Chapter 18, Problem 101AP

Interpretation Introduction

(a)

Interpretation:

The given oxidation-reduction 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 101AP

${\text{2MnO}}_4^{-}\left(\text{a}\text{q}\right)+16{\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{10I}}^{-}\left(\text{a}\text{q}\right)\to 2{\text{Mn}}^{2+}\left(\text{a}\text{q}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+5{\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)$.

Explanation of Solution

The given reaction is as follows:

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

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

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

And,

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

In reaction (1), give coefficient 2 to ${\text{I}}^{-}$ to balance the number of iodine atoms.

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

Add two electrons to the right to balance the charge thus,

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

In reaction (2), add 4 water molecules to right side of the reaction to balance the number of oxygen atom,

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

Now, add 8 hydrogen ions to the left, to balance the hydrogen atoms.

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

Last step is to balance the charge, the net charge on left side is + 7 and that on right side is + 2 thus, add 5 electrons to the left to balance the charge,

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

To get the overall reaction, add reaction (3) and (4)

$\begin{array}{l}5\times \left({\text{2I}}^{-}\left(\text{a}\text{q}\right)\to {\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\right)\\ 2\times \left({\text{MnO}}_4^{-}\left(\text{a}\text{q}\right)+8{\text{H}}^{+}\left(\text{a}\text{q}\right)+5{\text{e}}^{-}\to {\text{Mn}}^{2+}\left(\text{a}\text{q}\right)+4{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\right)\\ \overline{\underset{¯}{{\text{2MnO}}_4^{-}\left(\text{a}\text{q}\right)+16{\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{10I}}^{-}\left(\text{a}\text{q}\right)\to 2{\text{Mn}}^{2+}\left(\text{a}\text{q}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+5{\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)}}\end{array}$

Thus, the balance oxidation-reduction reaction is as follows:

${\text{2MnO}}_4^{-}\left(\text{a}\text{q}\right)+16{\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{10I}}^{-}\left(\text{a}\text{q}\right)\to 2{\text{Mn}}^{2+}\left(\text{a}\text{q}\right)+8{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+5{\text{I}}_{\text{2}}\left(\text{a}\text{q}\right)$.

Interpretation Introduction

(b)

Interpretation:

The given oxidation-reduction 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 101AP

${\text{4Cr}}_{}^{3+}\left(\text{a}\text{q}\right)+14\text{H}\text{O}\left(\text{l}\right)+3{\text{S}}_{\text{2}}{\text{O}}_8{}^{2-}\left(\text{a}\text{q}\right)\to 2{\text{Cr}}_2\text{O}{}^{2-}\left(\text{a}\text{q}\right)+28{\text{H}}^{+}\left(\text{a}\text{q}\right)+6{\text{SO}}_4{}^{3-}$.

Explanation of Solution

The given reaction is as follows:

${\text{S}}_{\text{2}}{\text{O}}_8{}^{2-}\left(\text{a}\text{q}\right)+{\text{Cr}}_{}^{3+}\left(\text{a}\text{q}\right)\to {\text{SO}}_4{}^{3-}\left(\text{a}\text{q}\right)+{\text{Cr}}_2\text{O}{}^{2-}\left(\text{a}\text{q}\right)$

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

${\text{S}}_{\text{2}}{\text{O}}_8{}^{2-}\left(\text{a}\text{q}\right)\to {\text{SO}}_4{}^{3-}\left(\text{a}\text{q}\right)$...... (1)

And,

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

In reaction (1), give coefficient 2 to ${\text{SO}}_4{}^{3-}\left(\text{a}\text{q}\right)$ to balance the number of sulfur atoms.

${\text{S}}_{\text{2}}{\text{O}}_8{}^{2-}\left(\text{a}\text{q}\right)\to 2{\text{SO}}_4{}^{3-}\left(\text{a}\text{q}\right)$

Add 4 electrons to the left to balance the charge thus,

${\text{S}}_{\text{2}}{\text{O}}_8{}^{2-}\left(\text{a}\text{q}\right)+4{\text{e}}^{-}\to 2{\text{SO}}_4{}^{3-}\left(\text{a}\text{q}\right)$...... (3)

In reaction (2), give coefficient 2 to ${\text{Cr}}^{3+}$ to balance the chromium atom thus,

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

Add 7 water molecules to left side of the reaction to balance the number of oxygen atom,

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

Balance the hydrogen atoms to add 14 hydrogen ions to the right:

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

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

${\text{2Cr}}_{}^{3+}\left(\text{a}\text{q}\right)+7\text{H}\text{O}\left(\text{l}\right)\to {\text{Cr}}_2\text{O}{}^{2-}\left(\text{a}\text{q}\right)+14{\text{H}}^{+}\left(\text{a}\text{q}\right)+6{\text{e}}^{-}$...... (4)

To get the overall reaction, add reaction (3) and (4)

