Chapter 4, Problem 82E

(a)

Interpretation Introduction

Interpretation:The reaction $\text{Al}\left(s\right)+{\text{MnO}}_{{}_4}^{-}\left(aq\right)\to {\text{MnO}}_2\left(s\right)+\text{Al}{\left(\text{OH}\right)}_{{}_4}^{-}\left(aq\right)$ occurring in basic 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 the interacting species are present in aqueous medium.

Answer to Problem 82E

The balanced reaction is $\text{Al}\left(s\right)+{\text{MnO}}_4^{-}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)\to {\text{MnO}}_2\left(s\right)+\text{Al}{\left(\text{OH}\right)}_4^{-}\left(aq\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  $\text{Al}\left(s\right)+{\text{MnO}}_4^{-}\left(aq\right)\to {\text{MnO}}_2\left(s\right)+\text{Al}{\left(\text{OH}\right)}_4^{-}\left(aq\right)$

The reduction half reaction for manganese is shown below.

  ${\text{MnO}}_{{}_4}^{-}\left(aq\right)\to {\text{MnO}}_2\left(s\right)$

To balance the oxygen atoms, add $2{\text{H}}_2\text{O}\left(l\right)$ on product side andthen the equation becomes,

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

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

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

For reaction in basic medium, ${\text{OH}}^{\text{-}}$ ion is added to each side of ${\text{H}}^{+}$ ion in the equation. In order to balance the charge and atoms, the ${\text{OH}}^{\text{-}}$ ion is added to both the sides of the equation so, the equation becomes ( ${\text{H}}^{+}$ ion and ${\text{OH}}^{\text{-}}$ ion present on both the sides are combined to form water):

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

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

  ${\text{MnO}}_{{}_4}^{-}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)+3{\text{e}}^{-}\to {\text{MnO}}_2\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)+4{\text{OH}}^{\text{-}}(aq)$  (1)

The oxidation half reaction for aluminum is shown below.

  $\text{Al}\left(s\right)\to \text{Al}{\left(\text{OH}\right)}_{{}_4}^{-}\left(aq\right)$

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

  $\text{Al}\left(s\right)+4{\text{H}}_2\text{O}\left(l\right)\to \text{Al}{\left(\text{OH}\right)}_{{}_4}^{-}\left(aq\right)$

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

  $\text{Al}\left(s\right)+4{\text{H}}_2\text{O}\left(l\right)\to \text{Al}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)$

For reaction in basic medium,

  ${\text{OH}}^{\text{-}}$ ion is added to each side of ${\text{H}}^{+}$ ion in the equation. In order to balance the charge and atoms, the

  ${\text{OH}}^{\text{-}}$ ion is added to both the sides of the equation so, the equation becomes ( ${\text{H}}^{+}$ ion and ${\text{OH}}^{\text{-}}$ ion present on both the sides are combined to form water):

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

To balance the charge, add ${\text{3e}}^{-}$ on product side then the equationbecomes,

  $\text{Al}\left(s\right)+4{\text{H}}_2\text{O}\left(l\right)+4{\text{OH}}^{-}\left(aq\right)\to \text{Al}{\left(\text{OH}\right)}_4^{-}\left(aq\right)+4{\text{H}}_{2}O(l)+{\text{3e}}^{-}$   (2)

Add equation (1) and (2) to obtain the final equation.

Thus, the final balanced equation in basic medium becomes,

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

(b)

Interpretation Introduction

Interpretation:The reaction ${\text{Cl}}_2\left(g\right)\to {\text{Cl}}^{-}\left(aq\right)+{\text{ClO}}^{-}\left(aq\right)$ occurring in basic 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 the interacting species are present in aqueous medium.

Answer to Problem 82E

The balanced reaction is ${\text{Cl}}_2\left(g\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{Cl}}^{-}\left(aq\right)+{\text{ClO}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$ .

Explanation of Solution

The given chemical equation is shown below.

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

The reduction half reaction for chlorine is shown below.

