Chapter 17, Problem 14PE

(a)

Interpretation Introduction

Interpretation:

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

  $\text{Cu}+{\text{AgNO}}_3\to \text{Ag}+\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

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:

  $\text{Cu}+{\text{AgNO}}_3\to \text{Ag}+\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$        (1)

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

  $\stackrel{0}{\text{Cu}}+\stackrel{+1}{\text{Ag}}\stackrel{+5}{\text{N}}\stackrel{-2}{{\text{O}}_3}\to \stackrel{0}{\text{Ag}}+\stackrel{+2}{\text{Cu}}{\left(\stackrel{+5}{\text{N}}{\stackrel{-2}{\text{O}}}_3\right)}_2$

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

Oxidation half-reaction for copper is as follows:

  $\stackrel{0}{\text{Cu}}\to \stackrel{+2}{\text{Cu}}{\left(\stackrel{+5}{\text{N}}{\stackrel{-2}{\text{O}}}_3\right)}_2+2{\text{e}}^{-}$        (2)

Reduction half-reaction for silver is as follows:

  $\stackrel{+1}{\text{Ag}}\stackrel{+5}{\text{N}}\stackrel{-2}{{\text{O}}_3}+1{\text{e}}^{-}\to \stackrel{0}{\text{Ag}}$        (3)

Multiply equation (3) by 2 to make same number of electron gained and loses.

  $\stackrel{+1}{\text{2Ag}}\stackrel{+5}{\text{N}}\stackrel{-2}{{\text{O}}_3}+2{\text{e}}^{-}\to \stackrel{0}{\text{2Ag}}$        (4)

Coefficient of atoms in equation (2) and equation (4) of half reactions are transferred to equation (1). Left atoms are balanced by equalizing its number on both sides. Thus balanced equation is as follows:

  $\text{Cu}+2{\text{AgNO}}_3\to 2\text{Ag}+\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$

(b)

Interpretation Introduction

Interpretation:

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

  ${\text{MnO}}_2+\text{HCl}\to {\text{MnCl}}_2+{\text{Cl}}_2+{\text{H}}_{\text{2}}\text{O}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction is as follows:

  ${\text{MnO}}_2+\text{HCl}\to {\text{MnCl}}_2+{\text{Cl}}_2+{\text{H}}_{\text{2}}\text{O}$        (5)

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

  $\stackrel{+4}{\text{Mn}}\stackrel{-2}{{\text{O}}_2}+\stackrel{+1}{\text{H}}\stackrel{-1}{\text{Cl}}\to \stackrel{+2}{\text{Mn}}\stackrel{-1}{{\text{Cl}}_2}+\stackrel{0}{{\text{Cl}}_2}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

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

Oxidation half-reaction for chlorine is as follows:

  $\text{2}\stackrel{+1}{\text{H}}\stackrel{-1}{\text{Cl}}\to \stackrel{0}{{\text{Cl}}_2}+2{\text{e}}^{-}$        (6)

Reduction half-reaction for manganese is as follows:

  $\stackrel{+4}{\text{Mn}}\stackrel{-2}{{\text{O}}_2}+2{\text{e}}^{-}\to \stackrel{+2}{\text{Mn}}\stackrel{-1}{{\text{Cl}}_2}$        (7)

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

  ${\text{MnO}}_2+4\text{HCl}\to {\text{MnCl}}_2+{\text{Cl}}_2+2{\text{H}}_{\text{2}}\text{O}$

(c)

Interpretation Introduction

Interpretation:

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

  $\text{HCl}+{\text{O}}_2\to {\text{Cl}}_2+{\text{H}}_{\text{2}}\text{O}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction is as follows:

  $\text{HCl}+{\text{O}}_2\to {\text{Cl}}_2+{\text{H}}_{\text{2}}\text{O}$        (9)

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

  $\stackrel{+1}{\text{H}}\stackrel{-1}{\text{Cl}}+{\stackrel{0}{\text{O}}}_2\to {\stackrel{0}{\text{Cl}}}_2+{\stackrel{+1}{\text{H}}}_{\text{2}}\stackrel{-2}{\text{O}}$

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

Balanced oxidation half-reaction is as follows:

  $\text{2}\stackrel{+1}{\text{H}}\stackrel{-1}{\text{Cl}}\to {\stackrel{0}{\text{Cl}}}_2+2{\text{e}}^{-}$        (10)

Balanced reduction half-reaction is as follows:

  ${\stackrel{0}{\text{O}}}_2+4{\text{e}}^{-}\to 2{\stackrel{+1}{\text{H}}}_{\text{2}}\stackrel{-2}{\text{O}}$        (11)

Multiply equation (10) by 2 so that number of electrons gained and lost becomes same and cancels each other. Thus, equation (10) is as follows:

  $\text{4}\stackrel{+1}{\text{H}}\stackrel{-1}{\text{Cl}}\to {\stackrel{0}{\text{2Cl}}}_2+4{\text{e}}^{-}$        (12)

Coefficient of atoms in equation (11) and equation (12) of half reactions are transferred to equation (9). Remaining atoms are balanced by equalizing its number on both sides. Thus balanced equation is as follows:

  $\text{4HCl}+{\text{O}}_2\to 2{\text{Cl}}_2+2{\text{H}}_{\text{2}}\text{O}$

(d)

Interpretation Introduction

Interpretation:

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

  $\text{Ag}+{\text{H}}_{\text{2}}\text{S}+{\text{O}}_2\to {\text{Ag}}_2\text{S}+{\text{H}}_{\text{2}}\text{O}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction is as follows:

  $\text{Ag}+{\text{H}}_{\text{2}}\text{S}+{\text{O}}_2\to {\text{Ag}}_2\text{S}+{\text{H}}_{\text{2}}\text{O}$        (13)

