Chapter 6, Problem 6K.5E

(a)

Interpretation Introduction

Interpretation:

The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.

  ${\text{O}}_{\text{3}}{\text{(g)+Br}}^{\text{-}}\text{(aq)}\to {\text{O}}_{\text{2}}{\text{(g)+BrO}}_{\text{3}}{}^{\text{-}}\text{(aq)}$

Concept Introduction:

Net ionic equation:

Net ionic equation is defined as the specific species that only involves to a particular reaction. This type of equations is generally used in acid-base neutralization reactions and redox reactions.

Oxidizing agent:

The material which gains electron in a chemical reaction is called oxidizing agent. In this reaction, the oxidation number will be decreased.

Reducing agent:

The material, which loses electrons in a chemical reaction, is called reducing agent. In this reaction, the oxidation number will be increased.

Answer to Problem 6K.5E

The balanced reaction of ozone with bromide ions is given below,

  ${\text{Br}}^{\text{-}}{\text{(aq)+3O}}_{\text{3}}\text{(g)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+3O}}_{\text{2}}\text{(g)}$

Here, the oxidizing agent is ${\text{O}}_{\text{3}}$ and reducing agent is ${\text{Br}}^{\text{-}}$.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  ${\text{O}}_{\text{3}}{\text{(g)+Br}}^{\text{-}}\text{(aq)}\to {\text{O}}_{\text{2}}{\text{(g)+BrO}}_{\text{3}}{}^{\text{-}}\text{(aq)}$

Oxidation half-reaction:

The oxidation number of $\text{Br}$ is increased from $-1$ to $+5$, therefore this is oxidation reaction.

  ${\text{Br}}^{\text{-}}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}\text{(aq)}$

Balance the equation except $\text{H}$ and $\text{O}$.

  ${\text{Br}}^{\text{-}}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}\text{(aq)}$

Balance the $\text{O}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ left side

  ${\text{Br}}^{\text{-}}{\text{(aq)+3H}}_{\text{2}}\text{O(l)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}\text{(aq)}$

Balance the $\text{H}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ molecules to the right side and add the same amount ${\text{OH}}^{\text{-}}$ ions to the left side

  ${\text{Br}}^{\text{-}}{\text{(aq)+3H}}_{\text{2}}{\text{O(l)+6OH}}^{\text{-}}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+6H}}_{\text{2}}\text{O(l)}$

Balance the net charges by adding the electrons.

Here, in left side, the net charge is $-7$ and the right side the net charge is $-1$, therefore it required $6$ electrons on the right side to reduce the charge from $-1$ to $-7$.

  ${\text{Br}}^{\text{-}}{\text{(aq)+3H}}_{\text{2}}{\text{O(l)+6OH}}^{\text{-}}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+6H}}_{\text{2}}{\text{O(l)+6e}}^{-}$

Therefore, the balanced oxidation half-reaction is

  ${\text{Br}}^{\text{-}}{\text{(aq)+3H}}_{\text{2}}{\text{O(l)+6OH}}^{\text{-}}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+6H}}_{\text{2}}{\text{O(l)+6e}}^{-}$

Here, ${\text{Br}}^{\text{-}}$ act as loses of electrons, so it is a reducing agent.

Reduction half-reaction:

The oxidation number of $\text{O}$ is $0$.

  ${\text{O}}_{\text{3}}\text{(g)}\to {\text{O}}_{\text{2}}\text{(g)}$

Balance the $\text{O}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ right side

  ${\text{O}}_{\text{3}}\text{(g)}\to {\text{O}}_{\text{2}}{\text{(g)+H}}_{\text{2}}\text{O(l)}$

Balance the $\text{H}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ molecules to the left side and add the same amount ${\text{OH}}^{\text{-}}$ ions to the right side

  ${\text{O}}_{\text{3}}{\text{(g)+2H}}_{\text{2}}\text{O(l)}\to {\text{O}}_{\text{2}}{\text{(g)+H}}_{\text{2}}{\text{O(l)+2OH}}^{\text{-}}\text{(aq)}$

Balance the net charges by adding the electrons.

