Chapter 17, Problem 13PE

(a)

Interpretation Introduction

Interpretation:

The below equation through change in oxidation number method has to be balanced.

  $\text{Cu}+{\text{O}}_2\to \text{CuO}$

Concept Introduction:

Steps for change in oxidation number method to balance redox reactions are as follows:

1 Oxidation number of each element has to be assigned and change in oxidation number has to be identified. Then add electrons to balance charge.

2 Two half-reactions with only elements that have changed oxidation numbers have to be formed.

3 Both reactions multiplied by smallest whole number that can make electrons lost equal to electron gained.

4 Coefficient should transfer to original equation.

5 Remaining oxygen atoms are balanced through water molecules.

6 For acidic medium, charge is balanced by addition of ${\text{H}}^{+}$ ion.

Explanation of Solution

Given reaction is as follows:

  $\text{Cu}+{\text{O}}_2\to \text{CuO}$        (1)

Oxidation number of each element in equation (1) can be assigned as follows:

  $\stackrel{0}{\text{Cu}}+{\stackrel{0}{\text{O}}}_2\to \stackrel{+2}{\text{Cu}}\stackrel{-2}{\text{O}}$

Change in oxidation number occurred in copper and oxygen thus two half-reactions can be formed as follows:

Oxidation half-reaction for copper is as follows:

  $\stackrel{0}{\text{Cu}}\to \stackrel{+2}{\text{Cu}}\stackrel{-2}{\text{O}}+2{\text{e}}^{-}$        (2)

Reduction half-reaction for oxygen is as follows:

  ${\stackrel{0}{\text{O}}}_2+4{\text{e}}^{-}\to 2\stackrel{+2}{\text{Cu}}\stackrel{-2}{\text{O}}$        (3)

Multiply equation (2) by 2 so that number of electrons gained and lost becomes same and cancels each other. Thus, equation (2) is as follows:

  $2\stackrel{0}{\text{Cu}}\to 2\stackrel{+2}{\text{Cu}}\stackrel{-2}{\text{O}}+4{\text{e}}^{-}$        (4)

Coefficient of atoms in equation (3) and equation (4) of half reactions gets transfer to equation (1). Remaining atoms are balanced by equalizing its number on both sides. Thus balanced equation is as follows:

  $\text{2Cu}+{\text{O}}_2\to 2\text{CuO}$

(b)

Interpretation Introduction

Interpretation:

The below equation through change in oxidation number method has to be balanced.

  ${\text{KClO}}_3\to \text{KCl}+{\text{O}}_2$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction is as follows:

  ${\text{KClO}}_3\to \text{KCl}+{\text{O}}_2$        (5)

Oxidation number of each element in equation (5) can be assigned as follows:

  $\stackrel{+1}{\text{K}}\stackrel{+5}{\text{Cl}}{\stackrel{-2}{\text{O}}}_3\to \stackrel{+1}{\text{K}}\stackrel{-1}{\text{Cl}}+{\stackrel{0}{\text{O}}}_2$

Change in oxidation number occurred in oxygen and chlorine thus two half-reactions can be formed as follows:

Oxidation half-reaction for oxygen is as follows:

  $\stackrel{+1}{\text{2K}}\stackrel{+5}{\text{Cl}}{\stackrel{-2}{\text{O}}}_3\to 3{\stackrel{0}{\text{O}}}_2+4{\text{e}}^{-}$        (6)

Reduction half-reaction for chlorine is as follows:

  $\stackrel{+1}{\text{K}}\stackrel{+5}{\text{Cl}}{\stackrel{-2}{\text{O}}}_3+4{\text{e}}^{-}\to \stackrel{+1}{\text{K}}\stackrel{-1}{\text{Cl}}$        (7)

Coefficient of atoms in both half reactions gets transfer to equation (5). Remaining atoms are balanced by equalizing its number on both sides. Thus balanced equation is as follows:

  ${\text{2KClO}}_3\to 2\text{KCl}+3{\text{O}}_2$

(c)

Interpretation Introduction

Interpretation:

The below equation through change in oxidation number method has to be balanced.

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction is as follows:

  $\text{Ca}+{\text{H}}_{\text{2}}\text{O}\to \text{Ca}{\left(\text{OH}\right)}_2+{\text{H}}_{\text{2}}$        (9)

Oxidation number of each element in equation (9) can be assigned as follows:

  $\stackrel{0}{\text{Ca}}+{\stackrel{+1}{\text{H}}}_{\text{2}}\stackrel{-2}{\text{O}}\to \stackrel{+2}{\text{Ca}}{\left(\stackrel{-2}{\text{O}}\stackrel{+1}{\text{H}}\right)}_2+\stackrel{0}{{\text{H}}_{\text{2}}}$

Change in oxidation number occurred in calcium and hydrogen thus two balanced half-reactions can be formed as follows:

Balanced oxidation half-reaction for calcium is as follows:

  $\stackrel{0}{\text{Ca}}\to \stackrel{+2}{\text{Ca}}{\left(\stackrel{-2}{\text{O}}\stackrel{+1}{\text{H}}\right)}_2+2{\text{e}}^{-}$        (10)

Balanced reduction half-reaction for hydrogen is as follows:

  ${\stackrel{+1}{\text{H}}}_{\text{2}}\stackrel{-2}{\text{O}}+2{\text{e}}^{-}\to \stackrel{0}{{\text{H}}_{\text{2}}}$        (11)

Coefficient of atoms in equation (10) and equation (11) of half reactions gets transfer to equation (9). Remaining atoms are balanced by equalizing its number on both sides. Thus balanced equation is as follows:

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

(d)

Interpretation Introduction

Interpretation:

The below equation through change in oxidation number method has to be balanced.