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

  $\stackrel{0}{\text{Ag}}+{\stackrel{+1}{\text{H}}}_{\text{2}}\stackrel{-2}{\text{S}}+{\stackrel{0}{\text{O}}}_2\to {\stackrel{+1}{\text{Ag}}}_2\stackrel{-2}{\text{S}}+{\stackrel{+1}{\text{H}}}_{\text{2}}\stackrel{-2}{\text{O}}$

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

Balanced oxidation half-reaction is as follows:

  $\text{2}\stackrel{0}{\text{Ag}}\to {\stackrel{+1}{\text{Ag}}}_2\stackrel{-2}{\text{S}}+2{\text{e}}^{-}$        (14)

Balanced reduction half-reaction is as follows:

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

Multiply equation (14) by 2 so that number of electrons gained and lost becomes same and cancels each other. Thus, equation (14) becomes as follows:

  $\text{4}\stackrel{0}{\text{Ag}}\to {\stackrel{+1}{\text{2Ag}}}_2\stackrel{-2}{\text{S}}+4{\text{e}}^{-}$        (16)

Coefficient of atoms in equation (15) and equation (16) of half reactions are transferred to equation (13). Remaining atoms are balanced by equalizing its number on both sides. Thus balanced equation is as follows:

  $\text{4Ag}+2{\text{H}}_{\text{2}}\text{S}+{\text{O}}_2\to 2{\text{Ag}}_2\text{S}+2{\text{H}}_{\text{2}}\text{O}$

(e)

Interpretation Introduction

Interpretation:

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

  ${\text{KMnO}}_4+{\text{CaC}}_{\text{2}}{\text{O}}_4+{\text{H}}_{\text{2}}{\text{SO}}_4\to {\text{K}}_{\text{2}}{\text{SO}}_4+{\text{MnSO}}_4+{\text{CaSO}}_4+{\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction is as follows:

  ${\text{KMnO}}_4+{\text{CaC}}_{\text{2}}{\text{O}}_4+{\text{H}}_{\text{2}}{\text{SO}}_4\to \left[\begin{array}{c}{\text{K}}_{\text{2}}{\text{SO}}_4+{\text{MnSO}}_4+{\text{CaSO}}_4\\ +{\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\end{array}\right]$        (17)

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

  $\stackrel{+1}{\text{K}}\stackrel{+7}{\text{Mn}}{\stackrel{-2}{\text{O}}}_4+\stackrel{+2}{\text{Ca}}{\stackrel{+3}{\text{C}}}_{\text{2}}{\stackrel{-2}{\text{O}}}_4+{\stackrel{+1}{\text{H}}}_{\text{2}}\stackrel{+6}{\text{S}}{\stackrel{-2}{\text{O}}}_4\to \left[\begin{array}{c}{\stackrel{+1}{\text{K}}}_{\text{2}}\stackrel{+6}{\text{S}}{\stackrel{-2}{\text{O}}}_4+\stackrel{+2}{\text{Mn}}\stackrel{+6}{\text{S}}{\stackrel{-2}{\text{O}}}_4+\stackrel{+2}{\text{Ca}}\stackrel{+6}{\text{S}}{\stackrel{-2}{\text{O}}}_4\\ +\stackrel{+4}{\text{C}}{\stackrel{-2}{\text{O}}}_{\text{2}}+{\stackrel{+1}{\text{H}}}_{\text{2}}\stackrel{-2}{\text{O}}\end{array}\right]$

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

Balanced oxidation half-reaction for carbon is as follows:

  $\stackrel{+2}{\text{Ca}}{\stackrel{+3}{\text{C}}}_{\text{2}}{\stackrel{-2}{\text{O}}}_4\to \stackrel{+4}{\text{2C}}{\stackrel{-2}{\text{O}}}_{\text{2}}+2{\text{e}}^{-}$        (18)

Balanced reduction half-reaction for manganese is as follows:

  $\stackrel{+1}{\text{K}}\stackrel{+7}{\text{Mn}}{\stackrel{-2}{\text{O}}}_4+5{\text{e}}^{-}\to \stackrel{+2}{\text{Mn}}\stackrel{+6}{\text{S}}{\stackrel{-2}{\text{O}}}_4$        (19)

Multiply equation (18) by 5 and equation (19) by 2 so that number of electrons gained and lost becomes same and cancels each other. Thus, equation (18) becomes as follows:

  $\stackrel{+2}{\text{5Ca}}{\stackrel{+3}{\text{C}}}_{\text{2}}{\stackrel{-2}{\text{O}}}_4\to \stackrel{+4}{\text{10C}}{\stackrel{-2}{\text{O}}}_{\text{2}}+5{\text{e}}^{-}$        (20)

Equation (19) becomes as follows:

  $\stackrel{+1}{\text{2K}}\stackrel{+7}{\text{Mn}}{\stackrel{-2}{\text{O}}}_4+10{\text{e}}^{-}\to 2\stackrel{+2}{\text{Mn}}\stackrel{+6}{\text{S}}{\stackrel{-2}{\text{O}}}_4$        (21)

Coefficient of atoms in equation (20) and equation (21) of half reactions are transferred to equation (17). Remaining atoms are balanced by equalizing its number on both sides. Thus balanced equation is as follows:

  ${\text{2KMnO}}_4+5{\text{CaC}}_{\text{2}}{\text{O}}_4+8{\text{H}}_{\text{2}}{\text{SO}}_4\to {\text{K}}_{\text{2}}{\text{SO}}_4+2{\text{MnSO}}_4+5{\text{CaSO}}_4+10{\text{CO}}_{\text{2}}+8{\text{H}}_{\text{2}}\text{O}$