Chapter 17, Problem 31E

(a)

Interpretation Introduction

Interpretation:

Three oxidation-reduction reactions are given. The balancing of all the reactions in basic media using half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

Answer to Problem 31E

The balanced equations are as follows,

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

Explanation of Solution

The balanced equation is defined as follows,

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

The reduction half cell reaction is,

${\text{MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{MnO}}_{\text{2}}{}_{\left(\text{s}\right)}$

The change in the oxidation number of manganese is from $+7$ to $+4$.

The oxidation half reaction is,

${\text{Al}}_{\left(\text{s}\right)}\stackrel{}{\to }\text{Al}{\left(\text{OH}\right)}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

The change in oxidation number of aluminum is from zero to $+3$.

As the atoms other than hydrogen and oxygen are already balanced, so directly balance the oxygen atom in the reduction half reaction by adding water to right hand side,

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

Balance the hydrogen atoms in the reduction half reaction by adding ${\text{H}}^{+}$ ions to the left hand side,

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

Balance the charge by adding appropriate number of electrons to the left hand side,

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

As the atoms other than hydrogen and oxygen are already balanced in the oxidation half reaction. Therefore directly balance the oxygen by adding water molecule to the left hand side.

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

Balance the hydrogen atoms by adding ${\text{H}}^{+}$ to the right hand side,

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

Balance the charge by adding electrons to the right hand side,

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

Add the oxidation and reduction half reaction to get the final equation,

$\begin{array}{c}{\text{MnO}}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+4H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3e}}^{\text{-}}\stackrel{}{\to }{\text{MnO}}_{\text{2}}{}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{Al}}_{\left(\text{s}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }\text{Al}{\left(\text{OH}\right)}_{\text{4}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+4H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+3e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides,

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

To balance the reaction in basic medium, ${\text{OH}}^{-}$ are added to neutralize ${\text{H}}^{+}$ in the reaction. But there are no ${\text{H}}^{+}$ in this reaction, so the balanced equation in the basic medium is,

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

(b)

Interpretation Introduction

Interpretation:

Three oxidation-reduction reactions are given. The balancing of all the reactions in basic media using half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

Answer to Problem 31E

The balanced equations are as follows,

${\text{Cl}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+2OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+OCl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Explanation of Solution

The balanced equation is defined as follows,

${\text{Cl}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+2OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+OCl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The reduction half cell reaction is,

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

The change in the oxidation number of chlorine is from zero to $-1$.

The oxidation half reaction is,

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

The change in oxidation number of chlorine is from zero to $+1$.

Balance all the atoms except hydrogen and oxygen in the reduction half reaction,

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

As there are no oxygen or hydrogen atoms in the reaction. So, directly balance the charge by adding appropriate number of electrons to the left hand side,

${\text{Cl}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$ (3)

Balance all the atoms except hydrogen and oxygen in the oxidation half,

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

Balance the oxygen atom by adding water molecule to the left hand side,

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

Balance hydrogen atoms by adding ${\text{H}}^{+}$ to the right hand side,

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

Balance the charge by adding electrons to the right hand side,

${\text{Cl}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{2OCl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+4H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}$ (4)

Add the oxidation and reduction half reaction,

$\begin{array}{c}{\text{Cl}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ {\text{Cl}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{2OCl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+4H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides,

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

In basic medium, ${\text{OH}}^{-}$ are added to neutralize ${\text{H}}^{+}$. Add ${\text{OH}}^{-}$ on both sides,

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

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

Simplify the equation,

${\text{Cl}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{+2OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+OCl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

(c)

Interpretation Introduction

Interpretation:

Three oxidation-reduction reactions are given. The balancing of all the reactions in basic media using half-reaction method is to be done.

Concept introduction:

The reaction in which both oxidation and reduction reaction occur simultaneously is called a redox reaction. According to the law of conservation of mass, the mass of all the species in a reaction must be balanced. Therefore balancing is necessary to conserve the mass and even the charge must be balanced to maintain the overall charge of the reaction.

Answer to Problem 31E

The balanced equations are as follows,

${\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2Al}}_{\left(\text{s}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}{\text{+OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{NH}}_{\text{3}}{}_{\left(\text{g}\right)}{\text{+2AlO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

Explanation of Solution

The balanced equation is defined as follows,

${\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2Al}}_{\left(\text{s}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}{\text{+OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{NH}}_{\text{3}}{}_{\left(\text{g}\right)}{\text{+2AlO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

The reduction half cell reaction is,

${\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{NH}}_{\text{3}}{}_{\left(\text{g}\right)}$

The change in the oxidation number of nitrogen is from $+3$ to $-3$.

The oxidation half reaction is,

${\text{Al}}_{\left(\text{s}\right)}\stackrel{}{\to }{\text{AlO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$

The change in oxidation number of aluminum is from zero to $+3$.

All the elements except hydrogen and oxygen are already balanced so directly balance the oxygen in the reduction half reaction by adding water to right hand side,

${\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{NH}}_{\text{3}}{}_{\left(\text{g}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the hydrogen atoms in the reduction half reaction by adding ${\text{H}}^{+}$ ions to the left hand side,

${\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+7H}}^{\text{+}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{NH}}_{\text{3}}{}_{\left(\text{g}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Balance the charge by adding appropriate number of electrons to the left hand side,

${\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+7H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+6e}}^{\text{-}}\stackrel{}{\to }{\text{NH}}_{\text{3}}{}_{\left(\text{g}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ (5)

All the atoms except hydrogen and oxygen in oxidation half reaction are already balanced. So, directly balance oxygen atom by adding water molecule to the left hand side,

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

Balance the hydrogen atom by adding ${\text{H}}^{+}$ to right hand side,

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

Balance the charge by adding electrons to the right hand side,

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

Multiply equation (6) by $2$ to cancel out the electrons and then add the oxidation and reduction half reaction,

$\begin{array}{c}{\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+7H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+6e}}^{\text{-}}\stackrel{}{\to }{\text{NH}}_{\text{3}}{}_{\left(\text{g}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{2Al}}_{\left(\text{s}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{2AlO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+6e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides,

${\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2Al}}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{NH}}_{\text{3}}{}_{\left(\text{g}\right)}{\text{+2AlO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+H}}^{\text{+}}{}_{\left(\text{aq}\right)}$

In basic medium, ${\text{OH}}^{-}$ are added to neutralize ${\text{H}}^{+}$. Add ${\text{OH}}^{-}$ on both sides,

$\begin{array}{l}{\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2Al}}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}{\text{+OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }\\ {\text{NH}}_{\text{3}}{}_{\left(\text{g}\right)}{\text{+2AlO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\end{array}$

The final equation is,

${\text{NO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{+2Al}}_{\left(\text{s}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}{\text{+OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{NH}}_{\text{3}}{}_{\left(\text{g}\right)}{\text{+2AlO}}_{\text{2}}{{}^{\text{-}}}_{\left(\text{aq}\right)}$