Chapter 11, Problem 41E

(a)

Interpretation Introduction

Interpretation: The value of $\Delta G^0$ and $K$ for the galvanic reaction containing Br and Cl needs to be determined.

Concept Introduction : Cell potential is the difference in potential between its cathode and anode.

The overall cell potential can be calculated by using the equation,

. ${\text{E}}^{\text{0}}{}_{\text{Cell}}{\text{= E}}^{\text{0}}{}_{\text{cathode}}-{\text{ E}}^{\text{0}}{}_{\text{anode}}$

The K value for equilibrium can be calculated by the following relation:

  $\log \text{K}=\frac{n{\text{E}}^{\text{0}}}{0.0591}$

Here, n is the number of electrons involved in overall reaction.

  ${\text{E}}^{\text{0}}$ is the cell constant.

Answer to Problem 41E

The value of $\Delta G^0$ = -52.1 kJ and $\text{K}=1.37\times {10}^9$ for the galvanic reaction.

Explanation of Solution

The oxidation half-reaction at the anode is:

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

The reduction half-reaction at the cathode is:

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

Thus, the overall reaction is:

  $\begin{array}{l}2{\text{Br}}^{-}(aq)+{\text{Cl}}_{\text{2}}(g)\to {\text{Br}}_{\text{2}}(aq)+{\text{2Cl}}^{-}{E}^0=+0.27\text{V}\\ =0.27\text{J/C}\end{array}$

Now, calculate the $\Delta {\text{G}}^0$ value using the equation,

  $\begin{array}{c}\Delta G^0=-nF{E}^0\\ =-(2\text{mol)}\text{(96}\text{.485}\frac{\text{C}}{\text{mol}})\left(0.27\frac{\text{J}}{\text{C}}\right)\\ =-52.102\text{J}\\ \text{=}\text{-52}\text{.1 kJ}\end{array}$

Using Nernst equation we calculate K.

  $\begin{array}{c}\log \text{K}=\frac{n{\text{E}}^{\text{0}}}{0.0591}\\ =\frac{(2\text{mol)}\left(0.27\text{V}\right)}{0.0591}\\ =9.14\end{array}$

Or,

  $\begin{array}{c}\text{K=}{10}^{9.14}\\ =1.37\times {10}^9\end{array}$

(b)

Interpretation Introduction

Interpretation: The value of $\Delta G^0$ and $K$ for the galvanic reaction needs to be determined.

Concept Introduction : The standard free energy is proportional to the standard cell potential.

  $\Delta G^0=-nF{E}^0$

Here, nis the number of electrons involved in overall reaction.

  $E^0$ is the cell constant,

And F = Faraday’s constant

Answer to Problem 41E

The value of $\Delta G^0$ = -86.8 kJ and $\text{K}=1.7\times {10}^{15}$ for the galvanic reaction.

Explanation of Solution

The oxidation half-reaction at the anode is:

  $5{\text{IO}}^{-}\left(aq\right)+10{\text{H}}^{\text{+}}\left(aq\right)+10{\text{e}}^{-}\to 5{\text{IO}}_3{}^{-}\left(aq\right)+5{\text{H}}_{\text{2}}\text{O}\left(l\right)E^0=1.60\text{V}$

The reduction half-reaction at the anode is:

  $2{\text{Mn}}^{\text{2+}}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{MnO}}_{\text{4}}{}^{-}\left(aq\right)+16{\text{H}}^{\text{+}}\left(aq\right)+10e^{-}{E}^0=-1.51\text{V}$

The overall reaction and its cell potential is,

  $2{\text{Mn}}^{2+}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)+5{\text{IO}}_{\text{4}}{}^{-}\left(aq\right)+6{\text{H}}^{\text{+}}\left(aq\right)\to 2{\text{MnO}}_{\text{4}}{}^{-}\left(aq\right)+{\text{5IO}}_{\text{3}}{}^{-}\left(aq\right)E^0=+0.09\text{V}$

Now we calculate $\Delta G^0$

  $\begin{array}{l}\Delta G^0=-nF{E}^0\\ =-\left(10\text{mol}\right)\left(96.485\frac{\text{C}}{\text{mol}}\times 0.09\frac{\text{J}}{\text{C}}\right)\\ =-86.840\text{J}\\ \text{=}\text{-86}\text{.8}\text{kJ}\end{array}$

Now we calculate the value of $K$

  $\begin{array}{c}\log \text{K}=\frac{n{\text{E}}^0}{0.0591}\\ =\frac{\left(10\text{mol}\right)\left(0.09\text{V}\right)}{0.0591}\\ =15.2\end{array}$

And,

  $\begin{array}{l}\text{K=}{\text{10}}^{\text{15}\text{.2}}\\ \text{K}\text{=}\text{1}\text{.7}\text{×}{\text{10}}^{\text{15}}\end{array}$

(c)

Interpretation Introduction

Interpretation: The value of $\Delta G^0$ and $K$ for the galvanic reaction needs to be determined.