Here, in left side, the net charge is $0$ and the right side the net charge is $-2$, therefore it required $2$ electrons on the left side to reduce the charge from $0$ to $-2$.

  ${\text{O}}_{\text{3}}{\text{(g)+2H}}_{\text{2}}{\text{O(l)+2e}}^{-}\to {\text{O}}_{\text{2}}{\text{(g)+H}}_{\text{2}}{\text{O(l)+2OH}}^{\text{-}}\text{(aq)}$

Therefore, the balanced reduction half-reaction is

  ${\text{O}}_{\text{3}}{\text{(g)+2H}}_{\text{2}}{\text{O(l)+2e}}^{-}\to {\text{O}}_{\text{2}}{\text{(g)+H}}_{\text{2}}{\text{O(l)+2OH}}^{\text{-}}\text{(aq)}$

Here, the ${\text{O}}_{\text{3}}$ act as gaining of electrons, so it is a oxidizing agent.

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction 6 electrons are lost and in reduction half reaction $2$ electron is gained. Therefore, multiply the oxidation half reaction by $2$ and the reduction half reaction by $6$.

  $\begin{array}{l}{\text{2Br}}^{\text{-}}{\text{(aq)+6H}}_{\text{2}}{\text{O(l)+12OH}}^{\text{-}}\text{(aq)}\to 2{\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+12H}}_{\text{2}}{\text{O(l)+12e}}^{-}\\ {\text{6O}}_{\text{3}}{\text{(g)+12H}}_{\text{2}}{\text{O(l)+12e}}^{-}\to 6{\text{O}}_{\text{2}}{\text{(g)+6H}}_{\text{2}}{\text{O(l)+12OH}}^{\text{-}}\text{(aq)}\end{array}$

Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.

  ${\text{2Br}}^{\text{-}}{\text{(aq)+6O}}_{\text{3}}\text{(g)}\to 2{\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+6O}}_{\text{2}}\text{(g)}$

Divide by 2 each side of the arrow.

  ${\text{Br}}^{\text{-}}{\text{(aq)+3O}}_{\text{3}}\text{(g)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+3O}}_{\text{2}}\text{(g)}$

Therefore, the balanced net ionic equation the above reaction is

  ${\text{Br}}^{\text{-}}{\text{(aq)+3O}}_{\text{3}}\text{(g)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+3O}}_{\text{2}}\text{(g)}$

(b)

Interpretation Introduction

Interpretation:

The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.

  ${\text{Br}}_{\text{2}}\text{(l)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+Br}}^{\text{-}}\text{(aq)}$

Concept introduction:

Refer to part (a).

Answer to Problem 6K.5E

The balanced reaction of bromine with itself in aqueous solution is given below,

  ${\text{3Br}}_{\text{2}}{\text{(l)+6OH}}^{\text{-}}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+5Br}}^{\text{-}}\text{(aq)}+3{\text{H}}_{\text{2}}\text{O(l)}$

Here, the oxidizing agent and reducing agent is ${\text{Br}}_{\text{2}}$.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  ${\text{Br}}_{\text{2}}\text{(l)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+Br}}^{\text{-}}\text{(aq)}$

Oxidation half-reaction:

The oxidation number of $\text{Br}$ is increased from $0$ to $+5$, therefore this is oxidation reaction.

  ${\text{Br}}_{\text{2}}\text{(l)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}\text{(aq)}$

Balance the equation except $\text{H}$ and $\text{O}$.

  ${\text{Br}}_{\text{2}}\text{(l)}\to 2{\text{BrO}}_{\text{3}}{}^{\text{-}}\text{(aq)}$

Balance the $\text{O}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ left side

  ${\text{Br}}_{\text{2}}{\text{(l)+6H}}_{\text{2}}\text{O(l)}\to 2{\text{BrO}}_{\text{3}}{}^{\text{-}}\text{(aq)}$

Balance the $\text{H}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ molecules to the left side and add the same amount ${\text{OH}}^{\text{-}}$ ions to the right side

  ${\text{Br}}_{\text{2}}{\text{(l)+6H}}_{\text{2}}{\text{O(l)+12OH}}^{\text{-}}\text{(aq)}\to 2{\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+12H}}_{\text{2}}\text{O(l)}$

Balance the net charges by adding the electrons.

