Chapter 8, Problem 8.10E

For each of the following reactions, determine the overall balanced electrochemical reaction, its standard electric potential, and the standard Gibbs energy of the reaction.

(a) $\text{Co}+{\text{F}}_2\to {\text{Co}}^{2+}+2{\text{F}}^{-}$

(b) $\text{Zn}+{\text{Fe}}^{2+}\to {\text{Zn}}^{2+}+\text{Fe}$

(c) $\text{Zn}+{\text{Fe}}^{3+}\to {\text{Zn}}^{2+}+\text{Fe}$

(d) ${\text{Hg}}^{2+}+\text{Hg}\to {\text{Hg}}_2{}^{2+}$

Interpretation Introduction

(a)

Interpretation:

The overall balanced electrochemical reaction and the values of $E°$ and $\Delta G°$ for the given reaction are to be determined.

Concept introduction:

Standard Gibbs free energy of a reaction is used check whether the reaction is spontaneous or not. If the value of $\Delta G°$ is positive, then the reaction is non spontaneous. If the value of $\Delta G°$ is negative, then the reaction is spontaneous.

Answer to Problem 8.10E

The overall balanced electrochemical reaction is as follows,

$\text{Co}+{\text{F}}_2\to {\text{Co}}^{2+}+2{\text{F}}^{-}$

The value of $E°$ and $\Delta G°$ for the given reaction is $3.145\text{V}$ and $-607.08\text{kJ}$ respectively.

Explanation of Solution

The given reaction is represented as,

$\text{Co}+{\text{F}}_2\to {\text{Co}}^{2+}+2{\text{F}}^{-}$

From Table $8.2$, the reduction half reaction of ${\text{F}}_{\text{2}}$ and the standard reduction potential of ${\text{F}}_{\text{2}}$ is represented as,

${\text{F}}_{\text{2}}+2{\text{e}}^{-}\to 2{\text{F}}^{-}E°=2.866\text{V}\mathrm{...}\text{(1)}$

The number of moles of electrons transferred in the above reaction is $2\text{mol}$.

The number of moles of electrons transferred in both reactions is equal. Therefore, the given reaction is a balanced electrochemical reaction.

From Table $8.2$, the reduction half reaction of ${\text{Co}}^{2+}$ and the standard reduction potential of ${\text{Co}}^{2+}$ is represented as,

${\text{Co}}^{2+}+2{\text{e}}^{-}\to \text{Co }E°=-0.28\text{V}$

The above equation is reversed and the value of $E°$ is multiplied by $-1$ to form an oxidation half reaction. The oxidation half reaction is represented as,

$\text{Co}\to {\text{Co}}^{2+}+2{\text{e}}^{-}E°=+0.28\text{V}\mathrm{...}\text{(2)}$

The number of moles of electrons transferred in the above reaction is $2\text{mol}$.

The relation between standard Gibbs free energy and standard electrical potential is represented as,

$\Delta G°=-nFE°\mathrm{...}(3)$

Where,

• $\Delta G°$ represents the standard Gibbs free energy of the reaction.

• $n$ represents the number of moles.

• $F$ represents the Faraday’s constant with value $96,485\text{C}/\text{mol}$.

• $E°$ represents the standard electrical potential.

Substitute the values of the standard reduction potential of ${\text{F}}_2$, $F$ and $n$ in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(2.866\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(553052.02\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -553.052\text{ kJ}\end{array}$

The value $\Delta G°$ for the reaction (1) is $-553.052\text{ kJ}$.

Substitute the values of the standard oxidation potential of $\text{Co}$, $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.28\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(54031.6\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -54.0316\text{ kJ}\end{array}$

The value $\Delta G°$ for the reaction (2) is $-54.0316\text{ kJ}$.

The value of $\Delta G°$ of overall reaction is calculated by adding the $\Delta G°$ of two half reactions. The standard Gibbs free energy of the given reaction is calculated as,

$\Delta G°=\Delta G{°}_1+\Delta G{°}_2$

Where,

• $\Delta G{°}_1$ represents the standard Gibbs free energy of the reaction (1).

• $\Delta G{°}_2$ represents the standard Gibbs free energy of the reaction (2).

Substitute the value of $\Delta G{°}_1$ and $\Delta G{°}_2$ in the above equation.

$\begin{array}{rll}\Delta G°& =& -553.052\text{ kJ}+\left(-54.0316\text{ kJ}\right)\\ & =& -607.08\text{kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction is $-607.08\text{kJ}$.

The number of electrons transferred in the overall reaction is $2\text{ mol}$.

Rearrange the equation (3) for the value of $E°$.

$E°=\frac{-\Delta G°}{nF}$

Substitute the values of $\Delta G°$, $F$ and number of moles of electrons transfer in the equation (3).