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

To balance the chlorine atoms both the sides add coefficient $2$ with ${\text{Cl}}^{-}$ then the equation becomes,

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

To balance the charge, add two electrons on reactant sidethen the equation becomes,

  ${\text{Cl}}_2\left(g\right)+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}\left(aq\right)$  (1)

The oxidation half reaction for chlorine is shown below.

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

To balance the chlorine atoms both the sides add coefficient $2$ with ${\text{ClO}}^{-}$ then the equation becomes,

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

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

  ${\text{Cl}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)\to 2{\text{ClO}}^{-}\left(aq\right)$

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

  ${\text{Cl}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)\to 2{\text{ClO}}^{-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)$

For reaction in basic medium, ${\text{OH}}^{\text{-}}$ ion is added to each side of ${\text{H}}^{+}$ ion in the equation. In order to balance the charge and atoms, the ${\text{OH}}^{\text{-}}$ ion is added to both the sides of the equation so, the equation becomes ( ${\text{H}}^{+}$ ion and ${\text{OH}}^{\text{-}}$ ion present on both the sides are combined to form water):

  ${\text{Cl}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)+4{\text{OH}}^{-}\left(aq\right)\to 2{\text{ClO}}^{-}\left(aq\right)+4H_{2}O(l)$

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

  ${\text{Cl}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)+4{\text{OH}}^{-}\left(aq\right)\to 2{\text{ClO}}^{-}\left(aq\right)+4H_{2}O(l)+{\text{2e}}^{-}$  (2)

Add equation (1) and (2) to obtain the final equation.

  ${\text{2Cl}}_2\left(g\right)+4{\text{OH}}^{-}\left(aq\right)\to 2{\text{ClO}}^{-}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

Therefore, the final balanced equation is ${\text{Cl}}_2\left(g\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{Cl}}^{-}\left(aq\right)+{\text{ClO}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$ .

(c)

Interpretation Introduction

Interpretation:The reaction ${\text{NO}}_{{}_2}^{-}\left(aq\right)+\text{Al}\left(s\right)\to {\text{NH}}_3\left(g\right)+{\text{AlO}}_{{}_2}^{-}\left(aq\right)$ occurring in basic 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 the interacting species are present in aqueous medium.

Answer to Problem 82E

The balanced reaction is ${\text{NO}}_{{}_2}^{-}\left(aq\right)+2\text{Al}\left(s\right)+{\text{H}}_2\text{O}\left(l\right)+{\text{OH}}^{-}\left(aq\right)\to {\text{NH}}_3\left(g\right)+2{\text{AlO}}_{{}_2}^{-}\left(aq\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{NO}}_{{}_2}^{-}\left(aq\right)+\text{Al}\left(s\right)\to {\text{NH}}_3\left(g\right)+{\text{AlO}}_{{}_2}^{-}\left(aq\right)$

The reduction half reaction for chlorine is shown below.

  ${\text{NO}}_{{}_2}^{-}\left(aq\right)\to {\text{NH}}_3\left(g\right)$

To balance the oxygen atoms both the sides add two ${\text{H}}_2\text{O}$ on product side then theequation becomes,

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

To balance the hydrogen atoms, add ${\text{7H}}^{+}$ on reactant side then the equation becomes,

  ${\text{NO}}_{{}_2}^{-}\left(aq\right)+7{\text{H}}^{+}\left(aq\right)\to {\text{NH}}_3\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

For reaction in basic medium, ${\text{OH}}^{\text{-}}$ ion is added to each side of ${\text{H}}^{+}$ ion in the equation. In order to balance the charge and atoms, the ${\text{OH}}^{\text{-}}$ ion is added to both the sides of the equation so, the equation becomes ( ${\text{H}}^{+}$ ion and ${\text{OH}}^{\text{-}}$ ion present on both the sides are combined to form water):

  ${\text{NO}}_{{}_2}^{-}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)\to {\text{NH}}_3\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)+7{\text{OH}}^{-}(aq)$

To balance the charge, add six electrons on reactant sideare added then the equation becomes,

  ${\text{NO}}_{{}_2}^{-}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)+6{\text{e}}^{-}\to {\text{NH}}_3\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)+7{\text{OH}}^{-}(aq)$  (1)

The oxidation half reaction for aluminum is shown below.

