}\text{q}\right)$.

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 47QAP

${\text{2Cu}}_{}^{2+}\left(\text{a}\text{q}\right)+4{\text{I}}^{-}\left(\text{a}\text{q}\right)\to 2\text{CuI}\left(\text{s}\right)+{\text{I}}_2\left(\text{a}\text{q}\right)$.

Explanation of Solution

The given reaction is as follows:

${\text{Cu}}_{}^{2+}\left(\text{a}\text{q}\right)+{\text{I}}^{-}\left(\text{a}\text{q}\right)\to \text{CuI}\left(\text{s}\right){\text{+I}}_2\left(\text{a}\text{q}\right)$

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

${\text{Cu}}_{}^{2+}\left(\text{a}\text{q}\right)+{\text{I}}^{-}\left(\text{a}\text{q}\right)\to \text{CuI}\left(\text{s}\right)$..... (1)

And,

${\text{I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)$..... (2)

In reaction (1), charge can be balanced by adding electrons. Since, there is + 1 charge on left thus, 1 electron is added to left:

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

Now, in reaction (2), iodine atom can be balanced by giving coefficient 2 to ${\text{I}}^{-}\left(\text{a}\text{q}\right)$ thus,

${\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)$

The charge can be balanced by adding 2 electrons to the right thus,

${\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)+2{\text{e}}^{-}$..... (4)

To get the net reaction, add reaction (3) and (4) as follows:

$\begin{array}{l}2\left({\text{Cu}}_{}^{2+}\left(\text{a}\text{q}\right)+{\text{I}}^{-}\left(\text{a}\text{q}\right)+{\text{e}}^{-}\to \text{CuI}\left(\text{s}\right)\right)\\ {\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_2\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\\ \overline{\underset{¯}{{\text{2Cu}}_{}^{2+}\left(\text{a}\text{q}\right)+4{\text{I}}^{-}\left(\text{a}\text{q}\right)\to 2\text{CuI}\left(\text{s}\right)+{\text{I}}_2\left(\text{a}\text{q}\right)}}\end{array}$

Thus, the balanced chemical reaction is as follows:

${\text{2Cu}}_{}^{2+}\left(\text{a}\text{q}\right)+4{\text{I}}^{-}\left(\text{a}\text{q}\right)\to 2\text{CuI}\left(\text{s}\right)+{\text{I}}_2\left(\text{a}\text{q}\right)$.