$\begin{array}{rll}E°& =& \frac{-\left(-607.08\text{kJ}\right)\left(\frac{1000\text{ J}}{1\text{ kJ}}\right)}{\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)}\\ & =& 3.145\text{V}\end{array}$

The value of $E°$ for the given reaction is $3.145\text{V}$.

Conclusion

The overall balanced electrochemical reaction is as follows,

$\text{Co}+{\text{F}}_2\to {\text{Co}}^{2+}+2{\text{F}}^{-}$

The value of $E°$ and $\Delta G°$ for the given reaction is $3.145\text{V}$ and $-607.08\text{kJ}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The overall balanced electrochemical reaction and the values of $E°$ and $\Delta G°$ for the given reaction are to be determined.

Concept introduction:

Standard Gibbs free energy of a reaction is used check whether the reaction is spontaneous or not. If the value of $\Delta G°$ is positive, then the reaction is non spontaneous. If the value of $\Delta G°$ is negative, then the reaction is spontaneous.

Answer to Problem 8.10E

The overall balanced electrochemical reaction is as follows,

$\text{Zn}+{\text{Fe}}^{2+}\to {\text{Zn}}^{2+}+\text{Fe}$

The value of $E°$ and $\Delta G°$ for the given reaction is $0.315\text{V}$ and $-60.746\text{kJ}$ respectively.

Explanation of Solution

The given reaction is represented as,

$\text{Zn}+{\text{Fe}}^{2+}\to {\text{Zn}}^{2+}+\text{Fe}$

From Table $8.2$, the reduction half reaction of ${\text{Zn}}^{2+}$ and the standard reduction potential of ${\text{Zn}}^{2+}$ is represented as,

${\text{Zn}}^{2+}+2{\text{e}}^{-}\to \text{Zn }E°=-0.7618\text{V}$

The above equation is reversed and the value of $E°$ is multiplied by $-1$, to from an oxidation half reaction. The oxidation half reaction is represented as,

$\text{Zn}\to {\text{Zn}}^{2+}+2{\text{e}}^{-}E°=0.7618\text{V }\mathrm{...}\text{(4)}$

The number of moles of electrons transferred in the above reaction is $2\text{mol}$.

From Table $8.2$, the reduction half reaction of ${\text{Fe}}^{2+}$ and the standard reduction potential of ${\text{Fe}}^{2+}$ is represented as,

${\text{Fe}}^{2+}+2{\text{e}}^{-}\to \text{Fe }E°=-0.447\text{V }\mathrm{...}\text{(5)}$

The number of moles of electrons transferred in the above reaction is $2\text{mol}$.

The number of moles of electrons transferred in both reactions is equal. Therefore, the given reaction is balanced electrochemical reaction.

The relation between standard Gibbs free energy and standard electrical potential is represented as,

$\Delta G°=-nFE°\mathrm{...}(3)$

Where,

• $\Delta G°$ represents the standard Gibbs free energy of the reaction.

• $n$ represents the number of moles.

• $F$ represents the Faraday’s constant with value $96,485\text{C}/\text{mol}$.

• $E°$ represents the standard electrical potential.

Substitute the values of the standard oxidation potential of $\text{Zn}$, $F$ and number of moles of electrons transfer in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.7618\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(147004.546\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -147.004\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (4) is $-147.004\text{kJ}$.

Substitute the values of the standard reduction potential of ${\text{Fe}}^{2+}$, $F$ and number of moles of electrons transfer in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(-0.447\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& \left(86257.59\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& 86.258\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (5) is $86.258\text{kJ}$.

The value of $\Delta G°$ of overall reaction is calculated by adding the $\Delta G°$ of two half reactions. The standard Gibbs free energy of the given reaction is calculated as,

$\Delta G°=\Delta G{°}_1+\Delta G{°}_2$

Where,

• $\Delta G{°}_1$ represents the standard Gibbs free energy of the reaction (4).

• $\Delta G{°}_1$ represents the standard Gibbs free energy of the reaction (5).

Substitute the value of ${\Delta }_{\text{rxn}}G{°}_1$ and ${\Delta }_{\text{rxn}}G{°}_1$ in the above equation.

$\begin{array}{rll}\Delta G°& =& -147.004\text{kJ}+86.258\text{kJ}\\ & =& -60.746\text{kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction is $-60.746\text{kJ}$.

The number of electrons transfer in the overall reaction is $2\text{ mol}$.

Rearrange the equation (3) for the value of $E°$.

$E°=\frac{-\Delta G°}{nF}$

Substitute the values of $\Delta G°$, $F$ and number of moles of electrons transfer in the equation (3).

$\begin{array}{rll}E°& =& \frac{-\left(-60.746\text{kJ}\right)\left(\frac{1000\text{ J}}{1\text{ kJ}}\right)}{\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)}\\ & =& 0.315\text{V}\end{array}$

The value of $E°$ for the given reaction is $0.315\text{V}$.

