Chapter 4, Problem 81E

(a)

Interpretation Introduction

Interpretation:The reaction $\text{Cu}\left(s\right)+{\text{NO}}_{{}_3}^{-}\left(aq\right)\to {\text{Cu}}^{2+}\left(aq\right)+\text{NO}\left(g\right)$ occurring in acidic medium by using half reaction method is to be balanced.

Concept introduction:The redox reactions are associated with the change in oxidation number of the interacting elements. These reactions are balanced by using half reaction method. The half reaction method is used either in acidic or basic medium. This method works better when medium of interacting species is aqueous medium.

Answer to Problem 81E

The balanced reaction is $3\text{Cu}\left(s\right)+2{\text{NO}}_{{}_3}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to 3{\text{Cu}}^{2+}\left(aq\right)+2\text{NO}\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  $\text{Cu}\left(s\right)+{\text{NO}}_{{}_3}^{-}\left(aq\right)\to {\text{Cu}}^{2+}\left(aq\right)+\text{NO}\left(g\right)$

The reduction half reaction for nitrogen is shown below.

  ${\text{NO}}_{{}_3}^{-}\left(aq\right)\to \text{NO}\left(g\right)$

To balance the oxygen atoms, add $2{\text{H}}_2\text{O}$ on product side then the equation becomes,

  ${\text{NO}}_{{}_3}^{-}\left(aq\right)\to \text{NO}\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

To balance the hydrogen atoms, add $4{\text{H}}^{+}$ on reactant side then the equation becomes,

  ${\text{NO}}_{{}_3}^{-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)\to \text{NO}\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

To balance the charge, add $3{\text{e}}^{-}$ on reactant side then the equation becomes,

  ${\text{NO}}_{{}_3}^{-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)+3{\text{e}}^{-}\to \text{NO}\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$  (1)

The oxidation half reaction for copper is shown below.

  $\text{Cu}\left(s\right)\to {\text{Cu}}^{2+}\left(aq\right)$

To balance the charge, add ${\text{2e}}^{-}$ on product side then the equation becomes,

  $\text{Cu}\left(s\right)\to {\text{Cu}}^{2+}\left(aq\right)+{\text{2e}}^{-}$  (2)

As the number of electrons gained in reduction half reaction must be equal to the number of electrons lost in oxidation half reaction.

To balance the electron transfer in half reactions, multiply equation (1) with $2$ this become,

  ${\text{2NO}}_{{}_3}^{-}\left(aq\right)+8{\text{H}}^{+}+6{\text{e}}^{-}\to 2\text{NO}\left(g\right)+4{\text{H}}_2\text{O}$  (3)

And, multiply equation (2) with $3$ this become,

  $\text{3Cu}\left(s\right)\to 3{\text{Cu}}^{2+}\left(aq\right)+6{\text{e}}^{-}$  (4)

Add equation (3) and (4) as shown below.

  $\text{3Cu}\left(s\right)+2{\text{NO}}_{{}_3}^{-}\left(aq\right)+8{\text{H}}^{+}+6{\text{e}}^{-}\to 3{\text{Cu}}^{2+}\left(aq\right)+2\text{NO}\left(g\right)+4{\text{H}}_2\text{O}+6{\text{e}}^{-}$

Thus, the final balanced equation becomes,

  $\text{3Cu}\left(s\right)+2{\text{NO}}_{{}_3}^{-}\left(aq\right)+8{\text{H}}^{+}\to 3{\text{Cu}}^{2+}\left(aq\right)+2\text{NO}\left(g\right)+4{\text{H}}_2\text{O}$

(b)

Interpretation Introduction

Interpretation:The reaction ${\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\to {\text{Cr}}^{3+}\left(aq\right)+{\text{Cl}}_2\left(g\right)$ occurring in acidic medium by using half reaction method is to be balanced.

Concept introduction:The redox reactions are associated with the change in oxidation number of the interacting elements. These reactions are balanced by using half reaction method. The half reaction method is used either in acidic or basic medium. This method works better when medium of interacting species is aqueous medium.

