Chapter 19, Problem 19.24QP

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant and Gibb’s free energy for the following cell reaction has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: E^o_cell is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Thermodynamics of redox reactions:

The change in free-energy represents the maximum amount of useful work that can be obtained in a reaction: $\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0$

Relation between ${\text{E}}_{cell}^0$ and equilibrium constant (K) of a redox reaction: ${\text{E}}_{cell}^0=\frac{RT}{nF}\ln K$

Relation between $\Delta {\text{G}}^0$ and K:$\Delta {\text{G}}^0\text{=}\text{-}RT\ln K$

Explanation of Solution

The given reaction is,

$P{b}^{2+}\left(aq\right)+Mg\left(s\right)\rightleftharpoons Pb\left(s\right)+M{g}^{2+}\left(aq\right)$

Lead ion is reduced and Magnesium is oxidized; hence oxidation occurs at anode electrode and reduction occurs at cathode electrode.

Calculation of Standard emf:

The standard reduction potential of Mg (+2) as follows,

$M{g}^{2+}\left(aq\right)+2e^{-}\to Mg\left(s\right)E^0=-2.37V$

The standard reduction potential of Pb (+2) as follows,

$P{b}^{2+}\left(aq\right)+2e^{-}\to Pb\left(s\right)E^0=-0.13V$

Calculated standard emf for galvanic cell as follows,

$\begin{array}{l}oxidation:Mg\left(s\right)\to M{g}^{2+}\left(aq\right)+2e^{-}{E}_{anode}^0=+2.37V\left(\text{half-reaction}\text{change}\text{the}\text{sign}\text{but not magnitude}\right)\\ \underset{\_}{\mathrm{Re}duction:P{b}^{2+}\left(aq\right)+2e^{-}\to Pb\left(s\right)E_{cathode}^0=-0.13V}\\ \underset{\_}{overall:P{b}^{2+}\left(aq\right)+Mg\left(s\right)\to Pb\left(s\right)+M{g}^{2+}\left(aq\right)}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{o}}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{+}{\text{E}}_{\text{anode}}^{\text{o}}\\ \text{=}\left(\text{-0}\text{.13}\right)\text{V}\text{+}\left(\text{+2}\text{.37}\right)\text{V}\\ \text{=}\text{+2}\text{.24}\text{V}\end{array}$

Therefore, Standard emf of a galvanic cell is $\text{+2}\text{.24V}$

Equilibrium constant calculation:

Using Gibb’s free energy equation as follows,

$\begin{array}{l}\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0\\ =-\left(2e^{-}\right)\left(96,500J/V.mol\right)\left(+2.24V\right)\\ =-432kJ/mol\end{array}$

Using equation as follows,

$\begin{array}{l}\Delta {\text{G}}^0\text{=}\text{-}RT\ln K\\ -432kJ/mol=-\left(8.314J/K.mol\right)\left(298K\right)\ln K\\ \ln K=\frac{-432\times {10}^{3}J/mol}{-\left(8.314J/K.mol\right)\left(298K\right)}=174.4\\ K=e^{174.4}=5\times {10}^{75}\end{array}$

Therefore, the equilibrium constant obtained is $5\times {10}^{75}$

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant and Gibb’s free energy for the following cell reaction has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: E^o_cell is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Thermodynamics of redox reactions:

The change in free-energy represents the maximum amount of useful work that can be obtained in a reaction: $\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0$

Relation between ${\text{E}}_{cell}^0$ and equilibrium constant (K) of a redox reaction: ${\text{E}}_{cell}^0=\frac{RT}{nF}\ln K$

Relation between $\Delta {\text{G}}^0$ and K:$\Delta {\text{G}}^0\text{=}\text{-}RT\ln K$

Explanation of Solution

The given reaction is,

$B{r}_2\left(l\right)+2I^{-}\left(aq\right)\rightleftharpoons 2B{r}^{-}\left(aq\right)+I_2\left(s\right)$

Bromium is reduced and Iodide ion is oxidized; hence oxidation occurs at anode electrode and reduction occurs at cathode electrode.

