Chapter 11, Problem 20E

(a)

Interpretation Introduction

Interpretation: The ${\epsilon }^o$ value for the following reaction needs to be calculated and if the reaction is spontaneous or not needs to be determined.

  ${\text{MnO}}_{\text{4}}{}^{\text{-}}(aq)\text{+}{\text{I}}^{\text{-}}(aq)\to {\text{I}}_{\text{2}}\text{+}\text{}{\text{Mn}}^{\text{+2}}(aq)$

Concept Introduction : Spontaneous reaction are those which is a spontaneous process under given conditions without any intervention. Such reactions are accompanied by certain distortion or change in entropy.

Cell potential under standard conditions is calculated as follows:

  $E^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}$

The reaction is spontaneous if the value of $E^0{}_{cell}$ is positive and the reaction is non spontaneous if value of $E^0{}_{cell}$ is negative.

Answer to Problem 20E

  $E^0{}_{cell}$ of the reaction = 0.97 V

As $E^0{}_{cell}$ is positive, hence the reaction is spontaneous.

Explanation of Solution

The given reaction with oxidation state is:

  $\text{(}\stackrel{+7}{\text{Mn}}{\stackrel{-2}{{\text{O}}_{\text{4}}}}^{\text{-}})(aq)\text{+}{\text{I}}^{\text{-}}(aq)\to {\stackrel{0}{\text{I}}}_{\text{2}}\text{+}\text{}{\text{Mn}}^{\text{+2}}(aq)$

From the reaction is it clear that Mn undergo reduction and I undergoes oxidation.

So, the half cells can be balanced as,

Reduction half cell:

  $\begin{array}{l}{\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}\\ {\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}\text{(Mn}\text{and}\text{O}\text{balanced)}\\ {\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{5e}}^{\text{-}}{\text{+8H}}^{\text{+}}\to {\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}\text{(balanced)}\end{array}$

Oxidation half cell:

  $\begin{array}{l}{\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}\\ {\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}\text{(Mn}\text{and}\text{O}\text{balanced)}\\ {\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{5e}}^{\text{-}}{\text{+8H}}^{\text{+}}\to {\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}\text{(balanced)}\end{array}$

Multiply oxidation half with 2 and reduction half with 5 to balance the electron gained and electron lost we get:

  $\begin{array}{l}{\text{2MnO}}_{\text{4}}{}^{\text{-}}\text{+}10{\text{e}}^{\text{-}}\text{+}{\text{16H}}^{\text{+}}\to 2{\text{Mn}}^{\text{2+}}\text{+}8{\text{H}}_{\text{2}}\text{O}\\ {\text{10I}}^{-}\to 5{\text{I}}_{\text{2}}\text{+}10{\text{e}}^{\text{-}}\end{array}$

Adding both the reaction we get the balanced equation:

  ${\text{2MnO}}_{\text{4}}{}^{\text{-}}\text{+}10{\text{I}}^{\text{-}}\text{+}{\text{16H}}^{\text{+}}\to 2{\text{Mn}}^{\text{2+}}\text{+}5{\text{I}}_2+8{\text{H}}_{\text{2}}\text{O}$

Now calculate cell potential using values from Table 11.1 as,

  $\begin{array}{c}{E}^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}\\ =1.51-(0.54)\\ =0.97\text{V}\end{array}$

As $E^0{}_{cell}$ is positive, hence the reaction is spontaneous.

(b)

Interpretation Introduction

Interpretation: The ${\epsilon }^o$ value for the following reaction needs to be calculated and if the reaction is spontaneous or not needs to be determined.

  $\stackrel{}{\text{Mn}}\stackrel{}{{\text{O}}_{\text{4}}}(aq)\text{+}{\text{F}}^{\text{-}}(aq)\rightleftharpoons {\stackrel{}{\text{F}}}_{\text{2}}\text{+}\text{}{\text{Mn}}^{\text{+2}}(aq)$

Concept Introduction : Spontaneous reaction are those which is a spontaneous process under given conditions without any intervention. Such reactions are accompanied by certain distortion or change in entropy.

Cell potential under standard conditions is calculated as follows:

  $E^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}$

The reaction is spontaneous if the value of $E^0{}_{cell}$ is positive and the reaction is non-spontaneous if value of $E^0{}_{cell}$ is negative.

Answer to Problem 20E

  $E^0{}_{cell}$ of the reaction = -1.36 V

As $E^0{}_{cell}$ is negative, hence the reaction is non-spontaneous.

Explanation of Solution

The given reaction with oxidation state is:

  $\text{(}\stackrel{+7}{\text{Mn}}{\stackrel{-2}{{\text{O}}_{\text{4}}}}^{\text{-}})(aq)\text{+}{\text{F}}^{\text{-}}(aq)\rightleftharpoons {\stackrel{0}{\text{F}}}_{\text{2}}\text{+}\text{}{\text{Mn}}^{\text{+2}}(aq)$

From the reaction is it clear that Mn undergo reduction and F undergoes oxidation.