Answer to Problem 81E

The balanced reaction is ${\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)+14{\text{H}}^{+}\left(aq\right)+6{\text{Cl}}^{-}\left(aq\right)\to 2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)+3{\text{Cl}}_2\left(g\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\to {\text{Cr}}^{3+}\left(aq\right)+{\text{Cl}}_2\left(g\right)$

The reduction half reaction for chromium is shown below.

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

To balance the chromium atoms both sides add coefficient $2$ with ${\text{Cr}}^{3+}\left(aq\right)$ equation becomes,

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

To balance the oxygen atoms, add ${\text{7H}}_2\text{O}$ on product side then the equation becomes,

  ${\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)\to 2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)$

To balance the hydrogen atoms, add $14{\text{H}}^{+}$ on reactant side then the equation becomes,

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

To balance the charge, add ${\text{6e}}^{-}$ on reactant side then the equation becomes,

  ${\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)+14{\text{H}}^{+}\left(aq\right)+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)$  (1)

The oxidation half reaction for chlorine is shown below.

  ${\text{Cl}}^{-}\left(aq\right)\to {\text{Cl}}_2\left(g\right)$

To balance the chlorine atoms both side add coefficient $2$ with ${\text{Cl}}^{-}\left(aq\right)$ then the equation becomes,

  ${\text{2Cl}}^{-}\left(aq\right)\to {\text{Cl}}_2\left(g\right)$

To balance the charge, add ${\text{2e}}^{-}$ on product side then the equation becomes,

  ${\text{2Cl}}^{-}\left(aq\right)\to {\text{Cl}}_2\left(g\right)+2{\text{e}}^{-}$  (2)

As, the number of electrons gained in reduction half reaction must be equal to the number of electrons lost in oxidation half reaction.

To balance the electron transfer in half reactions, multiply equation (2) with $3$ this become,

  ${\text{6Cl}}^{-}\left(aq\right)\to 3{\text{Cl}}_2\left(g\right)+6{\text{e}}^{-}$  (3)

Add equation (3) and (1) as shown below.

  ${\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)+14{\text{H}}^{+}\left(aq\right)+6{\text{Cl}}^{-}\left(aq\right)+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)+3{\text{Cl}}_2\left(g\right)+6{\text{e}}^{-}$

Thus, the final balanced equation becomes,

  ${\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)+14{\text{H}}^{+}\left(aq\right)+6{\text{Cl}}^{-}\left(aq\right)+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)+3{\text{Cl}}_2\left(g\right)+6{\text{e}}^{-}$

(c)

Interpretation Introduction

Interpretation:The reaction $\text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{PbSO}}_4\left(s\right)$ occurring in acidic medium by using half reaction method is to be balanced.

Concept introduction:The redox reactions are associated with the change in oxidation number of the interacting elements. These reactions are balanced by using half reaction method. The half reaction method is used either in acidic or basic medium. This method works better when medium of interacting species is aqueous medium.

Answer to Problem 81E

The balanced reaction is $\text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+2{\text{H}}_2{\text{SO}}_4\left(aq\right)\to 2{\text{PbSO}}_4\left(s\right)+{\text{H}}_2\text{O}\left(l\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  $\text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{PbSO}}_4\left(s\right)$

The reduction half reaction for lead is shown below.

  ${\text{PbO}}_2\left(s\right)\to {\text{PbSO}}_4\left(s\right)$

The above equation in ionic form would be re-written as shown below.

  ${\text{PbO}}_2\left(s\right)\to {\text{Pb}}^{2+}\left(s\right)$

To balance the oxygen atoms, add ${\text{2H}}_2\text{O}$ on product side then the equation becomes,

  ${\text{PbO}}_2\left(s\right)\to {\text{Pb}}^{2+}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)$

To balance the hydrogen atoms, add $4{\text{H}}^{+}$ on reactant side then the equation becomes,

  ${\text{PbO}}_2\left(s\right)+4{\text{H}}^{+}\left(aq\right)\to {\text{Pb}}^{2+}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)$

To balance the charge, add ${\text{2e}}^{-}$ on reactant side then the equation becomes,

  ${\text{PbO}}_2\left(s\right)+4{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}\to {\text{Pb}}^{2+}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)$  (1)

The oxidation half reaction for chlorine is shown below.