Here, in left side, the net charge is $-12$ and the right side the net charge is $\text{-2}$, therefore it required $10$ electron on the right side to reduce the charge from $-2$ to $-12$.

  ${\text{Br}}_{\text{2}}{\text{(l)+6H}}_{\text{2}}{\text{O(l)+12OH}}^{\text{-}}\text{(aq)}\to 2{\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+12H}}_{\text{2}}{\text{O(l)+10e}}^{-}$

Therefore, the balanced oxidation half-reaction is

  ${\text{Br}}_{\text{2}}{\text{(l)+6H}}_{\text{2}}{\text{O(l)+12OH}}^{\text{-}}\text{(aq)}\to 2{\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+12H}}_{\text{2}}{\text{O(l)+10e}}^{-}$

Here, ${\text{Br}}_{\text{2}}$ lost $10$ electrons, so it is a reducing agent.

Reduction half-reaction:

The oxidation number of $\text{Br}$ is decreased from $0$ to $-1$, therefore this is reduction reaction.

  ${\text{Br}}_{\text{2}}\text{(l)}\to {\text{Br}}^{\text{-}}\text{(aq)}$

Balance the equation except $\text{H}$ and $\text{O}$.

  ${\text{Br}}_{\text{2}}\text{(l)}\to 2{\text{Br}}^{\text{-}}\text{(aq)}$

Balance the net charges by adding the electrons.

Here, in left side, the net charge is $0$ and the right side the net charge is $\text{-2}$, therefore it required $2$ electrons on the left side to reduce the charge from $0$ to $\text{-2}$.

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

Here, the ${\text{Br}}_{\text{2}}$ gains $2$ electrons, so it is an oxidizing agent.

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction $10$ electrons are lost and in reduction half reaction $10$ electrons are gained. Therefore, multiply the oxidation half reaction by $2$ and the reduction half reaction by $2$.

  $\begin{array}{c}{\text{2Br}}_{\text{2}}{\text{(l)+12H}}_{\text{2}}{\text{O(l)+24OH}}^{\text{-}}\text{(aq)}\to 4{\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+24H}}_{\text{2}}{\text{O(l)+20e}}^{-}\\ {\text{10Br}}_{\text{2}}{\text{(l)+20e}}^{-}\to 20{\text{Br}}^{\text{-}}\text{(aq)}\end{array}$

Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.

  ${\text{12Br}}_{\text{2}}{\text{(l)+24OH}}^{\text{-}}\text{(aq)}\to 4{\text{BrO}}_{\text{3}}{}^{\text{-}}\text{(aq)+}20{\text{Br}}^{\text{-}}\text{(aq)}+{\text{12H}}_{\text{2}}\text{O(l)}$

Divide by $4$ each side of the arrow.

  ${\text{3Br}}_{\text{2}}{\text{(l)+6OH}}^{\text{-}}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+5Br}}^{\text{-}}\text{(aq)}+3{\text{H}}_{\text{2}}\text{O(l)}$

Therefore, the balanced net ionic equation the above reaction is

  ${\text{3Br}}_{\text{2}}{\text{(l)+6OH}}^{\text{-}}\text{(aq)}\to {\text{BrO}}_{\text{3}}{}^{\text{-}}{\text{(aq)+5Br}}^{\text{-}}\text{(aq)}+3{\text{H}}_{\text{2}}\text{O(l)}$

(c)

Interpretation Introduction

Interpretation:

The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.

  ${\text{Cr}}^{\text{3+}}{\text{(aq)+MnO}}_{\text{2}}\text{(s)}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+CrO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

Concept introduction:

Refer to part (a).