Conclusion

The overall balanced electrochemical reaction is as follows,

$\text{Zn}+{\text{Fe}}^{2+}\to {\text{Zn}}^{2+}+\text{Fe}$

The value of $E°$ and $\Delta G°$ for the given reaction is $0.315\text{V}$ and $-60.746\text{kJ}$ respectively.

Interpretation Introduction

(c)

Interpretation:

The overall balanced electrochemical reaction and the values of $E°$ and $\Delta G°$ for the given reaction are to be determined.

Concept introduction:

Standard Gibbs free energy of a reaction is used check whether the reaction is spontaneous or not. If the value of $\Delta G°$ is positive, then the reaction is non spontaneous. If the value of $\Delta G°$ is negative, then the reaction is spontaneous.

Answer to Problem 8.10E

The overall balanced electrochemical reaction is as follows,

$\text{3Zn}+2{\text{Fe}}^{3+}\to 3{\text{Zn}}^{2+}+2\text{Fe}$

The value of $E°$ and $\Delta G°$ for the given reaction is $0.7248\text{V}$ and $-419.592\text{ kJ}$ respectively.

Explanation of Solution

The given reaction is represented as,

$\text{Zn}+{\text{Fe}}^{3+}\to {\text{Zn}}^{2+}+\text{Fe}$

From Table $8.2$, the reduction half reaction of ${\text{Zn}}^{2+}$ and the standard reduction potential of ${\text{Zn}}^{2+}$ is represented as,

${\text{Zn}}^{2+}+2{\text{e}}^{-}\to \text{Zn }E°=-0.7618\text{V}$

The above equation is reversed and the value of $E°$ is multiplied by $-1$, to from an oxidation half reaction. The oxidation half reaction is represented as,

$\text{Zn}\to {\text{Zn}}^{2+}+2{\text{e}}^{-}E°=0.7618\text{V }\mathrm{...}\text{(6)}$

The number of moles of electrons transferred in the above reaction is $2\text{mol}$.

From Table $8.2$, the reduction half reaction of ${\text{Fe}}^{2+}$ and the standard reduction potential of ${\text{Fe}}^{2+}$ is represented as,

${\text{Fe}}^{3+}+3{\text{e}}^{-}\to \text{Fe }E°=-0.037\text{V }\mathrm{...}\text{(7)}$

The number of moles of electrons transferred in the above reaction is $3\text{mol}$.

The relation between standard Gibbs free energy and standard electrical potential is represented as,

$\Delta G°=-nFE°\mathrm{...}(3)$

Where,

• $\Delta G°$ represents the standard Gibbs free energy of the reaction.

• $n$ represents the number of moles.

• $F$ represents the Faraday’s constant with value $96,485\text{C}/\text{mol}$.

• $E°$ represents the standard electrical potential.

Substitute the values of the standard oxidation potential of $\text{Zn}$, $F$ and number of moles of electrons transfer in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.7618\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(147004.546\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -147.004\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (6) is $-147.004\text{kJ}$.

Substitute the values of the standard reduction potential of ${\text{Fe}}^{3+}$, $F$ and number of moles of electrons transfer in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{3 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(-0.037\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& \left(10709.835\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& 10.7098\text{ kJ}\end{array}$

The value $\Delta G°$ for the reaction (7) is $10.7098\text{ kJ}$.

The balanced overall electron chemical reaction is obtained by multiplying chemical equation (7) by $3$ and chemical equation (8) by $2$ and then adding these equations. The formation of overall balanced chemical equation is represented as,

$\begin{array}{l}\text{3Zn}\to 3{\text{Zn}}^{2+}+6{\text{e}}^{-}\text{ 3}\times \Delta G°=\text{3}\times \left(-147.004\text{kJ}\right)\\ {\text{2Fe}}^{3+}+6{\text{e}}^{-}\to 2\text{Fe 2}\times \Delta G°=2\times 10.7098\text{ kJ}\\ \text{3Zn}+2{\text{Fe}}^{3+}\to 3{\text{Zn}}^{2+}+2\text{Fe}\Delta G°=-419.592\text{ kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction is $-419.592\text{ kJ}$.

The number of electrons transferred in the overall reaction is $6\text{ mol}$.

Rearrange the equation (3) for the value of $E°$.

$E°=\frac{-\Delta G°}{nF}$

Substitute the values of $\Delta G°$, $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}E°& =& \frac{-\left(-419.592\text{ kJ}\right)\left(\frac{1000\text{ J}}{1\text{ kJ}}\right)}{\left(\text{6 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)}\\ & =& 0.7248\text{V}\end{array}$

The value of $E°$ for the given reaction is $0.7248\text{V}$.

