Chapter 18, Problem 18.30QP

(a)

Interpretation Introduction

Interpretation: The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the given reactions are to be calculated.

Concept introduction: The potential of a reduction half cell is called standard reduction potential if reactants and products are in their standard states at $\text{25}{}^{\text{ο}}\text{C}$.

To determine: The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the given reaction.

Answer to Problem 18.30QP

Solution

The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the given reaction are $\underset{\_}{-284.6kJ}$ and $\underset{\_}{2.71V}$ respectively.

Explanation of Solution

Explanation

The given reaction is,

$\text{2Na}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{2NaOH}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$ (1)

The standard change in free energy ( $\Delta G^{\text{ο}}$) is calculated by the formula,

$\begin{array}{c}\Delta G^{\text{ο}}=\Sigma \Delta G_{{}_{\text{Products}}}^{\text{ο}}-\Delta G_{{}_{\text{Reactants}}}^{\text{ο}}\\ =\left[2\times \Delta G_{{}_{\text{NaOH}\left(aq\right)}}^{\text{ο}}+\Delta G_{{}_{{\text{H}}_{\text{2}}\left(g\right)}}^{\text{ο}}\right]-\left[2\times \Delta G_{{}_{\text{Na}\left(s\right)}}^{\text{ο}}+2\times \Delta G_{{}_{{\text{H}}_{\text{2}}\text{O}\left(l\right)}}^{\text{ο}}\right]\end{array}$

Where,

• $\Delta G_{{}_{\text{NaOH}\left(aq\right)}}^{\text{ο}}$ is the standard free energy change for $\text{NaOH}$.
• $\Delta G_{{}_{{\text{H}}_{\text{2}}\left(g\right)}}^{\text{ο}}$ is the standard free energy change for ${\text{H}}_2$.
• $\Delta G_{{}_{\text{Na}\left(s\right)}}^{\text{ο}}$ is the standard free energy change for $\text{Na}$.
• $\Delta G_{{}_{{\text{H}}_{\text{2}}\text{O}\left(l\right)}}^{\text{ο}}$ is the standard free energy change for ${\text{H}}_{\text{2}}\text{O}$.

The value of $\Delta G_{{}_{{\text{H}}_{\text{2}}\left(g\right)}}^{\text{ο}}$ and $\Delta G_{{}_{\text{Na}\left(s\right)}}^{\text{ο}}$ is $0$.

The value of $\Delta G_{{}_{\text{NaOH}\left(aq\right)}}^{\text{ο}}$ is $-\text{419}\text{.2}\text{kJ/mol}$.

The value of $\Delta G_{{}_{{\text{H}}_{\text{2}}\text{O}\left(l\right)}}^{\text{ο}}$ is $-237.2\text{kJ/mol}$.

Substitute the value of $\Delta G_{{}_{{\text{H}}_{\text{2}}\left(g\right)}}^{\text{ο}}$, $\Delta G_{{}_{\text{Na}\left(s\right)}}^{\text{ο}}$, $\Delta G_{{}_{\text{NaOH}\left(aq\right)}}^{\text{ο}}$ and $\Delta G_{{}_{{\text{H}}_{\text{2}}\text{O}\left(l\right)}}^{\text{ο}}$ in the above equation.

$\begin{array}{c}\Delta G^{\text{ο}}=\left[2\times \left(-\text{419}\text{.2}\text{kJ/mol}\right)+0\right]-\left[2\times 0+2\times \left(-237.2\text{kJ/mol}\right)\right]\\ =\underset{\_}{-284.6kJ}\end{array}$

The oxidation half cell reaction at anode is,

$\text{Na}\to {\text{Na}}^{\text{+}}+{\text{e}}^{-}{E}_1{}^{\text{ο}}=-2.71\text{V}$ (2)

The reduction half cell reaction at cathode is,

${\text{2H}}^{\text{+}}+2{\text{e}}^{-}\to {\text{H}}_{\text{2}}{E}_2{}^{\text{ο}}=0.00\text{V}$ (3)

Where,

• $E_1{}^{\text{ο}}$ is the standard electrode potential of equation (2).
• $E_2{}^{\text{ο}}$ is the standard electrode potential of equation (3).

The standard electrode potential of cell ( $E_{\text{cell}}^{\text{ο}}$) is calculated by the formula,

$E_{\text{cell}}^{\text{ο}}=E_2{}^{\text{ο}}-E_1{}^{\text{ο}}$

Substitute the value of $E_1{}^{\text{ο}}$ and $E_2{}^{\text{ο}}$ in the above equation.

$\begin{array}{c}{E}_{\text{cell}}^{\text{ο}}=0.00\text{V}-\left(-2.71\text{V}\right)\\ =\underset{\_}{2.71V}\end{array}$

The value of $E_{\text{cell}}^{\text{ο}}$ is $\underset{\_}{2.71V}$.

(b)

Interpretation Introduction

To determine: The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the given reaction.

Answer to Problem 18.30QP

Solution

The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the given reaction are $\underset{\_}{-611.4kJ}$ and $\underset{\_}{3.16V}$ respectively.

