Chapter 8, Problem 8.14E

Interpretation Introduction

Interpretation:

The reaction that can provide $5.00\times {10}^2\text{ kJ}$ of 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 surroundings.

Answer to Problem 8.14E

Correct answer:

The correct option is (b).

Explanation of Solution

Reason for correct answer:

(b)

The given chemical equation (b) is represented as,

$\text{Ca}\left(\text{s}\right)+{\text{H}}^{+}\to {\text{Ca}}^{2+}+{\text{H}}_2$

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

${\text{Ca}}^{2+}+2{\text{e}}^{-}\to \text{Ca }E°=-2.868\text{V}$

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,

$\text{Ca}\to {\text{Ca}}^{2+}+2{\text{e}}^{-}E°=2.868\text{V}$ …(1)

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

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

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

$\Delta G°=-nFE°$ …(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 $\text{Ca}$, $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(2.868\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(553437.96\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -553.438\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (1) is $-553.438\text{kJ}$.

The standard electrical potential of the reaction (2) is $0.0\text{V}$. Therefore, the value $\Delta G°$ for the reaction (2) is $0\text{ kJ}$.

The balanced overall electron chemical reaction is obtained when chemical equation (1) is added in chemical equation (2). The formation of overall balanced chemical equation is represented as,

$\begin{array}{rll}\text{Ca}\to {\text{Ca}}^{2+}+2{\text{e}}^{-}\Delta G°& =& -553.438\text{kJ}\\ {\text{2H}}^{+}+2{\text{e}}^{-}\to {\text{H}}_2\Delta G°& =& 0\text{kJ}\\ \text{Ca}\left(\text{s}\right)+2{\text{H}}^{+}\to {\text{Ca}}^{2+}+{\text{H}}_2\Delta G°& =& -553.438\text{kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction is $-553.438\text{kJ}$.

The maximum amount of work that can be done by the given chemical reaction (b) is $-553.438\text{kJ}$.

The required work done is $5.00\times {10}^2\text{ kJ}$.

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

The correct option is (b).

Reason for incorrect options:

(a)

The given chemical equation (a) is represented as,

$\text{Zn}\left(\text{s}\right)+{\text{Cu}}^{2+}\to {\text{Zn}}^{2+}+\text{Cu}\left(\text{s}\right)$

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 form 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 ${\text{Cu}}^{+}$ and the standard reduction potential of ${\text{Cu}}^{+}$ is represented as,

${\text{Cu}}^{2+}+2{\text{e}}^{-}\to \text{Cu }E°=0.3419\text{V}$ …(5)

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

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 $\text{Cu}$, $F$ 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 (5) is $-65.9764\text{kJ}$.

The balanced overall electron chemical reaction is obtained when chemical equation (4) is added in chemical equation (5). The formation of overall balanced chemical equation is represented as,

$\begin{array}{rll}\text{Zn}\to {\text{Zn}}^{2+}+2{\text{e}}^{-}\Delta G°& =& -147.004\text{kJ}\\ {\text{Cu}}^{2+}+2{\text{e}}^{-}\to \text{Cu }\Delta G°& =& -65.9764\text{kJ}\\ \text{Zn}\left(\text{s}\right)+{\text{Cu}}^{2+}\to {\text{Zn}}^{2+}+\text{Cu}\left(\text{s}\right)\Delta G°& =& -212.9804\text{kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction (a) is $-212.9804\text{kJ}$.

(c)

The given chemical equation (c) is represented as,

$\text{Li}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\to {\text{Li}}^{+}+{\text{H}}_{\text{2}}+\text{OH}$

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

${\text{Li}}^{+}+{\text{e}}^{-}\to \text{Li }E°=-3.04\text{V}$

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,

$\text{Li}\to {\text{Li}}^{+}+{\text{e}}^{-}E°=3.04\text{V}$ …(6)

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

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

${\text{2H}}_{\text{2}}\text{O}+2{\text{e}}^{-}\to {\text{H}}_{\text{2}}{\text{+2OH}}^{-}E°=-0.8277\text{V}$ …(7)

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

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

$\begin{array}{rll}\Delta G°& =& -\left(\text{1 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(3.04\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(293314.4\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -293.3144\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (6) is $-293.3144\text{kJ}$.

