Chapter 8, Problem 8.15E

Interpretation Introduction

Interpretation:

The reaction that can provide sufficient energy to do required work is to be identified.

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. Standard Gibbs free energy of a redox reaction is also represented as the maximum amount of work done by the system on the surrounding.

Answer to Problem 8.15E

Correct answer:

The correct option is (b) and (d).

Explanation of Solution

Reason for correct answer:

(b)

The given chemical equation (b) is represented as,

$\text{Fe}+{\text{Ag}}^{+}\to {\text{Fe}}^{3+}+\text{Ag}$

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

${\text{Fe}}^{3+}+3{\text{e}}^{-}\to \text{Fe }E°=-0.037\text{V}$

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

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,

$\text{Fe}\to {\text{Fe}}^{3+}+3{\text{e}}^{-}E°=0.037\text{V}$ …(1)

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

${\text{Ag}}^{+}+{\text{e}}^{-}\to \text{Ag }E°=0.7996\text{V}$ …(2)

The number of moles of electrons transfer in the above reaction is $1\text{mol}$.

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

$\Delta G°=-nFE°$ …(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 $\text{Fe}$, $F$ and number of moles of electrons transferred 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 (1) is $-10.7098\text{ kJ}$.

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

$\begin{array}{rll}\Delta G°& =& -\left(\text{1 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.7996\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(77149.409\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -77.1494\text{ kJ}\end{array}$

The value $\Delta G°$ for the reaction (2) is $-77.1494\text{ kJ}$.

The balanced overall electrochemical reaction is obtained when chemical equation (2) is multiplied by $3$ and is added in chemical equation (1). The formation of overall balanced chemical equation is represented as,

$\begin{array}{rll}\text{Fe}\to {\text{Fe}}^{3+}+3{\text{e}}^{-}\Delta G°& =& -10.7098\text{ kJ}\\ {\text{3Ag}}^{+}+3{\text{e}}^{-}\to 3\text{Ag 3}\times \Delta G°& =& \text{3}\times \left(-77.1494\text{ kJ}\right)\\ \text{Fe}+3{\text{Ag}}^{+}\to {\text{Fe}}^{3+}+3\text{Ag}\Delta G°& =& -242.158\text{kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction is $-242.158\text{kJ}$.

The maximum amount of work that can be done by the given chemical reaction (b) is $-242.158\text{kJ}$.

(d)

The given chemical equation (d) is represented as,

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

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,

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

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

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

${\text{Au}}^{+}+{\text{e}}^{-}\to \text{Au }E°=1.692\text{V}$ …(5)

The number of moles of electrons transfer in the above reaction is $1\text{mol}$.

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 $\text{Au}$, $F$ and $n$ in the equation (3).

$\begin{array}{rll}{\Delta }_{\text{rxn}}G°& =& -\left(\text{1 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(1.692\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(163242.62\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -163.24\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (5) is $-163.24\text{kJ}$.

The balanced overall electrochemical reaction is obtained when chemical equation (5) is multiplied by $2$ and then added in chemical equation (4). The formation of overall balanced chemical equation is represented as,

$\begin{array}{rll}\text{Zn}\to {\text{Zn}}^{2+}+2{\text{e}}^{-}\Delta G°& =& -147.004\text{kJ}\\ {\text{2Au}}^{+}+2{\text{e}}^{-}\to 2\text{Au 2}\times \Delta G°& =& \text{2}\times \left(-163.24\text{kJ}\right)\\ {\text{2Au}}^{+}+\text{Zn}\to 2\text{Au}+{\text{Zn}}^{2+}\Delta G°& =& -473.484\text{kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction (d) is $-473.484\text{kJ}$.

The required work done is $196\text{kJ}$.

Therefore, the given reaction (b) and (d) can provide enough energy to perform the process.

The correct options are (b) and (d).

Reason for incorrect options:

(a)

The given chemical equation (a) is represented as,

${\text{Cu}}^{2+}+{\text{H}}_2\to \text{Cu}+{\text{H}}^{+}$

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

${\text{Cu}}^{2+}+2{\text{e}}^{-}\to \text{Cu }E°=0.3419\text{V}$ …(6)

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

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

${\text{2H}}^{+}+2{\text{e}}^{-}\to {\text{H}}_{2}E°=0\text{V}$

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

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,

${\text{H}}_2\to {\text{2H}}^{+}+2{\text{e}}^{-}E°=0\text{V}$ …(7)

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

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.3419\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(65976.443\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -65.9764\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (6) is $-65.9764\text{kJ}$.

The standard electrical potential of the reaction (7) is $0.0\text{V}$. Therefore, the value $\Delta G°$ for the reaction (7) is $0\text{ kJ}$.

The balanced overall electrochemical reaction is obtained when chemical equation (6) is added in chemical equation (7). The formation of overall balanced chemical equation is represented as,

$\begin{array}{rll}{\text{Cu}}^{2+}+2{\text{e}}^{-}\to \text{Cu }\Delta G°& =& -65.9764\text{kJ}\\ {\text{H}}_2\to {\text{2H}}^{+}+2{\text{e}}^{-}\Delta G°& =& 0\text{kJ}\\ {\text{Cu}}^{2+}+{\text{H}}_2\to \text{Cu}+2{\text{H}}^{+}\Delta G°& =& -65.9764\text{kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction (a) is $-65.9764\text{kJ}$.

(c)

The given chemical equation (c) is represented as,

$\text{Co}+{\text{Ni}}^{2+}\to {\text{Co}}^{2+}+\text{Ni}$

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 from an oxidation half reaction. The oxidation half reaction is represented as,

$\text{Co}\to {\text{Co}}^{2+}+2{\text{e}}^{-}E°=+0.28\text{V}$ …(8)

The chemical equation (7) is reversed and the value of $E°$ is multiplied by $-1$, to from an oxidation half reaction. The oxidation half reaction is represented as,

${\text{Ni}}^{2+}{\text{+2e}}^{-}\to \text{Ni }E°=-0.257\text{V}$ …(9)

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

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

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.28\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(154376\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -154.376\text{ kJ}\end{array}$

The value $\Delta G°$ for the reaction (8) is $-154.376\text{ kJ}$.

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

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(-0.257\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& \left(49593.29\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& 49.5933\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (9) is $49.5933\text{kJ}$.

The balanced overall electrochemical reaction is obtained when chemical equation (8) is added in chemical equation (9). The formation of overall balanced chemical equation is represented as,

$\begin{array}{rll}\text{Co}\to {\text{Co}}^{2+}+2{\text{e}}^{-}\Delta G°& =& -154.376\text{ kJ}\\ {\text{Ni}}^{2+}{\text{+2e}}^{-}\to \text{Ni }\Delta G°& =& 49.5933\text{kJ}\\ \text{Co}+{\text{Ni}}^{2+}\to {\text{Co}}^{2+}+\text{Ni }\Delta G°& =& -104.7827\text{kJ}\end{array}$

The value $\Delta G°$ for the given reaction (c) is $-104.7827\text{kJ}$.

The energy produce by the reaction (a) and (b) is less than the required energy.

Therefore, the option (a) and (c) are incorrect.

Conclusion

The reaction (b) and reaction (d) provides sufficient amount of energy to do required work. Therefore, the option (b) and (d) are correct.