  $\text{Pb}\left(s\right)\to {\text{Pb}}^{2+}\left(s\right)$

To balance the charge, add ${\text{2e}}^{-}$ on product side then the equation becomes,

  $\text{Pb}\left(s\right)\to {\text{Pb}}^{2+}\left(s\right)+2{\text{e}}^{-}$  (2)

Add equation (2) and (1) as shown below.

  ${\text{PbO}}_2\left(s\right)+4{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}+\text{Pb}\left(s\right)\to 2{\text{Pb}}^{2+}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)+2{\text{e}}^{-}$

Thus, the final balanced equation in ionic form becomes,

  ${\text{PbO}}_2\left(s\right)+4{\text{H}}^{+}\left(aq\right)+\text{Pb}\left(s\right)\to 2{\text{Pb}}^{2+}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)$

Therefore, the final balanced equation is shown below.

  ${\text{PbO}}_2\left(s\right)+2{\text{H}}_2{\text{SO}}_4\left(aq\right)+\text{Pb}\left(s\right)\to 2{\text{PbSO}}_4\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)$

(d)

Interpretation Introduction

Interpretation:The reaction ${\text{Mn}}^{2+}\left(aq\right)+{\text{NaBiO}}_3\left(s\right)\to {\text{Bi}}^{3+}\left(aq\right)+{\text{MnO}}_{{}_4}^{-}\left(aq\right)$ occurring in acidic medium by using half reaction method is to be balanced.

Concept introduction:The redox reactions are associated with the change in oxidation number of the interacting elements. These reactions are balanced by using half reaction method. The half reaction method is used either in acidic or basic medium. This method works better when medium of interacting species is aqueous medium.

Answer to Problem 81E

The balanced reaction is ${\text{2Mn}}^{2+}\left(aq\right)+5{\text{BiO}}_{{}_3}^{-}\left(s\right)+14{\text{H}}^{+}\left(aq\right)\to 5{\text{Bi}}^{3+}\left(aq\right)+2{\text{MnO}}_{{}_4}^{-}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{Mn}}^{2+}\left(aq\right)+{\text{NaBiO}}_3\left(s\right)\to {\text{Bi}}^{3+}\left(aq\right)+{\text{MnO}}_{{}_4}^{-}\left(aq\right)$

The ${\text{NaBiO}}_3\left(s\right)$ in ionic form would be written as ${\text{BiO}}_{{}_3}^{-}\left(s\right)$ .

The reaction is re-written as

  ${\text{Mn}}^{2+}\left(aq\right)+{\text{BiO}}_{{}_3}^{-}\left(s\right)\to {\text{Bi}}^{3+}\left(aq\right)+{\text{MnO}}_{{}_4}^{-}\left(aq\right)$

The reduction half reaction for bismuth is shown below.

  ${\text{BiO}}_{{}_3}^{-}\left(s\right)\to {\text{Bi}}^{3+}\left(aq\right)$

To balance the oxygen atoms, add ${\text{3H}}_2\text{O}$ on product side then the equation becomes,

  ${\text{BiO}}_{{}_3}^{-}\left(s\right)\to {\text{Bi}}^{3+}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)$

To balance the hydrogen atoms, add $6{\text{H}}^{+}$ on reactant side then the equation becomes,

  ${\text{BiO}}_{{}_3}^{-}\left(s\right)+6{\text{H}}^{+}\left(aq\right)\to {\text{Bi}}^{3+}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)$

To balance the charge, add ${\text{2e}}^{-}$ on reactant side then the equation becomes,

  ${\text{BiO}}_{{}_3}^{-}\left(s\right)+6{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}\to {\text{Bi}}^{3+}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)$  (1)

The oxidation half reaction for manganese is shown below.

