Chapter 19, Problem 68AP

Equations 18.10 and 19.3 to calculate the emf values of the Daniell cell at $\text{25ºC and 80ºC}$. Comment on your results. What assumptions are used in the derivation? (Hint: You need the thermodynamic data in Appendix 2.)

Interpretation Introduction

Interpretation:

The EMFvalues of the cell at the temperatures of ${25}^{\text{°}}\text{C}$ and ${80}^{\text{°}}\text{C}$ isto be calculated, and the assumption used in the derivation is to be explained.

Concept introduction:

$\Delta G$ is the actual free energy. It is a thermodynamic function and is also known as Gibb’s free energy.

Entropy $\left(\Delta S\right)$ is the randomness of the system, which refers tothe number of ways a substance can be arranged.

Enthalpy $\left(\Delta H\right)$ is equal to the sum of internal energy as well as the product of pressure and volume of the system.

Answer to Problem 68AP

Solution: The EMFof the cell at a temperature of ${25}^{\text{°}}\text{C}$ is $1.100\text{ V}$.

The EMFof the cell at a temperature of ${80}^{\text{°}}\text{C}$ is $1.095\text{ V}$.

Explanation of Solution

The standard free energy change as follows:

$\Delta G^{°}=-nF{E}^{\text{o}}$

Here, $\Delta G^{°}$ is the standard free energy change,$F$ is the Faradayconstant $\left(96500\text{ C/mol}{\text{.e}}^{-}\right)$, $E^{\text{o}}$ is the standard EMF,and $n$ is the number of electrons transferred.

Substitute the value of $\Delta G^{°}$ in theGibb’s free energy equation, as follows:

$\begin{array}{c}\Delta G^{\text{o}}=\Delta H^{\text{o}}-T\Delta S^{\text{o}}\\ -n\text{F}{E}^{\text{o}}=\Delta H^{\text{o}}-T\Delta S^{\text{o}}\end{array}$

$E^{\text{o}}=\frac{-\Delta H^{\text{o}}+T\Delta S^{\text{o}}}{n\text{F}}$…..(1)

Here, $\Delta G$ is Gibb’s free energy, $\Delta H$ is the enthalpy change, $T$ is the absolute temperature, and $\Delta S$ is the entropy change.

The considered reaction is:

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

The change in the enthalpy of the reaction is as follows:

$\begin{array}{l}\Delta H^{\text{o}}\text{=}{\displaystyle \sum \Delta H_{product}^{\circ }}-{\displaystyle \sum \Delta H_{reac\tan t}^{\circ }}\\ \Delta H^{\text{o}}=\Delta H_{\text{f}}^{\text{°}}\left(\text{Cu}\left(s\right)\right)+\Delta H_{\text{f}}^{\text{°}}\left({\text{Zn}}^{2+}\left(aq\right)\right)-[\Delta H_{\text{f}}^{\text{°}}\left(\text{Zn}\left(s\right)\right)+\Delta H_{\text{f}}^{\text{°}}\left({\text{Cu}}^{2+}\left(aq\right)\right)]\end{array}$

Here, $\Delta H_{\text{f}}^{°}$ is the standard enthalpy of formation.

The standard enthalpies of the formation of $\text{Cu(s)}$, ${\text{Zn}}^{2+}\left(aq\right)$, $\text{Zn}\left(s\right)$, and ${\text{Cu}}^{2+}\text{(aq)}$ are as follows:

$\begin{array}{c}\Delta H_f^{\circ }\text{[Cu(s)]}=\text{ 0}\\ \Delta H_f^{\circ }[{\text{Zn}}^{2+}\left(aq\right)]=-152.4kJ/mol\\ \Delta H_f^{\circ }[\text{Zn}\left(s\right)]=\text{ 0}\\ \Delta H_f^{\circ }[{\text{Cu}}^{2+}\text{(aq)}]=\text{ 64}\text{.39 }kJ/mol\end{array}$

Substitute the values in the above equation:

$\begin{array}{l}\Delta H^{\text{o}}=0+\left(-152.4\text{ kJ/mol}\right)-0+\left(\text{ 64}\text{.39kJ/mol}\right)\\ \Delta H^{\text{o}}=-216.8\text{ kJ/mol}\end{array}$

Thus, the standard enthalpy of formation is $-216.8\text{ kJ/mol}$.

