Chapter 25, Problem 25.55P

(a)

Interpretation Introduction

Interpretation:

The value of $\Delta G_{rxn}^o$ has to be calculated for the reaction $\text{2 Ag}\left(s\right){\text{ + F}}_{\text{2}}\left(aq\right)\underset{}{\rightleftharpoons }{\text{ 2 Ag}}^{\text{+}}\left(aq\right){\text{ + 2 F}}^{-}\left(aq\right)$.

Concept Introduction:

Standard potential ${\text{(E}}_{\text{cell}}^{\text{o}}\text{)}$ can be calculated by the following formula,

  ${\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{red}}^{\text{o}}\text{+}{\text{E}}_{\text{ox}}^{\text{o}}$

The relation between the standard cell voltage $\left(E_{cell}^o\right)$ and standard free energy $\left(\Delta G_{rxn}^o\right)$ is as follows,

  $\Delta G_{rxn}^o\text{ =}-{\nu }_{e}F{E}_{cell}^o$

Here, ${\nu }_e$ is the number of moles of electrons transferred in the two half reaction equations that gives the net reaction equation.

Answer to Problem 25.55P

The value of $\Delta G_{rxn}^o$ is $-\text{399 kJ}$.

Explanation of Solution

The given reaction is shown below,

  $\text{2 Ag}\left(s\right){\text{ + F}}_{\text{2}}\left(aq\right)\underset{}{\rightleftharpoons }{\text{ 2 Ag}}^{\text{+}}\left(aq\right){\text{ + 2 F}}^{-}\left(aq\right)$

Reduction half-cell is given below,

  ${\text{F}}_{\text{2}}\left(aq\right){\text{+ 2e}}^{-}\left(aq\right)\underset{}{\to }{\text{ 2 F}}^{-}\left(aq\right)E_{red}^o=\text{ 2}\text{.866 V}$

Oxidation half-cell is given below,

  $\begin{array}{l}\text{Ag}\left(s\right)\underset{}{\to }{\text{ Ag}}^{\text{+}}\left(aq\right)+{\text{ 1e}}^{-}\left(aq\right)\\ \\ {\text{E}}_{ox}^o\left[Ag|A{g}^{+}\right]\text{=}-{\text{E}}_{red}^o\left[A{g}^{+}|Ag\right]\text{=}-\text{0}\text{.7996 V}\end{array}$

Here, ${\nu }_e=2{\text{ mol e}}^{-}$

Calculate $E_{cell}^o$ for the following reaction as follows,

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{red}}^{\text{o}}\text{+}{\text{E}}_{\text{ox}}^{\text{o}}\\ \\ =\left(\text{2}\text{.866 V}\right)+\left(-\text{0}\text{.7996 V}\right)\\ \\ =\text{ 2}\text{.0664 V}\end{array}$

The value of $\Delta G_{rxn}^o$ is calculated as follows,

  $\begin{array}{c}\Delta G_{rxn}^o\text{ =}-{\nu }_{e}F{E}_{cell}^o\\ \\ =-2 mol e^{-}\times \text{96485 C/mol}\times \text{2}\text{.0664 V}\\ \\ \text{= }-\text{399 kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The value of $\Delta G_{rxn}^o$ has to be calculated for the reaction $\frac{1}{2}{\text{H}}_{\text{2}}\left(g\right){\text{ + Fe}}^{\text{3+}}\left(aq\right)\underset{}{\rightleftharpoons }{\text{ Fe}}^{\text{2+}}\left(aq\right){\text{ + H}}^{\text{+}}\left(aq\right)$.

Concept Introduction:

Refer to (a).

Answer to Problem 25.55P

The value of $\Delta G_{rxn}^o$ is $-\text{74}\text{.4 kJ}$.

Explanation of Solution

The given reaction is shown below,

  $\frac{1}{2}{\text{H}}_{\text{2}}\left(g\right){\text{ + Fe}}^{\text{3+}}\left(aq\right)\underset{}{\rightleftharpoons }{\text{ Fe}}^{\text{2+}}\left(aq\right){\text{ + H}}^{\text{+}}\left(aq\right)$

Reduction half-cell is given below,

  ${\text{Fe}}^{\text{3+}}\left(aq\right){\text{+ e}}^{-}\left(aq\right)\underset{}{\to }{\text{ Fe}}^{\text{2+}}\left(aq\right)E_{red}^o=\text{0}\text{.771 V}$

Oxidation half-cell is given below,

  $\begin{array}{l}\frac{1}{2}{\text{H}}_{\text{2}}\left(g\right)\underset{}{\to }{\text{ H}}^{\text{+}}\left(aq\right)+{\text{ 1e}}^{-}\left(aq\right)\\ \\ {\text{E}}_{ox}^o\left[H_2|H^{+}\right]\text{=}-{\text{E}}_{red}^o\left[H^{+}|H_2\right]\text{=}\text{0}\text{.00 V}\end{array}$

Here, ${\nu }_e=1{\text{ mol e}}^{-}$

Calculate $E_{cell}^o$ for the following reaction as follows,

  $\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{red}}^{\text{o}}\text{+}{\text{E}}_{\text{ox}}^{\text{o}}\\ \\ =\left(\text{0}\text{.771 V}\right)+\left(0.00\text{ V}\right)\\ \\ =\text{ 0}\text{.771 V}\end{array}$

The value of $\Delta G_{rxn}^o$ is calculated as follows,

  $\begin{array}{c}\Delta G_{rxn}^o\text{ =}-{\nu }_{e}F{E}_{cell}^o\\ \\ =-1 mol e^{-}\times \text{96485 C/mol}\times \text{0}\text{.771 V}\\ \\ \text{= }-\text{74}\text{.4 kJ}\end{array}$