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

To balance the oxygen atom, add ${\text{4H}}_2\text{O}$ on reactant side then the equation becomes,

  ${\text{Mn}}^{2+}\left(aq\right)+{\text{4H}}_2\text{O}\left(l\right)\to {\text{MnO}}_{{}_4}^{-}\left(aq\right)$

To balance the hydrogen atom, add $8{\text{H}}^{+}$ on productside then the equation becomes,

  ${\text{Mn}}^{2+}\left(aq\right)+{\text{4H}}_2\text{O}\left(l\right)\to {\text{MnO}}_{{}_4}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)$

To balance the charge, add ${\text{5e}}^{-}$ on product side then the equation becomes,

  ${\text{Mn}}^{2+}\left(aq\right)+{\text{4H}}_2\text{O}\left(l\right)\to {\text{MnO}}_{{}_4}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5{\text{e}}^{-}$  (2)

As, the number of electrons gained in reduction half reaction must be equal to the number of electrons lost in oxidation half reaction.

To balance the electron transfer in half reactions, multiply equation (1) with $5$ this become,

  ${\text{5BiO}}_{{}_3}^{-}\left(s\right)+30{\text{H}}^{+}\left(aq\right)+10{\text{e}}^{-}\to 5{\text{Bi}}^{3+}\left(aq\right)+15{\text{H}}_2\text{O}\left(l\right)$  (3)

And, multiply equation (2) with two this become,

  ${\text{2Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)\to 2{\text{MnO}}_{{}_4}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{e}}^{-}$  (4)

Add equation (3) and (4) as shown below.

  ${\text{2Mn}}^{2+}\left(aq\right)+5{\text{BiO}}_{{}_3}^{-}\left(s\right)+30{\text{H}}^{+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+10{\text{e}}^{-}\to 5{\text{Bi}}^{3+}\left(aq\right)+2{\text{MnO}}_{{}_4}^{-}\left(aq\right)+15{\text{H}}_2\text{O}\left(l\right)+10{\text{e}}^{-}+16{\text{H}}^{+}\left(aq\right)$

Thus, the final balanced equation becomes,

  ${\text{2Mn}}^{2+}\left(aq\right)+5{\text{BiO}}_{{}_3}^{-}\left(s\right)+14{\text{H}}^{+}\left(aq\right)\to 5{\text{Bi}}^{3+}\left(aq\right)+2{\text{MnO}}_{{}_4}^{-}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)$

(e)

Interpretation Introduction

Interpretation:The reaction ${\text{H}}_3{\text{AsO}}_4\left(aq\right)+\text{Zn}\left(s\right)\to {\text{AsH}}_3\left(g\right)+{\text{Zn}}^{2+}\left(aq\right)$ occurring in acidic medium by using half reaction method is to be balanced.

Concept introduction:The redox reactions are associated with the change in oxidation number of the interacting elements. These reactions are balanced by using half reaction method. The half reaction method is used either in acidic or basic medium. This method works better when medium of interacting species is aqueous medium.

Answer to Problem 81E

The balanced reaction is ${\text{H}}_3{\text{AsO}}_4\left(aq\right)+4\text{Zn}\left(s\right)+8{\text{H}}^{+}\left(aq\right)\to {\text{AsH}}_3\left(g\right)+4{\text{Zn}}^{2+}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{H}}_3{\text{AsO}}_4\left(aq\right)+\text{Zn}\left(s\right)\to {\text{AsH}}_3\left(g\right)+{\text{Zn}}^{2+}\left(aq\right)$

The reduction half reaction for arsenic is shown below.

  ${\text{H}}_3{\text{AsO}}_4\left(aq\right)\to {\text{AsH}}_3\left(g\right)$

To balance the oxygen atoms, add ${\text{4H}}_2\text{O}$ on product side then the equation becomes,

  ${\text{H}}_3{\text{AsO}}_4\left(aq\right)\to {\text{AsH}}_3\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)$

To balance the hydrogen atoms, add ${\text{8H}}^{+}$ on reactant side then the equation becomes,

  ${\text{H}}_3{\text{AsO}}_4\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to {\text{AsH}}_3\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)$

To balance the charge, add ${\text{8e}}^{-}$ on reactant side then the equation becomes,

  ${\text{H}}_3{\text{AsO}}_4\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+8{\text{e}}^{-}\to {\text{AsH}}_3\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)$  (1)

The oxidation half reaction for zinc is shown below.

  $\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)$

To balance the charge, add ${\text{2e}}^{-}$ on product side then the equation becomes,

  $\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)+{\text{2e}}^{-}$  (2)

As, the number of electrons gained in reduction half reaction must be equal to the number of electrons lost in oxidation half reaction.