$\begin{array}{l}3\times \left({\text{S}}_{\text{2}}{\text{O}}_8{}^{2-}\left(\text{a}\text{q}\right)+4{\text{e}}^{-}\to 2{\text{SO}}_4{}^{3-}\left(\text{a}\text{q}\right)\right)\\ 2\times \left({\text{2Cr}}_{}^{3+}\left(\text{a}\text{q}\right)+7\text{H}\text{O}\left(\text{l}\right)\to {\text{Cr}}_2\text{O}{}^{2-}\left(\text{a}\text{q}\right)+14{\text{H}}^{+}\left(\text{a}\text{q}\right)+6{\text{e}}^{-}\right)\\ \overline{\underset{¯}{{\text{4Cr}}_{}^{3+}\left(\text{a}\text{q}\right)+14\text{H}\text{O}\left(\text{l}\right)+3{\text{S}}_{\text{2}}{\text{O}}_8{}^{2-}\left(\text{a}\text{q}\right)\to 2{\text{Cr}}_2\text{O}{}^{2-}\left(\text{a}\text{q}\right)+28{\text{H}}^{+}\left(\text{a}\text{q}\right)+6{\text{SO}}_4{}^{3-}}}\end{array}$

Thus, the balance oxidation-reduction reaction is as follows:

${\text{4Cr}}_{}^{3+}\left(\text{a}\text{q}\right)+14\text{H}\text{O}\left(\text{l}\right)+3{\text{S}}_{\text{2}}{\text{O}}_8{}^{2-}\left(\text{a}\text{q}\right)\to 2{\text{Cr}}_2\text{O}{}^{2-}\left(\text{a}\text{q}\right)+28{\text{H}}^{+}\left(\text{a}\text{q}\right)+6{\text{SO}}_4{}^{3-}$.

Interpretation Introduction

(c)

Interpretation:

The given oxidation-reduction 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 101AP

${\text{5BiO}}_3{}^{-}\left(\text{a}\text{q}\right)+14{\text{H}}^{+}\left(\text{a}\text{q}\right){\text{+2Mn}}_{}^{2+}\left(\text{a}\text{q}\right)\to 5{\text{Bi}}^{3+}\left(\text{a}\text{q}\right)+7{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+2MnO}{}^{-}\left(\text{a}\text{q}\right)$.

Explanation of Solution

The given reaction is as follows:

${\text{BiO}}_3{}^{-}\left(\text{a}\text{q}\right)+{\text{Mn}}_{}^{2+}\left(\text{a}\text{q}\right)\to {\text{Bi}}^{3+}\left(\text{a}\text{q}\right)+\text{MnO}{}^{-}\left(\text{a}\text{q}\right)$

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

${\text{BiO}}_3{}^{-}\left(\text{a}\text{q}\right)\to {\text{Bi}}^{3+}\left(\text{a}\text{q}\right)$...... (1)

And,

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

In reaction (1), add 3 water molecules on right side to balance the number of oxygen atoms,

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

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

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

Now, balance the charge by adding 2 electrons to the left

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

In reaction (2), add 4 water molecules on left to balance the oxygen atoms.

${\text{Mn}}_{}^{2+}\left(\text{a}\text{q}\right)+4{\mathrm{H<mn>2</mn>}}_{}\text{O}\left(\text{l}\right)\to \text{MnO}{}^{-}\left(\text{a}\text{q}\right)$

Hydrogen atoms can be balanced by adding 8 hydrogen ions on the right:

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

Last step is to balance the charge, add 5 electrons to the right:

${\text{Mn}}_{}^{2+}\left(\text{a}\text{q}\right)+4{\mathrm{H<mn>2</mn>}}_{}\text{O}\left(\text{l}\right)\to \text{MnO}{}^{-}\left(\text{a}\text{q}\right)+8{\text{H}}^{+}\left(\text{a}\text{q}\right)+5{\text{e}}^{-}$..... (4)

To get the overall reaction, add reaction (3) and (4)

$\begin{array}{l}5\times \left({\text{BiO}}_3{}^{-}\left(\text{a}\text{q}\right)+6{\text{H}}^{+}\left(\text{a}\text{q}\right)+2{\text{e}}^{-}\to {\text{Bi}}^{3+}\left(\text{a}\text{q}\right)+3{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\right)\\ 2\times \left({\text{Mn}}_{}^{2+}\left(\text{a}\text{q}\right)+4{\mathrm{H<mn>2</mn>}}_{}\text{O}\left(\text{l}\right)\to \text{MnO}{}^{-}\left(\text{a}\text{q}\right)+8{\text{H}}^{+}\left(\text{a}\text{q}\right)+5{\text{e}}^{-}\right)\\ \overline{\underset{¯}{{\text{5BiO}}_3{}^{-}\left(\text{a}\text{q}\right)+14{\text{H}}^{+}\left(\text{a}\text{q}\right){\text{+2Mn}}_{}^{2+}\left(\text{a}\text{q}\right)\to 5{\text{Bi}}^{3+}\left(\text{a}\text{q}\right)+7{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+2MnO}{}^{-}\left(\text{a}\text{q}\right)}}\end{array}$

Thus, the balance oxidation-reduction reaction is as follows:

${\text{5BiO}}_3{}^{-}\left(\text{a}\text{q}\right)+14{\text{H}}^{+}\left(\text{a}\text{q}\right){\text{+2Mn}}_{}^{2+}\left(\text{a}\text{q}\right)\to 5{\text{Bi}}^{3+}\left(\text{a}\text{q}\right)+7{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+2MnO}{}^{-}\left(\text{a}\text{q}\right)$.