Answer to Problem 6K.5E

The balanced reaction for the formation of chromate ions from chromium (III) ions is given below,

  ${\text{2Cr}}^{\text{3+}}{\text{(aq)+3MnO}}_{\text{2}}{\text{(s)+4OH}}^{\text{-}}\text{(aq)}\to 2{\text{CrO}}_{\text{4}}{}^{\text{2-}}\text{(aq)+}3{\text{Mn}}^{\text{2+}}{\text{(aq)+2H}}_{\text{2}}\text{O(l)}$

Here, the oxidizing agent is ${\text{MnO}}_{\text{2}}$ and reducing agent is ${\text{Cr}}^{\text{3+}}$.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  ${\text{Cr}}^{\text{3+}}{\text{(aq)+MnO}}_{\text{2}}\text{(s)}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+CrO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

Oxidation half-reaction:

The oxidation number of $\text{S}$ is increased from $-2$ to $0$, therefore this is oxidation reaction.

  ${\text{Cr}}^{\text{3+}}\text{(aq)}\to {\text{CrO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

Balance the equation except $\text{H}$ and $\text{O}$.

  ${\text{Cr}}^{\text{3+}}\text{(aq)}\to {\text{CrO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

Balance the $\text{O}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ left side

  ${\text{Cr}}^{\text{3+}}{\text{(aq)+4H}}_{\text{2}}\text{O(l)}\to {\text{CrO}}_{\text{4}}{}^{\text{2-}}\text{(aq)}$

Balance the $\text{H}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ molecules to the right side and add the same amount ${\text{OH}}^{\text{-}}$ ions to the left side

  ${\text{Cr}}^{\text{3+}}{\text{(aq)+4H}}_{\text{2}}{\text{O(l)+8OH}}^{\text{-}}\text{(aq)}\to {\text{CrO}}_{\text{4}}{}^{\text{2-}}{\text{(aq)+8H}}_{\text{2}}\text{O(l)}$

Balance the net charges by adding the electrons.

Here, in left side, the net charge is $-5$ and the right side the net charge is $\text{-2}$, therefore it required $3$ electrons on the right side to reduce the charge from $-2$ to $-5$.

  ${\text{Cr}}^{\text{3+}}{\text{(aq)+4H}}_{\text{2}}{\text{O(l)+8OH}}^{\text{-}}\text{(aq)}\to {\text{CrO}}_{\text{4}}{}^{\text{2-}}{\text{(aq)+8H}}_{\text{2}}{\text{O(l)+3e}}^{-}$

Therefore, the balanced oxidation half-reaction is

  ${\text{Cr}}^{\text{3+}}{\text{(aq)+4H}}_{\text{2}}{\text{O(l)+8OH}}^{\text{-}}\text{(aq)}\to {\text{CrO}}_{\text{4}}{}^{\text{2-}}{\text{(aq)+8H}}_{\text{2}}{\text{O(l)+3e}}^{-}$

Here, ${\text{Cr}}^{\text{3+}}$ act as loses of electrons, so it is a reducing agent.

Reduction half-reaction:

The oxidation number of $\text{Mn}$ is decreased from $+4$ to $+2$, therefore this is reduction reaction.

  ${\text{MnO}}_{\text{2}}\text{(s)}\to {\text{Mn}}^{\text{2+}}\text{(aq)}$

Balance the equation except $\text{H}$ and $\text{O}$.

  ${\text{MnO}}_{\text{2}}\text{(s)}\to {\text{Mn}}^{\text{2+}}\text{(aq)}$

Balance the $\text{O}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ right side

  ${\text{MnO}}_{\text{2}}\text{(s)}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+2H}}_{\text{2}}\text{O(l)}$

Balance the $\text{H}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ molecules to the left side and add the same amount ${\text{OH}}^{\text{-}}$ ions to the right side

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

Balance the net charges by adding the electrons.

Here, in left side, the net charge is $0$ and the right side the net charge is $-2$, therefore it required $2$ electrons on the left side to reduce the charge from $0$ to $-2$.

  ${\text{MnO}}_{\text{2}}{\text{(s)+4H}}_{\text{2}}{\text{O(l)+2e}}^{-}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+2H}}_{\text{2}}{\text{O(l)+4OH}}^{\text{-}}\text{(aq)}$

Here, the ${\text{MnO}}_{\text{2}}$ act as gaining of electrons, so it is an oxidizing agent.