To balance the electron transfer in half reactions, multiply equation (2) with $4$ this become,

  $\text{4Zn}\left(s\right)\to 4{\text{Zn}}^{2+}\left(aq\right)+8{\text{e}}^{-}$  (3)

Add equation (2) and (3) as shown below.

  ${\text{H}}_3{\text{AsO}}_4\left(aq\right)+\text{4Zn}\left(s\right)+8{\text{H}}^{+}\left(aq\right)+8{\text{e}}^{-}\to {\text{AsH}}_3\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)+4{\text{Zn}}^{2+}\left(aq\right)+8{\text{e}}^{-}$

Thus, the final balanced equation becomes,

  ${\text{H}}_3{\text{AsO}}_4\left(aq\right)+\text{4Zn}\left(s\right)+8{\text{H}}^{+}\left(aq\right)\to {\text{AsH}}_3\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)+4{\text{Zn}}^{2+}\left(aq\right)$

(f)

Interpretation Introduction

Interpretation:The reaction ${\text{As}}_2{\text{O}}_3\left(s\right)+{\text{NO}}_3{}^{-}\left(aq\right)\to {\text{H}}_3{\text{AsO}}_4\left(aq\right)+\text{NO}\left(g\right)$ occurring in acidic medium by using half reaction method is to be balanced.

Concept introduction:The redox reactions are associated with the change in oxidation number of the interacting elements. These reactions are balanced by using half reaction method. The half reaction method is used either in acidic or basic medium. This method works better when medium of interacting species is aqueous medium.

Answer to Problem 81E

The balanced reaction is ${\text{3As}}_2{\text{O}}_3\left(s\right)+4{\text{NO}}_{{}_3}^{-}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)+4{\text{H}}^{+}\left(aq\right)\to 6{\text{H}}_3{\text{AsO}}_4\left(aq\right)+4\text{NO}\left(g\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{As}}_2{\text{O}}_3\left(s\right)+{\text{NO}}_{{}_3}^{-}\left(aq\right)\to {\text{H}}_3{\text{AsO}}_4\left(aq\right)+\text{NO}\left(g\right)$

The reduction half reaction for nitrogen is shown below.

  ${\text{NO}}_{{}_3}^{-}\left(aq\right)\to \text{NO}\left(g\right)$

To balance the oxygen atoms, add ${\text{2H}}_2\text{O}$ on product side then the equation becomes,

  ${\text{NO}}_{{}_3}^{-}\left(aq\right)\to \text{NO}\left(g\right)+2{\text{H}}_2\text{O}$

To balance the hydrogen atoms, add ${\text{4H}}^{+}$ on reactant side then the equation becomes,

  ${\text{NO}}_{{}_3}^{-}\left(aq\right)+4{\text{H}}^{+}\to \text{NO}\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

To balance the charge, add ${\text{3e}}^{-}$ on reactant side then the equation becomes,

  ${\text{NO}}_{{}_3}^{-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)+3{\text{e}}^{-}\to \text{NO}\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$  (1)

The oxidation half reaction for arsenic is shown below.

  ${\text{As}}_2{\text{O}}_3\left(s\right)\to {\text{H}}_3{\text{AsO}}_4\left(aq\right)$

To balance the arsenic atoms multiply ${\text{H}}_3{\text{AsO}}_4\left(aq\right)$ with $2$ .

  ${\text{As}}_2{\text{O}}_3\left(s\right)\to 2{\text{H}}_3{\text{AsO}}_4\left(aq\right)$

To balance the oxygen atoms, add ${\text{5H}}_2\text{O}$ on reactant side then the equation becomes,

  ${\text{As}}_2{\text{O}}_3\left(s\right)+5{\text{H}}_2\text{O}\left(l\right)\to 2{\text{H}}_3{\text{AsO}}_4\left(aq\right)$

To balance the hydrogen atoms, add ${\text{4H}}^{+}$ on productside then the equation becomes,

  ${\text{As}}_2{\text{O}}_3\left(s\right)+5{\text{H}}_2\text{O}\left(l\right)\to 2{\text{H}}_3{\text{AsO}}_4\left(aq\right)+4{\text{H}}^{+}\left(aq\right)$

To balance the charge, add ${\text{4e}}^{-}$ on product side then the equation becomes,

  ${\text{As}}_2{\text{O}}_3\left(s\right)+5{\text{H}}_2\text{O}\left(l\right)\to 2{\text{H}}_3{\text{AsO}}_4\left(aq\right)+4{\text{H}}^{+}\left(aq\right)+4{\text{e}}^{-}$  (2)

As, the number of electrons gained in reduction half reaction must be equal to the number of electrons lost in oxidation half reaction.