  $\text{Al}\left(s\right)\to {\text{AlO}}_{{}_2}^{-}\left(aq\right)$

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

  $\text{Al}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\to {\text{AlO}}_{{}_2}^{-}\left(aq\right)$

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

  $\text{Al}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\to {\text{AlO}}_{{}_2}^{-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)$

For reaction in basic medium, ${\text{OH}}^{\text{-}}$ ion is added to each side of ${\text{H}}^{+}$ ion in the equation. In order to balance the charge and atoms, the ${\text{OH}}^{\text{-}}$ ion is added to both the sides of the equation so, the equation becomes ( ${\text{H}}^{+}$ ion and ${\text{OH}}^{\text{-}}$ ion present on both the sides are combined to form water):

  $\text{Al}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)+4O{H}^{-}\left(aq\right)\to {\text{AlO}}_{{}_2}^{-}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$

To balance the charge, add three electrons on product side are addedthen the equation becomes,

  $\text{Al}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)+4O{H}^{-}\left(aq\right)\to {\text{AlO}}_{{}_2}^{-}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)+3{\text{e}}^{-}$   (2)

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

  $\text{2Al}\left(s\right)+4{\text{H}}_2\text{O}\left(l\right)+8O{H}^{-}\left(aq\right)\to 2{\text{AlO}}_{{}_2}^{-}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+6{\text{e}}^{-}$  (3)

Add equation (1) and (3) to obtain the final equation.

Therefore, the final balanced equation is:

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

(d)

Interpretation Introduction

Interpretation:The reaction ${\text{MnO}}_{{}_4}^{-}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)\to \text{MnS}\left(s\right)+\text{S}\left(s\right)$ occurring in basic 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 the interacting species are present in aqueous medium.

Answer to Problem 82E

The balanced reaction is ${\text{2MnO}}_{{}_4}^{-}\left(aq\right)+7{\text{S}}^{2-}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)\to 2\text{Mn}S\left(s\right)+5\text{S}\left(s\right)+16{\text{OH}}^{-}\left(aq\right)$ .

Explanation of Solution

The given chemical equation is shown below.

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

The reduction half reaction for manganese is shown below.

  ${\text{MnO}}_{{}_4}^{-}\left(aq\right)\to \text{MnS}\left(s\right)$

Balancing atoms other than O and H:

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

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

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

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

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

For reaction in basic medium, ${\text{OH}}^{\text{-}}$ ion is added to each side of ${\text{H}}^{+}$ ion in the equation. In order to balance the charge and atoms, the ${\text{OH}}^{\text{-}}$ ion is added to both the sides of the equation so, the equation becomes ( ${\text{H}}^{+}$ ion and ${\text{OH}}^{\text{-}}$

ion present on both the sides are combined to form water):

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

To balance the charge, add five electrons on reactant side then the equation becomes,

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

The oxidation half reaction for sulfur is shown below.

  ${\text{S}}^{2-}\left(aq\right)\to \text{S}\left(s\right)$

To balance the charge, add two electrons on product side then the equation becomes,

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

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

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

And multiply equation (2) with five then the equation becomes,

  ${\text{5S}}^{2-}\left(aq\right)\to 5\text{S}\left(s\right)+10{\text{e}}^{-}$  (4)

Add equation (3) and (4) to obtain the final equation.

Therefore, the final balanced equation is:

  ${\text{2MnO}}_{{}_4}^{-}\left(aq\right)+7{\text{S}}^{2-}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)\to 2\text{Mn}S\left(s\right)+5\text{S}\left(s\right)+16{\text{OH}}^{-}\left(aq\right)$

(e)

Interpretation Introduction

Interpretation:The reaction ${\text{CN}}^{-}\left(aq\right)+{\text{MnO}}_{{}_4}^{-}\left(aq\right)\to {\text{CNO}}^{-}\left(aq\right)+{\text{MnO}}_2\left(s\right)$ occurring in basic 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 the interacting species are present in aqueous medium.