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction $2$ electrons are lost and in reduction half reaction $2$ electrons is gained. Therefore, multiply the oxidation half reaction by $2$ and the reduction half reaction by $2$.

  $\begin{array}{c}{\text{2Cr}}^{\text{3+}}{\text{(aq)+8H}}_{\text{2}}{\text{O(l)+16OH}}^{\text{-}}\text{(aq)}\to 2{\text{CrO}}_{\text{4}}{}^{\text{2-}}{\text{(aq)+16H}}_{\text{2}}{\text{O(l)+6e}}^{-}\\ {\text{3MnO}}_{\text{2}}{\text{(s)+12H}}_{\text{2}}{\text{O(l)+6e}}^{-}\to 3{\text{Mn}}^{\text{2+}}{\text{(aq)+6H}}_{\text{2}}{\text{O(l)+12OH}}^{\text{-}}\text{(aq)}\end{array}$

Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.

  ${\text{2Cr}}^{\text{3+}}{\text{(aq)+3MnO}}_{\text{2}}{\text{(s)+4OH}}^{\text{-}}\text{(aq)}\to 2{\text{CrO}}_{\text{4}}{}^{\text{2-}}\text{(aq)+}3{\text{Mn}}^{\text{2+}}{\text{(aq)+2H}}_{\text{2}}\text{O(l)}$

Therefore, the balanced net ionic equation the above reaction is

  ${\text{2Cr}}^{\text{3+}}{\text{(aq)+3MnO}}_{\text{2}}{\text{(s)+4OH}}^{\text{-}}\text{(aq)}\to 2{\text{CrO}}_{\text{4}}{}^{\text{2-}}\text{(aq)+}3{\text{Mn}}^{\text{2+}}{\text{(aq)+2H}}_{\text{2}}\text{O(l)}$

(d)

Interpretation Introduction

Interpretation:

The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.

  ${\text{P}}_{\text{4}}\text{(S)}\to {\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}{\text{(aq)+PH}}_{\text{3}}\text{(aq)}$

Concept introduction:

Refer to part (a).

Answer to Problem 6K.5E

The balanced reaction for the formation of phosphine, a poisonous gas with odor of decaying fish is given below,

  ${\text{P}}_{\text{4}}{\text{(S)+2OH}}^{\text{-}}{\text{(aq)+2H}}_{\text{2}}\text{O(l)}\to 2{\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}{\text{(aq)+2PH}}_{\text{3}}\text{(aq)}$

Here, the oxidizing agent and reducing agent is ${\text{P}}_{\text{4}}$.

Explanation of Solution

The unbalanced skeletal equation for the reaction is

  ${\text{P}}_{\text{4}}\text{(S)}\to {\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}{\text{(aq)+PH}}_{\text{3}}\text{(aq)}$

Oxidation half-reaction:

The oxidation number of $\text{P}$ is increased from $0$ to $+1$, therefore this is oxidation reaction.

  ${\text{P}}_{\text{4}}\text{(S)}\to {\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}\text{(aq)}$

Balance the equation except $\text{H}$ and $\text{O}$.

  ${\text{P}}_{\text{4}}\text{(S)}\to 4{\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}\text{(aq)}$

Balance the $\text{O}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ left side

  ${\text{P}}_{\text{4}}{\text{(S)+8H}}_{\text{2}}\text{O(l)}\to 4{\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}\text{(aq)}$

Balance the $\text{H}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ molecules to the right side and add the same amount ${\text{OH}}^{\text{-}}$ ions to the left side

  ${\text{P}}_{\text{4}}{\text{(S)+8H}}_{\text{2}}{\text{O(l)+16OH}}^{\text{-}}\text{(aq)}\to 4{\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}{\text{(aq)+16H}}_{\text{2}}\text{O(l)}$

Balance the net charges by adding the electrons.

Here, in left side, the net charge is $-16$ and the right side the net charge is $\text{-4}$, therefore it required $12$ electrons on the right side to reduce the charge from $-4$ to $-16$.