Calculation of Standard emf:

The standard reduction potential of Br (-1) as follows,

$B{r}_2\left(l\right){\text{ + 2e}}^{-}\to 2B{r}^{-}\left(aq\right)E^0=+1.07V$

The standard reduction potential of Iodide ion as follows,

$I_2\left(s\right){\text{ + 2e}}^{-}\to 2I^{-}\left(aq\right)E^0=+0.53V$

Calculated standard emf for galvanic cell as follows,

$\begin{array}{l}oxidation:2I^{-}\left(aq\right)\to I_2\left(s\right){\text{ + 2e}}^{-}{E}_{anode}^0=-0.53V\left(\text{half-reaction}\text{change}\text{the}\text{sign}\text{but not magnitude}\right)\\ \underset{\_}{\mathrm{Re}duction:B{r}_2\left(l\right){\text{ + 2e}}^{-}\to 2B{r}^{-}\left(aq\right)E_{cathode}^0=+1.07V}\\ \underset{\_}{overall:B{r}_2\left(l\right)+2I^{-}\left(aq\right)\to 2B{r}^{-}\left(aq\right)+I_2\left(s\right)}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{o}}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{+}{\text{E}}_{\text{anode}}^{\text{o}}\\ \text{=}\left(\text{+1}\text{.07}\right)\text{V}\text{+}\left(\text{-0}\text{.53}\right)\text{V}\\ \text{=}\text{+0}\text{.54}\text{V}\end{array}$

Therefore, Standard emf of a galvanic cell is $\text{+0}\text{.54}\text{V}$

Equilibrium constant calculation:

Using Gibb’s free energy equation as follows,

$\begin{array}{l}\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0\\ =-\left(2e^{-}\right)\left(96,500J/V.mol\right)\left(\text{+0}\text{.54}\text{V}\right)\\ =-104kJ/mol\end{array}$

Using equation as follows,

$\begin{array}{l}\Delta {\text{G}}^0\text{=}\text{-}RT\ln K\\ -104kJ/mol=-\left(8.314J/K.mol\right)\left(298K\right)\ln K\\ \ln K=\frac{-104\times {10}^{3}J/mol}{-\left(8.314J/K.mol\right)\left(298K\right)}=41.98\\ K=e^{41.98}=1.8\times {10}^{18}\end{array}$

Therefore, the equilibrium constant obtained is $2\times {10}^{18}$

c)

Interpretation Introduction

Interpretation:

The equilibrium constant and Gibb’s free energy for the following cell reaction has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: E^o_cell is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Thermodynamics of redox reactions:

The change in free-energy represents the maximum amount of useful work that can be obtained in a reaction: $\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0$

Relation between ${\text{E}}_{cell}^0$ and equilibrium constant (K) of a redox reaction: ${\text{E}}_{cell}^0=\frac{RT}{nF}\ln K$

Relation between $\Delta {\text{G}}^0$ and K:$\Delta {\text{G}}^0\text{=}\text{-}RT\ln K$

Explanation of Solution

The given reaction is,

$O_2\left(g\right)+4H^{+}\left(aq\right)+4F{e}^{2+}\left(aq\right)\rightleftharpoons 2H_{2}O\left(l\right)+4F{e}^{3+}\left(aq\right)$

Oxygen and proton are reduced and Iron ion is oxidized; hence oxidation occurs at anode electrode and reduction occurs at cathode electrode.

Calculation of Standard emf:

The standard reduction potential of Oxygen as follows,

$O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-}\to 2H_{2}O\left(l\right)E^0=+1.23V$

The standard reduction potential of Iron ion as follows,

$F{e}^{3+}\left(aq\right)+e^{-}\to F{e}^{2+}\left(aq\right)E^0=+0.77V$

Calculated standard emf for galvanic cell as follows,

Reversing half-reaction changes the sign of $E^0$ but not the magnitude.

$\begin{array}{l}oxidation:4F{e}^{2+}\left(aq\right)\to 4F{e}^{3+}\left(aq\right)+4e^{-}{E}_{anode}^0=-0.77V\\ \underset{\_}{\mathrm{Re}duction:O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-}\to 2H_{2}O\left(l\right)E_{cathode}^0=+1.23V}\\ \underset{\_}{overall:O_2\left(g\right)+4H^{+}\left(aq\right)+4F{e}^{2+}\left(aq\right)\rightleftharpoons 2H_{2}O\left(l\right)+4F{e}^{3+}\left(aq\right)}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{o}}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{+}{\text{E}}_{\text{anode}}^{\text{o}}\\ \text{=}\left(\text{+1}\text{.23}\right)\text{V}\text{+}\left(\text{-0}\text{.77}\right)\text{V}\\ \text{=}\text{+0}\text{.46}\text{V}\end{array}$

