Chapter 21, Problem 21.152P

(a)

Interpretation Introduction

Interpretation:

For the given voltaic cell the $E^{\circ }{}_{cell},\Delta G^{\circ }andK$ values have to be calculated.

Concept Introduction:

The Standard Gibb’s free energy change and the standard cell potential are related as followed:

$\begin{array}{l}\text{Δ}°\text{G}\text{=}\text{-nFE}{°}_{\text{cell}}\\ where,\\ \boxed{\begin{array}{l}\text{n - Number of electrons involved per equivalent of the net redox reaction in the cell}\\ \text{F - Faraday’s Constant }\left(\text{96500 C}\right)\\ {\text{E°}}_{\text{cell}}\text{- Standard cell potential}\text{.}\end{array}}\end{array}$

The Nernst equation depicts the relationship between $E_{cell}$ and ${\text{E}}^{\text{o}}{}_{\text{cell}}$ as follows,

$\begin{array}{l}{\text{E}}_{\text{cell}}{\text{=E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592V}{n}\log Q\\ \\ \boxed{\begin{array}{c}where,{\text{E}}_{\text{cell}}=cellpotential\\ {\text{E}}^{\text{o}}{}_{\text{cell}}=\text{Standard}cellpotential\\ n=No.ofelectrons\\ Q=\text{Reaction}Quotient\end{array}}\end{array}$

Explanation of Solution

The half reactions for given cell with their ${\text{E}}^{\text{o}}$ values are noted by using Appendix D as follows,

$\begin{array}{l}\underset{\_}{\begin{array}{c}Cd\left(s\right)\to C{d}^{2+}\left(aq\right)+2e^{-}{E}^{\circ }\left(Oxidation\right)=-0.40V\\ C{u}^2{}^{+}\left(aq\right)+2e^{-}\to Cu\left(s\right)E^{\circ }\left(\text{Reduction}\right)=0.34V\end{array}}\\ Cd\left(s\right)+C{u}^2{}^{+}\left(aq\right)\to Cu\left(s\right)+C{d}^{2+}\left(aq\right)\end{array}$

Observing the standard reduction potentials values it is clear that $Cd$ oxidizes and $C{u}^{2+}$ reduces since, $Cd$ is better reducing agent.

The $E^{\circ }{}_{cell}$ value is calculated by using the equation ${\text{E°}}_{\text{cell}}{\text{= E°}}_{\text{cathode}}{\text{ - E°}}_{\text{anode}}$.

$\begin{array}{c}{\text{E°}}_{\text{cell}}{\text{= E°}}_{\text{cathode}}{\text{ - E°}}_{\text{anode}}\\ \text{=}\text{0}\text{.34}\text{V-}\left(\text{-0}\text{.40V}\right)\\ \text{=}\text{0}\text{.74V}\end{array}$

Next, using the $\text{Δ}°\text{G}\text{=}\text{-nFE}{°}_{\text{cell}}$ equation, the $\text{Δ}°\text{G}$ value is calculated.

$\begin{array}{c}\text{Δ}°\text{G}\text{=}\text{-nFE}{°}_{\text{cell}}\\ \text{= -}\text{2×}\left(\text{96485}\text{C/mol}{\text{e}}^{\text{-}}\right)\text{×}\left(\text{0}\text{.74}\text{J/C}\right)\\ =-1.427978\times {10}^5\\ =-1.4\times {10}^5\text{J}\end{array}$

Finally, using the number of electrons and the cell potential the K value is determined.

$\begin{array}{c}\log \text{K}\text{=}\frac{\text{nE}{°}_{\text{cell}}}{0.0592}\\ \log \text{K= }\frac{\text{2×}\left(\text{0}\text{.74}\right)}{0.0592}\\ =25\\ \text{K}=1\times {10}^{25}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The $E_{cell},and\Delta G$ values have to be calculated when the given voltaic cell started to operate with increased $\left[{\text{Cd}}^{\text{2+}}\right]$.

Concept Introduction:

The Standard Gibb’s free energy change and the standard cell potential are related as followed:

$\begin{array}{l}\text{Δ}°\text{G}\text{=}\text{-nFE}{°}_{\text{cell}}\\ where,\\ \boxed{\begin{array}{l}\text{n - Number of electrons involved per equivalent of the net redox reaction in the cell}\\ \text{F - Faraday’s Constant }\left(\text{96500 C}\right)\\ {\text{E°}}_{\text{cell}}\text{- Standard cell potential}\text{.}\end{array}}\end{array}$

The Nernst equation depicts the relationship between $E_{cell}$ and ${\text{E}}^{\text{o}}{}_{\text{cell}}$ as follows,

$\begin{array}{l}{\text{E}}_{\text{cell}}{\text{=E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592V}{n}\log Q\\ \\ \boxed{\begin{array}{c}where,{\text{E}}_{\text{cell}}=cellpotential\\ {\text{E}}^{\text{o}}{}_{\text{cell}}=\text{Standard}cellpotential\\ n=No.ofelectrons\\ Q=\text{Reaction}Quotient\end{array}}\end{array}$