The change in the entropy of the reaction is as follows:

$\begin{array}{l}\Delta S^{\text{o}}\text{=}{\displaystyle \sum \Delta S_{product}^{\circ }}-{\displaystyle \sum \Delta S_{reac\tan t}^{\circ }}\\ \Delta S^{\text{o}}=S^{\text{o}}\left(\text{Cu}\left(s\right)\right)+S^{\text{o}}\left({\text{Zn}}^{2+}\left(aq\right)\right)-[S^{\text{o}}\left(\text{Zn}\left(s\right)\right)+S^{\text{o}}\left({\text{Cu}}^{2+}\left(aq\right)\right)]\end{array}$

Here, $\Delta S_{\text{f}}^{°}$ is the standard entropy of formation.

The standard entropies of the formation of $\text{Cu(s)}$, ${\text{Zn}}^{2+}\left(aq\right)$, $\text{Zn}\left(s\right)$, and ${\text{Cu}}^{2+}\text{(aq)}$ are as follows:

$\begin{array}{c}\Delta S^{\text{o}}\text{[Cu(s)]}=\text{ 33}\text{.3 J/K}\text{.mol}\\ \Delta S^{\text{o}}[{\text{Zn}}^{2+}\left(aq\right)]=-106.48\text{ J/K}\text{.mol}\\ \Delta S^{\text{o}}[\text{Zn}\left(s\right)]=\text{ 41}\text{.6 J/K}\text{.mol}\\ \Delta S^{\text{o}}[{\text{Cu}}^{2+}\text{(aq)}]=-99.6\text{ J/K}\text{.mol}\end{array}$

Substitute the values in the above equation:

$\begin{array}{l}\Delta S^{\text{o}}=33.3\text{ J/K}\text{.mol}+\left(-106.48\text{ J/K}\text{.mol}\right)-\left(41.6\text{ J/K}\text{.mol }-99.6\text{ J/K}\text{.mol}\right)\\ \Delta S^{\text{o}}=-15.2\text{ J/K}\text{.mol}\end{array}$

Thus, the standard entropy of formation is $-15.2\text{ J/K}\text{.mol}$.

Substitute the values of the standard entropy of formation and the standard enthalpy of formation in equation $\left(1\right)$.

At a temperature of ${25}^{\text{°}}\text{C}$$\left(298\text{ K}\right)$,

$\begin{array}{l}{E}^{\text{o}}=\frac{-\left(\frac{-216.8\cancel{\text{kJ}}}{1\cancel{\text{mol}}}\times \frac{1000\cancel{\text{J}}}{1\cancel{\text{kJ}}}\right)+\left(298\cancel{\text{K}}\right)\left(-15.2\cancel{\text{J}}\text{/}\cancel{\text{K}}\text{.}\cancel{\text{mol}}\right)}{\left(2\right)\left(96500\cancel{\text{J}}\text{/V}\text{.}\cancel{\text{mol}}\right)}\\ E^{\text{o}}=1.100\text{ V}\end{array}$

Therefore, the EMFof the cell ata temperature of ${25}^{\text{°}}\text{C}$ is $1.100\text{ V}$.

Ata temperature of ${80}^{\text{°}}\text{C}$$\left(\text{353 K}\right)$,

Substitute the value of the standard entropy of formation and the standard enthalpy of formation in equation $\left(1\right)$.

$\begin{array}{l}{E}^{\text{o}}=\frac{-\left(\frac{-216.8\cancel{\text{kJ}}}{1\cancel{\text{mol}}}\times \frac{1000\cancel{\text{J}}}{1\cancel{\text{kJ}}}\right)+\left(\text{353 }\cancel{\text{K}}\right)\left(-15.2\cancel{\text{J}}\text{/}\cancel{\text{K}}\text{.}\cancel{\text{mol}}\right)}{\left(2\right)\left(96500\cancel{\text{J}}\text{/V}\text{.}\cancel{\text{mol}}\right)}\\ E^{\text{o}}=1.095\text{ V}\end{array}$

Therefore, the EMFof the cell ata temperature of ${80}^{\text{o}}\text{C}$ is $1.095\text{ V}$.

In this calculation, the value of theEMFof the cell does not depend on the temperature, but, in practice, the value of theEMFof the cell does decrease with temperature. This is because the assumption used in the derivation is that $\Delta S^{\text{o}}$ and $\Delta H^{\text{o}}$ are temperature independent, which is not correct.

Conclusion

The EMF valuesof the cell at the temperatures of ${25}^{\text{°}}\text{C}$ and ${80}^{\text{°}}\text{C}$ is $1.100\text{ V}$ and $1.095\text{ V}$, respectively, and the assumption used in the derivation is that $\Delta S^{\text{o}}$ and $\Delta H^{\text{o}}$ are temperature independent,which is not correct.