Concept Introduction : The overall cell potential can be calculated by using the equation,

. ${\text{E}}^{\text{0}}{}_{\text{Cell}}{\text{= E}}^{\text{0}}{}_{\text{reduced}}-{\text{ E}}^{\text{0}}{}_{\text{oxidised}}$

The standard free energy is proportional to the standard cell potential.

  $G^0=-nF{E}^0$

Here, n is the number of electrons involved in overall reaction.

  $E^0$ is the cell constant,

And F = Faraday’s constant

Under standard conditions, the equilibrium constant for the overall reaction can be found by using the Nernst equation at ${25}^0\text{C}$ .

  $\log \text{K}=\frac{n{\text{E}}^{\text{0}}}{0.0591}$

Here, n is the number of electrons involved in overall reaction.

  ${\text{E}}^{\text{0}}$ is the cell constant.

Answer to Problem 41E

The value of $\Delta G^0$ = -212.3 kJ and $\text{K}=1.7\times {10}^{27}$ for the galvanic reaction.

Explanation of Solution

The oxidation half-reaction at the anode is,

  ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+2{\text{H}}^{\text{+}}\left(aq\right)+2e^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)E^0=1.78\text{V}$

The reduction half-reaction at the cathode is,

  ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)\to {\text{O}}_2\left(g\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}{E}^0=-0.68\text{V}$

The overall reaction is,

  $\text{2}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)\to {\text{O}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)E^0=+1.1\text{V}$

Now calculate $\Delta G^0$

  $\begin{array}{c}\Delta G^0=-nF{E}^0\\ =-\left(2\text{mol}\right)\left(96.485\frac{\text{C}}{\text{mol}}\right)\left(1.144\frac{\text{J}}{\text{C}}\right)\\ =-212.270\text{J}\\ \text{=}\text{-212}\text{.3 kJ}\end{array}$

Now calculating K ,

  $\begin{array}{c}\log \text{K=}\frac{n{E}^0}{0.0591}\\ =\frac{\left(2\text{mol}\right)\left(1.1\text{V}\right)}{0.0591}\\ =37.225\end{array}$

  $\text{K=}\text{1}\text{.7}\times {\text{10}}^{27}$

(d)

Interpretation Introduction

Interpretation: The value of $\Delta G^0$ and $K$ for the galvanic reaction needs to be determined.

Concept Introduction:The Gibbs free energy helps in finding the maximum amount of work done in a thermodynamic system.

The standard free energy is proportional to the standard cell potential.

  $G^0=-nF{E}^0$

Here, n is the number of electrons involved in overall reaction.

  $E^0$ is the cell constant,

And F = Faraday’s constant

Answer to Problem 41E

The value of $\Delta G^0$ = -662 kJ and $\text{K}={10}^{116}$ for the galvanic reaction.

Explanation of Solution

The oxidation half-reaction at anode is,

  $3\text{Mn}\left(s\right)\to 3{\text{Mn}}^{2+}\left(aq\right)+6{\text{e}}^{-}{E}^0=1.18\text{V}$

The reduction half-reaction at cathode is,

  $2{\text{Fe}}^{3+}\left(aq\right)+6{\text{e}}^{-}\to 2\text{Fe}\left(s\right)E^0=-0.036\text{V}$

The overall reaction is,

  ${\text{2Fe}}^{3+}\left(aq\right)+3\text{Mn}\left(s\right)\to 3{\text{Mn}}^{2+}\left(aq\right)+2\text{Fe}\left(s\right)E^0=+\mathrm{1.1.44}\text{V}$

To calculate $\Delta G^0$ we use the following relation.

  $\begin{array}{c}\Delta G^0=-nF{E}^0\\ =-\left(6\text{mol}\right)\left(96.485\frac{\text{C}}{\text{mol}}\right)\left(1.1\frac{\text{J}}{\text{C}}\right)\\ =-662.270\text{J}\\ \text{= -662}\text{kJ}\end{array}$

To calculate K we use the following relation.

  $\begin{array}{c}\log \text{K=}\frac{nF{E}^0}{0.0591}\\ =\frac{\left(2\text{mol}\right)\left(1.144\text{V}\right)}{0.0591}\\ =116.14\end{array}$

  $\text{K}={10}^{116}$