Answer to Problem 82E

The balanced reaction is ${\text{3CN}}^{-}\left(aq\right)+2{\text{MnO}}_{{}_4}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\to 3{\text{CNO}}^{-}\left(aq\right)+2{\text{MnO}}_2\left(s\right)+2{\text{OH}}^{-}\left(aq\right)$ .

Explanation of Solution

The given chemical equation is shown below.

  ${\text{CN}}^{-}\left(aq\right)+{\text{MnO}}_{{}_4}^{-}\left(aq\right)\to {\text{CNO}}^{-}\left(aq\right)+{\text{MnO}}_2\left(s\right)$

The reduction half reaction for manganese is shown below.

  ${\text{MnO}}_{{}_4}^{-}\left(aq\right)\to {\text{MnO}}_2\left(s\right)$

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

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

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

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

For reaction in basic medium, ${\text{OH}}^{\text{-}}$ ion is added to each side of ${\text{H}}^{+}$ ion in the equation. In order to balance the charge and atoms, the ${\text{OH}}^{\text{-}}$ ion is added to both the sides of the equation so, the equation becomes ( ${\text{H}}^{+}$ ion and ${\text{OH}}^{\text{-}}$ ion present on both the sides are combined to form water):

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

To balance the charge, add three electrons on reactant side then the equation becomes,

  ${\text{MnO}}_{{}_4}^{-}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)+3{\text{e}}^{-}\to {\text{MnO}}_2\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)+4O{H}^{-}(aq)$  (1)

The oxidation half reaction for carbon is shown below.

  ${\text{CN}}^{-}\left(aq\right)\to {\text{CNO}}^{-}\left(aq\right)$

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

  ${\text{CN}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{CNO}}^{-}\left(aq\right)$

To balance the hydrogen atoms, add ${\text{2H}}^{+}$ on product side then the equation becomes,

  ${\text{CN}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{CNO}}^{-}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)$

For reaction in basic medium, ${\text{OH}}^{\text{-}}$ ion is added to each side of ${\text{H}}^{+}$ ion in the equation. In order to balance the charge and atoms, the ${\text{OH}}^{\text{-}}$ ion is added to both the sides of the equation so, the equation becomes ( ${\text{H}}^{+}$ ion and ${\text{OH}}^{\text{-}}$ ion present on both the sides are combined to form water):

  ${\text{CN}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)+2O{\text{H}}^{-}\left(aq\right)\to {\text{CNO}}^{-}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

To balance the charge, add two electrons on product side then the equation becomes,

  ${\text{CN}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)+2O{\text{H}}^{-}\left(aq\right)\to {\text{CNO}}^{-}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)+2{\text{e}}^{-}$  (2)

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

  ${\text{2MnO}}_{{}_4}^{-}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)+6{\text{e}}^{-}\to 2{\text{MnO}}_2\left(s\right)+4{\text{H}}_2\text{O}\left(l\right)+8O{H}^{-}(aq)$   (3)

And multiply equation (2) with threethen the equation becomes,

  ${\text{3CN}}^{-}\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)+6O{\text{H}}^{-}\left(aq\right)\to 3{\text{CNO}}^{-}\left(aq\right)+6{\text{H}}_2\text{O}\left(l\right)+6{\text{e}}^{-}$   (4)

Add equation (3) and (4) to obtain the final equation.

Therefore, the final balanced equation is:

  ${\text{3CN}}^{-}\left(aq\right)+2{\text{MnO}}_{{}_4}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\to 3{\text{CNO}}^{-}\left(aq\right)+2{\text{MnO}}_2\left(s\right)+2{\text{OH}}^{-}\left(aq\right)$