Therefore, Standard emf of a galvanic cell is $\text{+0}\text{.46}\text{V}$

Equilibrium constant calculation:

Using Gibb’s free energy equation as follows,

$\begin{array}{l}\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0\\ =-\left(4e^{-}\right)\left(96,500J/V.mol\right)\left(\text{+0}\text{.46}\text{V}\right)\\ =-177kJ/mol\end{array}$

Using equation as follows,

$\begin{array}{l}\Delta {\text{G}}^0\text{=}\text{-}RT\ln K\\ -177kJ/mol=-\left(8.314J/K.mol\right)\left(298K\right)\ln K\\ \ln K=\frac{-177\times {10}^{3}J/mol}{-\left(8.314J/K.mol\right)\left(298K\right)}=71.4\\ K=e^{71.4}=1\times {10}^{31}\end{array}$

Therefore, the equilibrium constant obtained is $1\times {10}^{31}$

d)

Interpretation Introduction

Interpretation:

The equilibrium constant and Gibb’s free energy for the following cell reaction has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: E^o_cell is composed of a contribution from the anode and a contribution from the cathode is given by,

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Where both $E_{cathode}^o$ and $E_{anode}^o$ are the standard reduction potentials of the electrodes.

Thermodynamics of redox reactions:

The change in free-energy represents the maximum amount of useful work that can be obtained in a reaction: $\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0$

Relation between ${\text{E}}_{cell}^0$ and equilibrium constant (K) of a redox reaction: ${\text{E}}_{cell}^0=\frac{RT}{nF}\ln K$

Relation between $\Delta {\text{G}}^0$ and K:$\Delta {\text{G}}^0\text{=}\text{-}RT\ln K$

Explanation of Solution

The given reaction is,

$2Al\left(s\right)+3I_2\left(s\right)\rightleftharpoons 2A{l}^{3+}\left(aq\right)+6I^{-}\left(aq\right)$

Iodine is reduced and Aluminum is oxidized; hence oxidation occurs at anode electrode and reduction occurs at cathode electrode.

Calculation of Standard emf:

The standard reduction potential of Al (+3) as follows,

$A{l}^{3+}\left(aq\right)+3e^{-}\to Al\left(s\right)E^0=-1.66V$

The standard reduction potential of Iodide ion as follows,

$I_2\left(s\right){\text{ + 2e}}^{-}\to 2I^{-}\left(aq\right)E^0=+0.53V$

Calculated standard emf for galvanic cell as follows,

Reversing oxidation half-reaction changes the sign $E^0$ but not the magnitude.

$\begin{array}{l}oxidation:2Al\left(s\right)\to 2A{l}^{3+}\left(aq\right)+6e^{-}{E}_{anode}^0=+1.66V\\ \underset{\_}{\mathrm{Re}duction:3I_2\left(s\right){\text{ + 6e}}^{-}\to 6I^{-}\left(aq\right)E_{cathode}^0=+0.53V}\\ \underset{\_}{overall:2Al\left(s\right)+3I_2\left(s\right)\rightleftharpoons 2A{l}^{3+}\left(aq\right)+6I^{-}\left(aq\right)}\end{array}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{o}}\text{=}{\text{E}}_{\text{cathode}}^{\text{o}}\text{+}{\text{E}}_{\text{anode}}^{\text{o}}\\ \text{=}\left(\text{+0}\text{.53}\right)\text{V}\text{+}\left(\text{+1}\text{.66}\right)\text{V}\\ \text{=}\text{+2}\text{.19V}\end{array}$

Therefore, Standard emf of a galvanic cell is $\text{+2}\text{.19V}$

Equilibrium constant calculation:

Using Gibb’s free energy equation as follows,

$\begin{array}{l}\Delta {\text{G}}^0\text{=}{\text{-nFE}}_{cell}^0\\ =-\left(6e^{-}\right)\left(96,500J/V.mol\right)\left(\text{+2}\text{.19V}\right)\\ =-1268kJ/mol\end{array}$

Using equation as follows,

$\begin{array}{l}\Delta {\text{G}}^0\text{=}\text{-}RT\ln K\\ -1268kJ/mol=-\left(8.314J/K.mol\right)\left(298K\right)\ln K\\ \ln K=\frac{-1268\times {10}^{3}J/mol}{-\left(8.314J/K.mol\right)\left(298K\right)}=511.8\\ K=e^{511.8}=1.8\times {10}^{222}\end{array}$

Therefore, the equilibrium constant obtained is $1.8\times {10}^{222}$