Explanation of Solution

Explanation

The given reaction is,

$\text{2Pb}\left(s\right)+{\text{O}}_2\left(g\right)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{2PbSO}}_{\text{4}}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$ (3)

The standard change in free energy ( $\Delta G^{\text{ο}}$) is calculated by the formula,

$\begin{array}{c}\Delta G^{\text{ο}}=\Sigma \Delta G_{{}_{\text{Products}}}^{\text{ο}}-\Delta G_{{}_{\text{Reactants}}}^{\text{ο}}\\ =\left[2\times \Delta G_{{}_{{\text{PbSO}}_{\text{4}}\left(s\right)}}^{\text{ο}}+2\times \Delta G_{{}_{{\text{H}}_{\text{2}}\text{O}\left(l\right)}}^{\text{ο}}\right]-\left[2\times \Delta G_{{}_{\text{Pb}\left(s\right)}}^{\text{ο}}+1\times \Delta G_{{}_{{\text{O}}_2\left(g\right)}}^{\text{ο}}+2\times \Delta G_{{}_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)}}^{\text{ο}}\right]\end{array}$

Where,

• $\Delta G_{{}_{{\text{PbSO}}_{\text{4}}\left(s\right)}}^{\text{ο}}$ is the standard free energy change for ${\text{PbSO}}_{\text{4}}\left(s\right)$.
• $\Delta G_{{}_{{\text{H}}_{\text{2}}\text{O}\left(l\right)}}^{\text{ο}}$ is the standard free energy change for ${\text{H}}_{\text{2}}\text{O}$.
• $\Delta G_{{}_{\text{Pb}\left(s\right)}}^{\text{ο}}$ is the standard free energy change for $\text{Pb}\left(s\right)$.
• $\Delta G_{{}_{{\text{O}}_2\left(g\right)}}^{\text{ο}}$ is the standard free energy change for ${\text{O}}_2\left(g\right)$
• $\Delta G_{{}_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)}}^{\text{ο}}$ is the standard free energy change for ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)$.

The value of $\Delta G_{{}_{\text{Pb}\left(s\right)}}^{\text{ο}}$ and $\Delta G_{{}_{{\text{O}}_2\left(g\right)}}^{\text{ο}}$ is $0$.

The value of $\Delta G_{{}_{{\text{PbSO}}_{\text{4}}\left(s\right)}}^{\text{ο}}$ is $-813\text{kJ/mol}$.

The value of $\Delta G_{{}_{{\text{H}}_{\text{2}}\text{O}\left(l\right)}}^{\text{ο}}$ is $-237.2\text{kJ/mol}$.

The value of $\Delta G_{{}_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)}}^{\text{ο}}$ is $-744.5\text{kJ/mol}$.

Substitute the value of $\Delta G_{{}_{{\text{H}}_{\text{2}}\left(g\right)}}^{\text{ο}}$, $\Delta G_{{}_{\text{Na}\left(s\right)}}^{\text{ο}}$, $\Delta G_{{}_{\text{NaOH}\left(aq\right)}}^{\text{ο}}$ and $\Delta G_{{}_{{\text{H}}_{\text{2}}\text{O}\left(l\right)}}^{\text{ο}}$ in the above equation.

$\begin{array}{c}\Delta G^{\text{ο}}=\left[2\times \left(-813\text{kJ/mol}\right)+2\times \left(-237.2\text{kJ/mol}\right)\right]-\left[\left(2\times 0\right)+\left(1\times 0\right)+2\times \left(-744.5\text{kJ/mol}\right)\right]\\ =\underset{\_}{-611.4kJ}\end{array}$

The conversion of $\text{kJ}$ to $\text{J}$ is done as,

$\text{1}\text{kJ}=1000\text{J}$

Therefore, the conversion of $\text{-611}\text{.4}\text{kJ}$ to $\text{J}$ is done as,

$\begin{array}{c}\text{-611}\text{.4}\text{kJ}=\text{-611}\text{.4}\times 1000\text{J}\\ \text{=-611400}\text{J}\end{array}$

The oxidation half cell reaction at anode is,

$\text{Pb}+{\text{SO}}_4^{2-}\to {\text{PbSO}}_{\text{4}}+2{\text{e}}^{-}$

The reduction half cell reaction at cathode is,

${\text{PbO}}_{\text{2}}+{\text{SO}}_4^{2-}+{\text{4H}}^{+}+2{\text{e}}^{-}\to {\text{PbSO}}_{\text{4}}+2{\text{H}}_{\text{2}}\text{O}$

The $E_{\text{cell}}^{\text{ο}}$ is calculated by the formula,

$\Delta G^{\text{ο}}=-nF{E}_{\text{cell}}^{\text{ο}}$

Where,

• $n$ is the number of electrons.
• $F$ is the faraday constant ( $96500\text{J}{\text{V}}^{-1}$).

The value of $n$ for the reaction is $2$.

Substitute the value of $n$, $F$ and $\Delta G^{\text{ο}}$ in the above equation.

$\begin{array}{c}\text{-611400}\text{J}=-2\times 96500\text{J}{\text{V}}^{-1}\times E_{\text{cell}}^{\text{ο}}\\ E_{\text{cell}}^{\text{ο}}=\underset{\_}{3.16V}\end{array}$

The value of $E_{\text{cell}}^{\text{ο}}$ is $\underset{\_}{3.16V}$.

Conclusion

The values of $\Delta G^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ for the first reaction are $\underset{\_}{-284.6kJ}$ and $\underset{\_}{2.71V}$ respectively