Chapter 17, Problem 32E

(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 32E

The balanced equations are as follows,

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

Explanation of Solution

The balanced equation is defined as follows,

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

The reduction half cell reaction is,

${\text{CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right)\stackrel{}{\to }\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right)$

The change in the oxidation number of chromium is from $+6$ to $+3$.

The oxidation half reaction is,

$\text{Cr}\left(\text{s}\right)\stackrel{}{\to }\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right)$

The change in oxidation number of chromium 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,

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

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

${\text{CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right){\text{+5H}}^{\text{+}}\left(\text{aq}\right)\stackrel{}{\to }\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)$

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

${\text{CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right){\text{+5H}}^{\text{+}}\left(\text{aq}\right){\text{+3e}}^{\text{-}}\stackrel{}{\to }\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)$ (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.

$\text{Cr}\left(\text{s}\right){\text{+3H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right)$

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

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

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

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

Add the oxidation and reduction half reaction,

$\begin{array}{c}{\text{CrO}}_{\text{4}}{}^{\text{2-}}\left(\text{aq}\right){\text{+5H}}^{\text{+}}\left(\text{aq}\right){\text{+3e}}^{\text{-}}\stackrel{}{\to }\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ \text{Cr}\left(\text{s}\right){\text{+3H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\text{Cr}{\left(\text{OH}\right)}_{\text{3}}\left(\text{s}\right){\text{+3H}}^{\text{+}}\left(\text{aq}\right){\text{+3e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides,

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

To balance the reaction in basic medium, ${\text{OH}}^{-}$ are added to neutralize ${\text{H}}^{+}$ in the reaction. Add ${\text{OH}}^{-}$ on both the sides to neutralize ${\text{H}}^{+}$ ions.

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

The final equation is,

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

(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 32E

The balanced equations are as follows,

${\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+7S}}^{\text{2-}}\left(\text{aq}\right){\text{+8H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\text{2MnS}\left(\text{s}\right)\text{+5S}\left(\text{s}\right){\text{+16OH}}^{\text{-}}\left(\text{aq}\right)$

Explanation of Solution

The balanced equation is defined as follows,

${\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+7S}}^{\text{2-}}\left(\text{aq}\right){\text{+8H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\text{2MnS}\left(\text{s}\right)\text{+5S}\left(\text{s}\right){\text{+16OH}}^{\text{-}}\left(\text{aq}\right)$

The reduction half cell reaction is,

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

The change in the oxidation number of manganese is from $+7$  to $+2$.

The oxidation half reaction is,

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

The change in oxidation number of sulphur is from $-2$  to zero.

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

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

Balance the oxygen atoms by adding water molecules to the right hand side,

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

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

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

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

${\text{MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right)\text{+S}\left(\text{s}\right){\text{+8H}}^{\text{+}}\left(\text{aq}\right){\text{+7e}}^{\text{-}}\stackrel{}{\to }\text{MnS}\left(\text{s}\right){\text{+4H}}_{\text{2}}\text{O}\left(\text{l}\right)$ (3)

All the atoms except hydrogen and oxygen are already balanced and there is no hydrogen or oxygen atom in the reaction. So, directly balance the charge by adding appropriate number of electrons.,

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

Multiply equation (3) by $2$ and equation (4) by $7$ to balance the charge and add the oxidation and reduction half reaction,

$\begin{array}{c}{\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right)\text{+2S}\left(\text{s}\right){\text{+16H}}^{\text{+}}\left(\text{aq}\right){\text{+14e}}^{\text{-}}\stackrel{}{\to }\text{2MnS}\left(\text{s}\right){\text{+8H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ {\text{7S}}^{\text{2-}}\left(\text{aq}\right)\stackrel{}{\to }\text{7S}\left(\text{s}\right){\text{+14e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides,

${\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+16H}}^{\text{+}}\left(\text{aq}\right){\text{+7S}}^{\text{2-}}\left(\text{aq}\right)\stackrel{}{\to }\text{2MnS}\left(\text{s}\right)\text{+5S}\left(\text{s}\right){\text{+8H}}_{\text{2}}\text{O}\left(\text{l}\right)$

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

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

$\begin{array}{l}{\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+16H}}^{\text{+}}\left(\text{aq}\right){\text{+7S}}^{\text{2-}}\left(\text{aq}\right){\text{+16OH}}^{\text{-}}\left(\text{aq}\right)\stackrel{}{\to }\\ \text{2MnS}\left(\text{s}\right)\text{+5S}\left(\text{s}\right){\text{+8H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+16OH}}^{\text{-}}\left(\text{aq}\right)\end{array}$

Simplify the equation,

${\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+7S}}^{\text{2-}}\left(\text{aq}\right){\text{+8H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\text{2MnS}\left(\text{s}\right)\text{+5S}\left(\text{s}\right){\text{+16OH}}^{\text{-}}\left(\text{aq}\right)$

(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 32E

The balanced equations are as follows,

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

Explanation of Solution

The balanced equation is defined as follows,

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

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,

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

The change in the oxidation number of carbon is from $+2$ to $+4$.

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,

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

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,

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

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

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

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

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

${\text{CN}}^{\text{-}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{CNO}}^{\text{-}}\left(\text{aq}\right){\text{+2H}}^{\text{+}}\left(\text{aq}\right){\text{+2e}}^{\text{-}}$ (6)

Multiply equation (6) by $3$ and equation (5) by $2$ to cancel out the electrons and then add the oxidation and reduction half reaction,

$\begin{array}{c}{\text{2MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+8H}}^{\text{+}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}\stackrel{}{\to }{\text{2MnO}}_{\text{2}}\left(\text{s}\right){\text{+4H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ {\text{3CN}}^{\text{-}}\left(\text{aq}\right){\text{+3H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{3CNO}}^{\text{-}}\left(\text{aq}\right){\text{+6H}}^{\text{+}}\left(\text{aq}\right){\text{+6e}}^{\text{-}}\end{array}$

Cancel similar terms on both the sides,

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

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

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

The final equation is,

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