Substitute the values of the standard reduction potential of ${\text{H}}_2\text{O}$, $F$ and number of moles of electrons transfer in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(-0.8277\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& \left(159721.269\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& 159.721\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (7) is $159.721\text{kJ}$.

The balanced overall electron chemical reaction is obtained when chemical equation (6) is multiplied by $2$ and then added in chemical equation (7). The formation of overall balanced chemical equation is represented as,

$\begin{array}{rll}\text{2Li}\to 2{\text{Li}}^{+}+2{\text{e}}^{-}\text{ 2}\times \Delta G°& =& \text{2}\times \left(-293.3144\text{kJ}\right)\\ {\text{2H}}_{\text{2}}\text{O}+2{\text{e}}^{-}\to {\text{H}}_{\text{2}}{\text{+2OH}}^{-}\Delta G°& =& 159.721\text{kJ}\\ \text{2Li}\left(\text{s}\right)+2{\text{H}}_{\text{2}}\text{O}\to 2{\text{Li}}^{+}+2{\text{OH}}^{-}+{\text{H}}_{\text{2}}\Delta G°& =& -426.9078\text{kJ}\end{array}$

Therefore, the value $\Delta G°$ for the given reaction (c) is $-426.9078\text{kJ}$.

(d)

The given chemical equation (d) is represented as,

${\text{H}}_2+{\text{OH}}^{-}+{\text{Hg}}_2{\text{Cl}}_2\to {\text{H}}_2\text{O}+\text{Hg}+{\text{Cl}}^{-}$

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

${\text{Hg}}_2{\text{Cl}}_2+2e^{-}\to 2\text{Hg}+{\text{2Cl}}^{-}E°=0.2682\text{V}$ …(8)

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

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,

${\text{H}}_{\text{2}}{\text{+2OH}}^{-}\to {\text{2H}}_{\text{2}}\text{O}+2{\text{e}}^{-}E°=0.8277\text{V}$ …(9)

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

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

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.2682\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(51754.554\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -51.7546\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (8) is $-51.7546\text{kJ}$.

Substitute the values of the standard oxidation potential of ${\text{H}}_2\text{O}$, $F$ and number of moles of electrons transferred in the equation (3).

$\begin{array}{rll}\Delta G°& =& -\left(\text{2 mol}\right)\left(96,485\text{C}/\text{mol}\right)\left(0.8277\text{V}\right)\left(\frac{1\text{J}/\text{C}}{1\text{ V}}\right)\\ & =& -\left(159721.269\text{J}\right)\left(\frac{1\text{ kJ}}{1000\text{ J}}\right)\\ & =& -159.721\text{kJ}\end{array}$

The value $\Delta G°$ for the reaction (7) is $159.721\text{kJ}$.

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

$\begin{array}{rll}{\text{Hg}}_2{\text{Cl}}_2+2e^{-}\to 2\text{Hg}+{\text{2Cl}}^{-}\Delta G°& =& -51.7546\text{kJ}\\ {\text{H}}_{\text{2}}{\text{+2OH}}^{-}\to {\text{2H}}_{\text{2}}\text{O}+2{\text{e}}^{-}\Delta G°& =& -159.721\text{kJ}\\ {\text{H}}_2+2{\text{OH}}^{-}+{\text{Hg}}_2{\text{Cl}}_2\to 2{\text{H}}_2\text{O}+2\text{Hg}+{\text{Cl}}^{-}\Delta G°& =& -211.4756\text{kJ}\end{array}$

The value $\Delta G°$ for the given reaction (d) is $-211.4756\text{kJ}$.

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

Conclusion

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