So, the half cells can be balanced as,

Reduction half cell:

  $\begin{array}{c}{\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}\\ {\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{5e}}^{\text{-}}\to {\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}\text{(Mn}\text{and}\text{O}\text{balanced)}\\ {\text{MnO}}_{\text{4}}{}^{\text{-}}\text{+}{\text{5e}}^{\text{-}}{\text{+8H}}^{\text{+}}\to {\text{Mn}}^{\text{2+}}\text{+}{\text{4H}}_{\text{2}}\text{O}\text{(balanced)}\\ \end{array}$

Oxidation half cell:

  $\begin{array}{c}{\text{F}}^{\text{-}}\to {\text{F}}_{\text{2}}\text{+}{\text{e}}^{\text{-}}\\ {\text{2F}}^{\text{-}}\to {\text{F}}_{\text{2}}\text{+}{\text{2e}}^{\text{-}}\text{(Balanced)}\end{array}$

Multiply oxidation half with 2 and reduction half with 5 to balance the electron gained and electron lost we get:

  $\begin{array}{c}{\text{2MnO}}_{\text{4}}{}^{\text{-}}\text{+}10{\text{e}}^{\text{-}}\text{+}{\text{16H}}^{\text{+}}\to 2{\text{Mn}}^{\text{2+}}\text{+}8{\text{H}}_{\text{2}}\text{O}\\ {\text{10F}}^{-}\to 5{\text{F}}_{\text{2}}\text{+}10{\text{e}}^{\text{-}}\end{array}$

Adding both the reaction we get the balanced equation:

  ${\text{2MnO}}_{\text{4}}{}^{\text{-}}\text{+}10{\text{F}}^{\text{-}}\text{+}{\text{16H}}^{\text{+}}\to 2{\text{Mn}}^{\text{2+}}\text{+}5{\text{F}}_2+8{\text{H}}_{\text{2}}\text{O}$

Now calculate cell potential using values from Table 11.1 as,

  $\begin{array}{c}{E}^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}\\ =1.51-(2.87)\\ =-1.36\text{V}\end{array}$

As $E^0{}_{cell}$ is negative, hence the reaction is non-spontaneous.

(c)

Interpretation Introduction

Interpretation: The ${\epsilon }^o$ value for the following reaction needs to be calculated and if the reaction is spontaneous or not needs to be determined.

  ${\text{H}}_{\text{2}}(g)\rightleftharpoons {\text{H}}^{\text{+}}(aq)+{\text{H}}^{-}(aq)$

Concept Introduction :: Spontaneous reaction are those which is a spontaneous process under given conditions without any intervention. Such reactions are accompanied by certain distortion or change in entropy.

Cell potential under standard conditions is calculated as follows:

  $E^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}$

The reaction is spontaneous if the value of $E^0{}_{cell}$ is positive and the reaction is non-spontaneous if value of $E^0{}_{cell}$ is negative.

Answer to Problem 20E

  $E^0{}_{cell}$ of the reaction = -2.23 V

As $E^0{}_{cell}$ is negative, hence the reaction is non-spontaneous.

Explanation of Solution

The given reaction with oxidation state is:

  ${\stackrel{0}{\text{H}}}_{\text{2}}(g)\rightleftharpoons {\text{H}}^{\text{+}}(aq)+{\text{H}}^{-}(aq)$

From the reaction is it clear that H undergo reduction as well as oxidation.

The reaction is already balanced.

Now calculate cell potential using values from Table 11.1 as,

  $\begin{array}{c}{E}^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}\\ =-2.23-(0.00)\\ =-2.23\text{V}\end{array}$

As $E^0{}_{cell}$ is negative, hence the reaction is non-spontaneous.

(d)

Interpretation Introduction

Interpretation: The ${\epsilon }^o$ value for the following reaction needs to be calculated and if the reaction is spontaneous or not needs to be determined.

  ${\text{Au}}^{\text{+3}}(aq)+\text{Ag}(s)\rightleftharpoons {\text{Ag}}^{\text{+}}(aq)+\text{Au}(s)$

Concept Introduction :: Spontaneous reaction are those which is a spontaneous process under given conditions without any intervention. Such reactions are accompanied by certain distortion or change in entropy.

Cell potential under standard conditions is calculated as follows:

  $E^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}$

The reaction is spontaneous if the value of $E^0{}_{cell}$ is positive and the reaction is non spontaneous if value of $E^0{}_{cell}$ is negative.

Answer to Problem 20E

  $E^0{}_{cell}$ of the reaction = 0.70 V

As $E^0{}_{cell}$ is positive, hence the reaction is spontaneous.

Explanation of Solution

The given reaction with oxidation state is:

  ${\text{Au}}^{\text{+3}}(aq)+\stackrel{0}{\text{Ag}}(s)\rightleftharpoons {\text{Ag}}^{\text{+}}(aq)+\stackrel{0}{\text{Au}}(s)$

From the reaction is it clear that Ag undergo oxidation and Au undergoes reduction.

Multiply oxidation half with 2 and reduction half with 5 to balance the electron gained and electron lost we get:

  $\begin{array}{l}3\stackrel{}{\text{Ag}}(s)\to 3{\text{Ag}}^{\text{+}}(aq)+3e^{=}\\ {\text{Au}}^{\text{3+}}(aq)+3e^{-}\to \text{Au}(s)\end{array}$

Adding both the reaction we get the balanced equation:

  ${\text{Au}}^{\text{+3}}(aq)+3\stackrel{}{\text{Ag}}(s)\rightleftharpoons 3{\text{Ag}}^{\text{+}}(aq)+\stackrel{}{\text{Au}}(s)$

Now calculate cell potential using values from Table 11.1 as,

As $E^0{}_{cell}$ is negative, hence the reaction is non-spontaneous.

  $\begin{array}{l}{E}^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}\\ =1.51-(0.80)\\ =0.70\text{V}\\ \end{array}$

As $E^0{}_{cell}$ is positive, hence the reaction is spontaneous.