Explanation of Solution

The half reactions for given cell with their ${\text{E}}^{\text{o}}$ values are noted by using Appendix D as follows,

$\begin{array}{l}\underset{\_}{\begin{array}{c}Cd\left(s\right)\to C{d}^{2+}\left(aq\right)+2e^{-}{E}^{\circ }\left(Oxidation\right)=-0.40V\\ C{u}^2{}^{+}\left(aq\right)+2e^{-}\to Cu\left(s\right)E^{\circ }\left(\text{Reduction}\right)=0.34V\end{array}}\\ Cd\left(s\right)+C{u}^2{}^{+}\left(aq\right)\to Cu\left(s\right)+C{d}^{2+}\left(aq\right)\end{array}$

Observing the standard reduction potentials values it is clear that $Cd$ oxidizes and $C{u}^{2+}$ reduces since, $Cd$ is better reducing agent.

$\begin{array}{c}{\text{E}}_{\text{cell}}{\text{=E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592V}{n}\log Q\\ {\text{E}}_{\text{cell}}\text{=0}\text{.74V}-\frac{0.0592V}{2}\log \frac{\left[{\text{Cd}}^{\text{2+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}\left(\text{since,}\text{Q=}\frac{\text{Product}\text{concentration}}{\text{Reactant}\text{concentration}}\right)\end{array}$

Observing the cell equation shows that the moles ratios of given ions is $1:1$ and cell operates the $\left[{\text{Cd}}^{\text{2+}}\right]$ concentration increases with equal decrease in $C{u}^{2+}$ concentration. Therefore, the reduced $\left[C{u}^{2+}\right]$ is $0.05M$ since, $\left[{\text{Cd}}^{\text{2+}}\right]$ increases from 1 to 1.95M.

$\begin{array}{c}\\ {\text{E}}_{\text{cell}}\text{=0}\text{.74V}-\frac{0.0592V}{2}\log \frac{\left[\text{1}\text{.95}\right]}{\left[\text{0}\text{.05}\right]}\\ =0.69V\end{array}$

(c)

Interpretation Introduction

Interpretation:

For the given voltaic cell the $E_{cell},and\Delta G$ values have to be calculated for $\left[{\text{Cu}}^{\text{2+}}\right]$ at equilibrium.

Concept Introduction:

The Standard Gibb’s free energy change and the standard cell potential are related as followed:

$\begin{array}{l}\text{Δ}°\text{G}\text{=}\text{-nFE}{°}_{\text{cell}}\\ where,\\ \boxed{\begin{array}{l}\text{n - Number of electrons involved per equivalent of the net redox reaction in the cell}\\ \text{F - Faraday’s Constant }\left(\text{96500 C}\right)\\ {\text{E°}}_{\text{cell}}\text{- Standard cell potential}\text{.}\end{array}}\end{array}$

The Nernst equation depicts the relationship between $E_{cell}$ and ${\text{E}}^{\text{o}}{}_{\text{cell}}$ as follows,

$\begin{array}{l}{\text{E}}_{\text{cell}}{\text{=E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592V}{n}\log Q\\ \\ \boxed{\begin{array}{c}where,{\text{E}}_{\text{cell}}=cellpotential\\ {\text{E}}^{\text{o}}{}_{\text{cell}}=\text{Standard}cellpotential\\ n=No.ofelectrons\\ Q=\text{Reaction}Quotient\end{array}}\end{array}$

Explanation of Solution

The half reactions for given cell with their ${\text{E}}^{\text{o}}$ values are noted by using Appendix D as follows,

$\begin{array}{l}\underset{\_}{\begin{array}{c}Cd\left(s\right)\to C{d}^{2+}\left(aq\right)+2e^{-}{E}^{\circ }\left(Oxidation\right)=-0.40V\\ C{u}^2{}^{+}\left(aq\right)+2e^{-}\to Cu\left(s\right)E^{\circ }\left(\text{Reduction}\right)=0.34V\end{array}}\\ Cd\left(s\right)+C{u}^2{}^{+}\left(aq\right)\to Cu\left(s\right)+C{d}^{2+}\left(aq\right)\end{array}$

Observing the standard reduction potentials values it is clear that $Cd$ oxidizes and $C{u}^{2+}$ reduces since, $Cd$ is better reducing agent.

$\begin{array}{c}{\text{E}}_{\text{cell}}\text{=}{\text{E}}^{\text{o}}{}_{\text{cell}}-\frac{0.0592V}{n}\log Q\\ \text{0V}\text{=}\text{0}\text{.74V}-\frac{0.0592V}{2}\log \frac{\left[\text{2-x}\right]}{\left[\text{x}\right]}\\ 25=\log \frac{\left[\text{2}\right]}{\left[\text{x}\right]}\\ 1\times {10}^{25}=\frac{\left[\text{2}\right]}{\left[\text{x}\right]}\\ x=2\times {10}^{-25}MC{u}^{2+}\end{array}$