To balance the electron transfer in half reactions, multiply equation (1) with $4$ this become,

  ${\text{4NO}}_{{}_3}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+12{\text{e}}^{-}\to 4\text{NO}\left(g\right)+8{\text{H}}_2\text{O}\left(l\right)$  (3)

And, equation (2) with three as shown below.

  ${\text{3As}}_2{\text{O}}_3\left(s\right)+15{\text{H}}_2\text{O}\left(l\right)\to 6{\text{H}}_3{\text{AsO}}_4\left(aq\right)+12{\text{H}}^{+}\left(aq\right)+12{\text{e}}^{-}$

(4)

Add equation (3) and (4) as shown below.

  ${\text{3As}}_2{\text{O}}_3\left(s\right)+{\text{4NO}}_{{}_3}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+12{\text{e}}^{-}+15{\text{H}}_2\text{O}\left(l\right)\to 6{\text{H}}_3{\text{AsO}}_4\left(aq\right)+4\text{NO}\left(g\right)+8{\text{H}}_2\text{O}\left(l\right)+12{\text{H}}^{+}\left(aq\right)+12{\text{e}}^{-}$

Thus, the final balanced equation becomes,

  ${\text{3As}}_2{\text{O}}_3\left(s\right)+{\text{4NO}}_{{}_3}^{-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)\to 6{\text{H}}_3{\text{AsO}}_4\left(aq\right)+4\text{NO}\left(g\right)$

(g)

Interpretation Introduction

Interpretation:The reaction ${\text{Br}}^{-}\left(aq\right)+{\text{MnO}}_{{}_4}^{-}\left(aq\right)\to {\text{Br}}_2\left(l\right)+{\text{Mn}}^{2+}\left(aq\right)$ occurring in acidic medium by using half reaction method is to be balanced.

Concept introduction:The redox reactions are associated with the change in oxidation number of the interacting elements. These reactions are balanced by using half reaction method. The half reaction method is used either in acidic or basic medium. This method works better when medium of interacting species is aqueous medium.

Answer to Problem 81E

The balanced reaction is ${\text{10Br}}^{-}\left(aq\right)+2{\text{MnO}}_{{}_4}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)\to 5{\text{Br}}_2\left(l\right)+2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{Br}}^{-}\left(aq\right)+{\text{MnO}}_{{}_4}^{-}\left(aq\right)\to {\text{Br}}_2\left(l\right)+{\text{Mn}}^{2+}\left(aq\right)$

The reduction half reaction for manganese is shown below.

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

To balance the oxygen atoms, add ${\text{4H}}_2\text{O}$ on product side then the equation becomes,

  ${\text{MnO}}_{{}_4}^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$

To balance the hydrogen atoms, add ${\text{8H}}^{+}$ on reactant side then the equation becomes,

  ${\text{MnO}}_{{}_4}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$

To balance the charge, add ${\text{5e}}^{-}$ on reactant side then the equation becomes,

  ${\text{MnO}}_{{}_4}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5{\text{e}}^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$  (1)

The oxidation half reaction for bromine is shown below.

  ${\text{Br}}^{-}\left(aq\right)\to {\text{Br}}_2\left(l\right)$

To balance the bromine atoms, add coefficient two with ${\text{Br}}^{-}\left(aq\right)$ as shown below,

  ${\text{2Br}}^{-}\left(aq\right)\to {\text{Br}}_2\left(l\right)$

To balance the charge, add ${\text{2e}}^{-}$ on product side equation becomes,

  ${\text{2Br}}^{-}\left(aq\right)\to {\text{Br}}_2\left(l\right)+{\text{2e}}^{-}$  (2)

As, the number of electrons gained in reduction half reaction must be equal to the number of electrons lost in oxidation half reaction.

