Chapter 22, Problem 57AP

Interpretation Introduction

Interpretation:

The standard Gibbs free energy and the equilibrium constant for the given reaction are to be calculated from the standard reduction potential.

Concept introduction:

The change in free energy is called Gibbs free energy and is denoted as $\Delta G^o$.

If $\Delta G^o$ is negative, it indicates that the electrical work has been done by the system on the surroundings and if it is positive then it means the electrical work has been done on the system by the surroundings.

The standard cell potential can be calculated by the expression as: $E_{cell}^o=E_{Cathode}^o-E_{Anode}^o$

The standard Gibbs free energy change is calculated by the expression as:

$\Delta G^o=-n\text{F}{E}_{cell}^o$

Here, $n$ is the moles of electrons are transferred per mole of reaction, $\Delta G^o$ is the standard Gibbs free energy change, $E_{cell}^o$ is the standard cell potential, and $\text{F}$ is the Faraday’s constant.

The equilibrium constant is calculated by the expression as:

$E_{cell}^o=\frac{0.0592}{n}\log \left(\text{K}\right)$

The relation between free energy change and standard free energy change is as:

$\Delta G=\Delta G^{\text{o}}+\text{R}T\ln Q$

Here, $\Delta G$ is free energy change, $\Delta G^o$ is standard free energy change, $Q$ is the reaction quotient, $\text{R}$ is the gas constant, and $T$ is the temperature.

At equilibrium the above equation is reducedto the expression: $\Delta G^{\text{o}}\text{=}-\text{R }T\text{lnK }$

Answer to Problem 57AP

Solution: $\Delta G°= -1.8\times 10^{2}kJ/mol$ and $K_c = 6\times 10^{30}$

Explanation of Solution

The overall reaction is obtained by adding the two half reactions as follows: $\begin{array}{c}\text{anode}\left(oxidation\right)\text{: Zn }\left(s\right)\to {\text{ Zn}}^{\text{2+}}\left(aq\right)\text{ + 2}{e}^{-}\\ \text{cathode}\left(oxidation\right){\text{: 2[Cu}}^{\text{2+}}\left(aq\right)+e^{-}\to {\text{ Cu}}^{\text{+}}\left(aq\right)\text{] }\\ \text{ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ \text{ overall: Zn}\left(s\right){\text{ + 2Cu}}^{\text{2+}}\left(aq\right) \to {\text{ Zn}}^{\text{2+}}\left(aq\right){\text{ + 2Cu}}^{\text{+}}\left(aq\right)\end{array}$

The standard electrode potential is given as:

$\begin{array}{c}{E}_{\text{cell}}^{\text{o}}\text{=}{E}_{\text{cathode}}^{\text{o}}-E_{\text{anode}}^{\text{o}}\\ =E_{{\text{Zn}}^{\text{2+}}\text{/Zn}}^{\text{o}}-E_{{\text{Cu}}^{\text{2+}}\text{/Cu}}^{\text{o}}\\ =0.15\text{ V}-\text{(}-0.76\text{ V) }\\ \text{= 0}\text{.91 V }\end{array}$

The Gibbs free energy for the given reaction is calculated as follows:

$\Delta G° = -nFE°$

Substitute 2 for $n$, $\text{96500 J}/\text{V}\text{.mol}$ for $F$, and $\text{0}\text{.91 V}$ for $E°$ in the above equation:

$\begin{array}{l}\Delta G° = -\left(2\right)\left(\text{96500 J/V}\text{.mol}\right)\left(\text{0}\text{.91 V}\right)\\ \text{ =} -1.756\times {10}^5\text{J/mol }\end{array}$

Since in one joule, there are 0.001 kilojoules present.

Then,

$\begin{array}{c} -1.8\times {10}^5\text{J/mol =}\left(-1.8\times {10}^5\frac{\cancel{\text{J}}}{\text{mol}}\right)\left(\frac{1\text{ kJ}}{1000\cancel{\text{J}}}\right)\\ =-1.8\times {10}^2\text{ kJ/mol}\end{array}$

Hence, the Gibbs free energy is $-1.8\times {10}^2\text{ kJ/mol}$.

The equilibrium constant for the given reaction is as follows:

$\begin{array}{c}\Delta G^{\text{o}}=-RT\text{ ln }{K}_c\\ \text{ln }{K}_c=\frac{\Delta G^{\text{o}}}{-RT}\end{array}$

Now, substitute $-1.756\times {10}^5\text{J/mol }$ for $\Delta G^{\text{o}}$, $\text{8}\text{.314}\text{J/Kmol}$ for $R$, and $298\text{ K}$ for $T$ in the above equation as follows:

$\begin{array}{c}\text{ln }{K}_c=-\frac{\left(-\text{1}\text{.8}\text{×}{\text{10}}^{\text{5}}\text{J/mol}\right)}{\left(\text{8}\text{.314}\text{J/Kmol}\right)\left(\text{298}\text{K}\right)}\\ \text{ln }{K}_c=70.88\end{array}$

Take antilog on both sides as follows:

$\begin{array}{c}{K}_c = e^{70.88} \\ = 6\times 10^{30}\end{array}$

Conclusion

The standard Gibbs free energy and the equilibrium constant for the given reaction are $-1.8\times {10}^2\text{ kJ/mol}$ and $6\times 10^{30}$, respectively.