  ${\text{P}}_{\text{4}}{\text{(S)+8H}}_{\text{2}}{\text{O(l)+16OH}}^{\text{-}}\text{(aq)}\to 4{\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}{\text{(aq)+16H}}_{\text{2}}{\text{O(l)+12e}}^{-}$

Therefore, the balanced oxidation half-reaction is

  ${\text{P}}_{\text{4}}{\text{(S)+8H}}_{\text{2}}{\text{O(l)+16OH}}^{\text{-}}\text{(aq)}\to 4{\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}{\text{(aq)+16H}}_{\text{2}}{\text{O(l)+12e}}^{-}$

Here, ${\text{P}}_{\text{4}}$ act as loses of electrons, so it is a reducing agent.

Reduction half-reaction:

The oxidation number of $\text{P}$ is decreased from $0$ to $-3$, therefore this is reduction reaction.

  ${\text{P}}_{\text{4}}\text{(S)}\to {\text{PH}}_{\text{3}}\text{(aq)}$

Balance the equation except $\text{H}$ and $\text{O}$.

  ${\text{P}}_{\text{4}}\text{(S)}\to 4{\text{PH}}_{\text{3}}\text{(aq)}$

Balance the $\text{H}$ atom by adding ${\text{H}}_{\text{2}}\text{O}$ molecules to the left side and add the same amount ${\text{OH}}^{\text{-}}$ ions to the right side

  ${\text{P}}_{\text{4}}{\text{(S)+12H}}_{\text{2}}\text{O(l)}\to 4{\text{PH}}_{\text{3}}{\text{(aq)+12OH}}^{\text{-}}\text{(aq)}$

Balance the net charges by adding the electrons.

Here, in left side, the net charge is $0$ and the right side the net charge is $-12$, therefore it required $12$ electrons on the left side to reduce the charge from $0$ to $-12$.

  ${\text{P}}_{\text{4}}{\text{(S)+12H}}_{\text{2}}{\text{O(l)+12e}}^{\text{-}}\to 4{\text{PH}}_{\text{3}}{\text{(aq)+12OH}}^{\text{-}}\text{(aq)}$

Here, the ${\text{P}}_{\text{4}}$ act as gaining of electrons, so it is an oxidizing agent.

Now add the two half reactions together. Match the number of electrons in each side.  Because in oxidation half reaction $12$ electrons are lost and in reduction half reaction $12$ electrons is gained. Therefore, multiply the oxidation half reaction by $12$ and the reduction half reaction by $12$.

  $\begin{array}{c}{\text{12P}}_{\text{4}}{\text{(S)+96H}}_{\text{2}}{\text{O(l)+192OH}}^{\text{-}}\text{(aq)}\to 48{\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}{\text{(aq)+192H}}_{\text{2}}{\text{O(l)+144e}}^{-}\\ {\text{12P}}_{\text{4}}{\text{(S)+144H}}_{\text{2}}{\text{O(l)+144e}}^{\text{-}}\to 48{\text{PH}}_{\text{3}}{\text{(aq)+144OH}}^{\text{-}}\text{(aq)}\end{array}$

Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.

  ${\text{24P}}_{\text{4}}{\text{(S)+48OH}}^{\text{-}}{\text{(aq)+48H}}_{\text{2}}\text{O(l)}\to 48{\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}\text{(aq)+}48{\text{PH}}_{\text{3}}\text{(aq)}$

Divide by $24$ each side of the arrow.

  ${\text{P}}_{\text{4}}{\text{(S)+2OH}}^{\text{-}}{\text{(aq)+2H}}_{\text{2}}\text{O(l)}\to 2{\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}{\text{(aq)+2PH}}_{\text{3}}\text{(aq)}$

Therefore, the balanced net ionic equation the above reaction is

  ${\text{P}}_{\text{4}}{\text{(S)+2OH}}^{\text{-}}{\text{(aq)+2H}}_{\text{2}}\text{O(l)}\to 2{\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}{}^{\text{-}}{\text{(aq)+2PH}}_{\text{3}}\text{(aq)}$