Chapter 18, Problem 48QAP

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 48QAP

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

Explanation of Solution

The given reaction is as follows:

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

The above reaction can be separated as follows:

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

And,

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

The reaction (1) can be balanced by adding 2 electrons to the right side of the reaction arrow:

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

Now, to balance reaction (2), first add 1 water molecule to the right thus,

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

Now, second step is to balance the hydrogen atoms, 1 hydrogen ion is added to left thus,

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

Last step is to balance the charge thus, 1 electron is added to left as follows:

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

Net reaction can be obtained by adding reaction (3) and (4) thus,

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

Thus, the balanced chemical reaction is as follows:

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

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 48QAP

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

Explanation of Solution

The given reaction is as follows:

$\text{Mg}\left(\text{s}\right)+{\text{HNO}}_{\text{3}}\left(\text{a}\text{q}\right)\to {\text{Mg}}^{2+}\left(\text{a}\text{q}\right)+{\text{NO}}_2\left(\text{g}\right)$

First separate the above reaction to two half reactions as follows:

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

And,

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

To balance reaction (1) add 2 electrons to the right thus,

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

Reaction (2) can be balanced by adding 1 water molecule to the right thus,

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

Hydrogen atom can be balanced by adding 1 hydrogen ion to the left thus,

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

Now, charge can be balanced by adding 1 electron to the left thus,

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

Net reaction can be obtained by adding reaction (3) and (4) thus,

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

Thus, the balanced chemical reaction is as follows:

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