Chapter 22, Problem 79AP

(a) The free Cu(I) ion is unstable in solution and has a tendency to disproportionate:

${\text{2Cu}}^{\text{+}}\left(aq\right)\underset{}{\rightleftarrows }{\text{ Cu}}^{\text{2+}}\left(aq\right)\text{+ Cu(}s\text{)}$

Use the information in Table 19.1 to calculate the equilibrium constant for the reaction. (b) Based on your results in part (a), explain why most Cu(I) compounds are insoluble.

Interpretation Introduction

Interpretation:

The equilibrium constant ($K_c$

) for the given reaction is to be calculated. The reason behind $\text{Cu}\left(\text{I}\right)$

compounds being insoluble is to be explained.

Concept introduction:

The change in free energy is called Gibb’s free energy and is represented as $\Delta G^o$.

If $\Delta G^o$ is negative, it indicates that the electrical work has been done by the system on the surroundings, and if it is positive then it means the electrical work has been done on the system by the surroundings.

The standard cell potential can be calculated by the expression as: $E_{cell}^o=E_{Cathode}^o-E_{Anode}^o$.

The standard Gibbs free energy change is calculated by the expression as:

$\Delta G^o=-n\text{F}{E}_{cell}^o$

Here, $n$ is the moles of electrons are transferred per mole of reaction, $\Delta G^o$ is the standard Gibbs free energy change, $E_{cell}^o$

is the standard cell potential, and $\text{F}$ is the Faraday’s constant.

The equilibrium constant is calculated by the expression as:

$E_{cell}^o=\frac{0.0592}{n}\log \left(\text{K}\right)$

The relation between free energy change and standard free energy change is as:

$\Delta G=\Delta G^{\text{o}}+\text{R}T\ln Q$

Here, $\Delta G$ is free energy change, $\Delta G^o$ is standard free energy change, $Q$ is the reaction quotient, $\text{R}$

is the gas constant, and $T$ is the temperature.

At equilibrium the above equation is reduced to the expression: $\Delta G^{\text{o}}\text{=}-\text{R }T\text{lnK }$

Answer to Problem 79AP

Solution:

a) $K_c = 2.7\times 10^6$

b) Stable compounds containing ${\text{Cu}}^{+}$ ions are insoluble.

Explanation of Solution

a) The equilibrium constant for the reaction

${\text{2Cu}}^{\text{+}}\text{(}aq\text{)}\rightleftarrows {\text{Cu}}^{\text{2+}}\text{(}aq\text{)+Cu(}s\text{)}$

From table 19.1, the electrode potential of two half-cell reactions are given as follows:

$\begin{array}{c}{\text{Cu}}^{2+}+2e^{-} \to \text{Cu } ,E°=\text{ 0}\text{.34 V}\\ {\text{ Cu}}^{2+}+e^{-} \to {\text{Cu}}^{+}, E°=0.15\text{ V}\end{array}$

The Gibbs free energy for first half of the reaction is given as follows:

$\text{ ΔG°}=-nF{E}^{\text{o}}{}_{\text{cell}}$ …… (1)

Substitute $\text{0}\text{.34 V}$ for $E^{\text{o}}{}_{\text{cell}}$, 2 for $n$ and $\text{96500 J}/\text{V}\text{.mol}$ for $F$ in the above equation as follows:

$\begin{array}{c}\text{ ΔG°}=-\left(2\right)\left(\text{96500 J}/\text{V}\text{.mol}\right)\left(\text{0}\text{.34 V}\right)\\ =-\text{6}{\text{.562 × 10}}^{\text{4}}\text{J}/\text{mol}\end{array}$

The Gibbs free energy for second half of the reaction is given as follows:

$\text{ ΔG°}=-nF{E}^{\text{o}}{}_{\text{cell}}$

Substitute $\text{0}\text{.15 V}$ for $E^{\text{o}}{}_{\text{cell}}$, 1 for $n$ and $\text{96500 J}/\text{V}\text{.mol}$ for $F$ in the above equation as follows:

$\begin{array}{c}\text{ ΔG°}=-\left(1\right)\left(\text{96500 J}/\text{V}\text{.mol}\right)\left(\text{0}\text{.15 V}\right)\\ =-1.448\times {10}^4\text{J/mol}\end{array}$

The overall reaction by the two half-cell reactions is given as follows: $\begin{array}{l}\text{anode}\left(oxidation\right){\text{: Cu}}^{2+}+2e^{-} \to \text{Cu }\\ \text{cathode}\left(oxidation\right)\text{: }2{\text{Cu}}^{+} \to 2{\text{Cu}}^{2+}+2e^{-}\\ \text{ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}\\ {\text{ overall: 2Cu}}^{\text{+}}\to {\text{Cu}}^{\text{2+}}\text{+ Cu } \end{array}$

The total standard Gibbs free energy is calculated as follows:

$\begin{array}{c}\Delta G=-6.562\times {10}^4\text{J/mol}+\left(2\right)\left(1.448\times {10}^4\text{J/mol}\right)\\ =-6.562\times {10}^4\text{J/mol }+2.896\times {10}^4\text{J/mol}\\ =-3.666\times {10}^4\text{J/mol}\end{array}$

Now, the equilibrium constant is calculated as follows:

$\begin{array}{c}\Delta G=-RT\text{ ln }{K}_c\\ K_c={\text{e}}^{{}^{-\frac{\text{ΔG°}}{\text{RT}}}}\end{array}$

Substitute $-3.666\times {10}^4\text{J/mol}$ for $\Delta G$, $\text{8}\text{.314}\text{J/mol}\cdot \text{K}$ for $R$, and $298\text{ K}$ for $T$ in the above equation as follows:

$\begin{array}{c}{K}_c=e^{-\frac{\text{(-3}\text{.666}\text{×}{\text{10}}^{\text{4}}\text{J/mol)}}{\text{(8}\text{.314}\text{J/mol}\cdot \text{K)(298}\text{K)}}}\\ K_c = 2.7\times 10^6\end{array}$

Therefore, the equilibrium constant for the reaction is $2.7\times 10^6$.

b) Most $\text{Cu(I)}$ compound are insoluble.

According to part (a), the free ${\text{Cu}}^{+}$ ions are unstable in solution. So, the stable compounds that contain cuprous ions are insoluble.