  $\text{PbS}+{\text{H}}_2{\text{O}}_2\to {\text{PbSO}}_4+{\text{H}}_{\text{2}}\text{O}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction is as follows:

  $\text{PbS}+{\text{H}}_2{\text{O}}_2\to {\text{PbSO}}_4+{\text{H}}_{\text{2}}\text{O}$        (12)

Oxidation number of each element in equation (12) can be assigned as follows:

  $\stackrel{+2}{\text{Pb}}\stackrel{-2}{\text{S}}+{\stackrel{+1}{\text{H}}}_2{\stackrel{-1}{\text{O}}}_2\to \stackrel{+2}{\text{Pb}}\stackrel{+6}{\text{S}}{\stackrel{-2}{\text{O}}}_4+{\stackrel{+1}{\text{H}}}_{\text{2}}\stackrel{-2}{\text{O}}$

Change in oxidation number occurred in sulfur and oxygen thus two balanced half-reactions can be formed as follows:

Balanced oxidation half-reaction for sulfur is as follows:

  $\stackrel{+2}{\text{Pb}}\stackrel{-2}{\text{S}}\to \stackrel{+2}{\text{Pb}}\stackrel{+6}{\text{S}}{\stackrel{-2}{\text{O}}}_4+8{\text{e}}^{-}$        (13)

Balanced reduction half-reaction for oxygen is as follows:

  ${\stackrel{+1}{\text{H}}}_2{\stackrel{-1}{\text{O}}}_2+2{\text{e}}^{-}\to 2{\stackrel{+1}{\text{H}}}_{\text{2}}\stackrel{-2}{\text{O}}$        (14)

Multiply equation (14) by 4 so that number of electrons gained and lost becomes same and cancels each other. Thus, equation (14) becomes as follows:

  ${\stackrel{+1}{\text{4H}}}_2{\stackrel{-1}{\text{O}}}_2+8{\text{e}}^{-}\to 4{\stackrel{+1}{\text{H}}}_{\text{2}}\stackrel{-2}{\text{O}}$        (15)

Coefficient of atoms in equation (13) and equation (15) of half reactions gets transfer to equation (12). Remaining atoms are balanced by equalizing its number on both sides. Thus balanced equation is as follows:

  $\text{PbS}+4{\text{H}}_2{\text{O}}_2\to {\text{PbSO}}_4+4{\text{H}}_{\text{2}}\text{O}$

(e)

Interpretation Introduction

Interpretation:

The below equation through change in oxidation number method has to be balanced.

  ${\text{CH}}_4+{\text{NO}}_2\to {\text{N}}_2+{\text{CO}}_2+{\text{H}}_{\text{2}}\text{O}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction is as follows:

  ${\text{CH}}_4+{\text{NO}}_2\to {\text{N}}_2+{\text{CO}}_2+{\text{H}}_{\text{2}}\text{O}$        (16)

Oxidation number of each element in equation (16) can be assigned as follows:

  $\stackrel{-4}{\text{C}}{\stackrel{+1}{\text{H}}}_4+\stackrel{+4}{\text{N}}{\stackrel{-2}{\text{O}}}_2\to {\stackrel{0}{\text{N}}}_2+\stackrel{+4}{\text{C}}{\stackrel{-2}{\text{O}}}_2+{\stackrel{+1}{\text{H}}}_{\text{2}}\stackrel{-2}{\text{O}}$

Change in oxidation number occurred in carbon and nitrogen thus two balanced half-reactions can be formed as follows:

Balanced oxidation half-reaction for carbon is as follows:

  $\stackrel{-4}{\text{C}}{\stackrel{+1}{\text{H}}}_4\to \stackrel{+4}{\text{C}}{\stackrel{-2}{\text{O}}}_2+8{\text{e}}^{-}$        (17)

Balanced reduction half-reaction for nitrogen is as follows:

  $\stackrel{+4}{\text{2N}}{\stackrel{-2}{\text{O}}}_2+8{\text{e}}^{-}\to {\stackrel{0}{\text{N}}}_2$        (18)

Coefficient of atoms in equation (17) and equation (18) of half reactions gets transfer to equation (19). Remaining atoms are balanced by equalizing its number on both sides. Thus balanced equation is as follows:

  ${\text{CH}}_4+2{\text{NO}}_2\to {\text{N}}_2+{\text{CO}}_2+2{\text{H}}_{\text{2}}\text{O}$