To balance the electron transfer in half reactions, multiply equation (1) with $2$ this become,

  ${\text{2MnO}}_{{}_4}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{e}}^{-}\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)$  (3)

And, multiply equation (2) with $5$ this become,

  ${\text{10Br}}^{-}\left(aq\right)\to 5{\text{Br}}_2\left(l\right)+10{\text{e}}^{-}$  (4)

Add equation (3) and (4) as shown below.

  ${\text{2MnO}}_{{}_4}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+10{\text{e}}^{-}+{\text{10Br}}^{-}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+5{\text{Br}}_2\left(l\right)+10{\text{e}}^{-}$

Thus, the final balanced equation becomes,

  ${\text{2MnO}}_{{}_4}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)+{\text{10Br}}^{-}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+5{\text{Br}}_2\left(l\right)$

(h)

Interpretation Introduction

Interpretation:The reaction ${\text{CH}}_3\text{OH}\left(aq\right)+{\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)\to {\text{CH}}_2\text{O}\left(aq\right)+{\text{Cr}}^{3+}\left(aq\right)$ occurring in acidic medium by using half reaction method is to be balanced.

Concept introduction:The redox reactions are associated with the change in oxidation number of the interacting elements. These reactions are balanced by using half reaction method. The half reaction method is used either in acidic or basic medium. This method works better when medium of interacting species is aqueous medium.

Answer to Problem 81E

The balanced reaction is ${\text{3CH}}_3\text{OH}\left(aq\right)+{\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to 3{\text{CH}}_2\text{O}\left(aq\right)+2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{CH}}_3\text{OH}\left(aq\right)+{\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)\to {\text{CH}}_2\text{O}\left(aq\right)+{\text{Cr}}^{3+}\left(aq\right)$

The reduction half reaction for chromium is shown below.

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

To balance the chromium atoms both the sides add coefficient $2$ with ${\text{Cr}}^{3+}\left(aq\right)$ equation becomes,

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

To balance the oxygen atoms, add ${\text{7H}}_2\text{O}$ on product side then the equation becomes,

  ${\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)\to 2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)$

To balance the hydrogen atoms, add $14{\text{H}}^{+}$ on reactant side then the equation becomes,

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

To balance the charge, add ${\text{6e}}^{-}$ on reactant side then the equation becomes,

  ${\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)+14{\text{H}}^{+}\left(aq\right)+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)$  (1)

The oxidation half reaction for carbon is shown below.

  ${\text{CH}}_3\text{OH}\left(aq\right)\to {\text{CH}}_2\text{O}\left(aq\right)$

To balance the hydrogenatoms, add $2{\text{H}}^{+}$ on reactant side then the equation becomes,

  ${\text{CH}}_3\text{OH}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{CH}}_2\text{O}\left(aq\right)$

To balance the charge, add ${\text{2e}}^{-}$ on reactantside then the equation becomes,

  ${\text{CH}}_3\text{OH}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}\to {\text{CH}}_2\text{O}\left(aq\right)$

  $$  (2)

As, the number of electrons gained in reduction half reaction must be equal to the number of electrons lost in oxidation half reaction.

To balance the electron transfer in half reactions, multiply equation (2) with $3$ this become,

  ${\text{3CH}}_3\text{OH}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)+6{\text{e}}^{-}\to 3{\text{CH}}_2\text{O}\left(aq\right)$  (3)

Add equation (3) and (1) as shown below.

  ${\text{3CH}}_3\text{OH}\left(aq\right)+{\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)+14{\text{H}}^{+}\left(aq\right)+6{\text{e}}^{-}\to 3{\text{CH}}_2\text{O}\left(aq\right)+2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)+6{\text{H}}^{+}\left(aq\right)+6{\text{e}}^{-}$

Thus, the final balanced equation becomes,

  ${\text{3CH}}_3\text{OH}\left(aq\right)+{\text{Cr}}_2{\text{O}}_{{}_7}^{2-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to 3{\text{CH}}_2\text{O}\left(aq\right)+2{\text{Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)$