Conclusion

The overall balanced electrochemical reaction is as follows,

$\text{3Zn}+2{\text{Fe}}^{3+}\to 3{\text{Zn}}^{2+}+2\text{Fe}$

The value of $E°$ and $\Delta G°$ for the given reaction is $0.7248\text{V}$ and $-419.592\text{ kJ}$ respectively.

Interpretation Introduction

(d)

Interpretation:

The overall balanced electrochemical reaction and the values of $E°$ and $\Delta G°$ for the given reaction are to be determined.

Concept introduction:

Standard Gibbs free energy of a reaction is used check whether the reaction is spontaneous or not. If the value of $\Delta G°$ is positive, then the reaction is non spontaneous. If the value of $\Delta G°$ is negative, then the reaction is spontaneous.

Answer to Problem 8.10E

The overall balanced electrochemical reaction is as follows,

${\text{Hg}}^{2+}+\text{Hg}\to {\text{Hg}}_2{}^{2+}$

The value of $E°$ and $\Delta G°$ for the given reaction is $1.771\text{V}$ and $-341.749\text{kJ}$ respectively.

Explanation of Solution

The given reaction is represented as,

${\text{Hg}}^{2+}+\text{Hg}\to {\text{Hg}}_2{}^{2+}$

From Table $8.2$, the reduction half reaction of ${\text{Hg}}^{2+}$ into ${\text{Hg}}_2{}^{2+}$ and the standard reduction potential of ${\text{Hg}}^{2+}$ into ${\text{Hg}}_2{}^{2+}$ is represented as,

${\text{2Hg}}^{2+}+2{\text{e}}^{-}\to {\text{Hg}}_2{}^{2+}E°=0.920\text{V }\mathrm{...}\text{(8)}$

The number of moles of electrons transfer in the above reaction is $2\text{mol}$.

From Table $8.2$, the reduction half reaction of ${\text{Hg}}^{2+}$ into $\text{Hg}$ and the standard reduction potential of ${\text{Hg}}^{2+}$ into $\text{Hg}$ is represented as,

${\text{Hg}}^{2+}+2{\text{e}}^{-}\to \text{Hg }E°=0.851\text{V }\mathrm{...}\text{(9)}$

The number of moles of electrons transferred in the above reaction is $2\text{mol}$.

The number of moles of electrons transferred in both reactions is equal. Therefore, the given reaction is balanced electrochemical reaction.

The relation between standard Gibbs free energy and standard electrical potential is represented as,

$\Delta G°=-nFE°\mathrm{...}(3)$

Where,

• $\Delta G°$ represents the standard Gibbs free energy of the reaction.

• $n$ represents the number of moles.

• $F$ represents the Faraday’s constant with value $96,485\text{C}/\text{mol}$.

• $E°$ represents the standard electrical potential.

Substitute the values of the standard reduction potential of ${\text{Hg}}^{2+}$, $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.920\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(177532.4\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -177.532\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (8) is $-177.532\text{kJ}$.

Substitute the values of the standard reduction potential of ${\text{Hg}}_2{}^{2+}$, $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.851\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(164217.47\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -164.217\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (9) is $-164.217\text{kJ}$.

The value of $\Delta G°$ of overall reaction is calculated by subtracting the $\Delta G°$ of two half reactions. The standard Gibbs free energy of the given reaction is calculated as,

$\Delta G°=\Delta G{°}_1-\Delta G{°}_2$

Where,

• $\Delta G{°}_1$ represents the standard Gibbs free energy of the reaction (4).

• $\Delta G{°}_1$ represents the standard Gibbs free energy of the reaction (5).

Substitute the value of $\Delta G{°}_1$ and $\Delta G{°}_1$ in the above equation.

$\begin{array}{rll}\Delta G°& =& -177.532\text{kJ+}\left(-164.217\text{kJ}\right)\\ & =& -341.749\text{kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction is $-341.749\text{kJ}$.

The number of electrons transferred in the overall reaction is $2\text{ mol}$.

Rearrange the equation (3) for the value of $E°$.

$E°=\frac{-\Delta G°}{nF}$

Substitute the values of $\Delta G°$, $F$ and number of moles of electrons transfer in the equation (3).

$\begin{array}{rll}E°& =& \frac{-\left(-341.749\text{kJ}\right)\left(\frac{1000\text{ J}}{1\text{ kJ}}\right)}{\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)}\\ & =& 1.771\text{V}\end{array}$

The value of $E°$ for the given reaction is $1.771\text{V}$.

Conclusion

The overall balanced electrochemical reaction is as follows,

${\text{Hg}}^{2+}+\text{Hg}\to {\text{Hg}}_2{}^{2+}$

The value of $E°$ and $\Delta G°$ for the given reaction is $1.771\text{V}$ and $-341.749\text{kJ}$ respectively.