Q21.55CP - Chemistry: Molecular Nature Of Matter And Change

Question

A voltaic cell has one half-cell with a $\mathbf{Cu}$ bar in a $\mathbf{1}\mathbf{.}\mathbf{00}{\mathbf{MCu}}^{\mathbf{2}\mathbf{+}}$salt, and the other half-cell with a Cd bar in the same volume of a $\mathbf{1}\mathbf{.}\mathbf{00}{\mathbf{MCd}}^{\mathbf{2}\mathbf{+}}$ salt.

(a) Find ${\mathit{E}}_{\mathbf{c}\mathbf{e}\mathbf{l}\mathbf{l}}^{\mathbf{o}}$ , $\mathbf{∆}{\mathit{G}}^{\mathbf{o}}$ and $\mathit{K}$ 

(b) As the cell operates, $\left[{\mathrm{Cd}}^{2+}\right]$ increases; find ${\mathbf{E}}_{\mathbf{cell}}$ and $\mathit{\Delta }\mathit{G}$ when $\left[{\mathrm{Cd}}^{2+}\right]$ is $\mathbf{1}\mathbf{.}\mathbf{95}\mathbf{M}$.

(c) Find${\mathbf{E}}_{\mathbf{cell}}\mathbf{,}\mathbf{\Delta G}$ and $\left[{\mathrm{Cu}}^{2+}\right]$ at equilibrium.

Step-by-Step Solution

Answer

(a) The values are $\mathrm{K}=1.08\times {10}^{25}$, $E_{cell}^0=0.74V$ and $\mathrm{\Delta G}°=-1.42\times {10}^5\mathrm{J}$.

(b) The values are $\mathrm{\Delta G}=-1.34\times {10}^5 \mathrm{J}$ and $E_{cell}^0=0.69 V$.

(c) The values are $\left[{\mathrm{Cu}}^{2+}\right]\cong 1.0\mathrm{M}$, $\mathrm{\Delta G}=0\mathrm{J}$, and $∆E_{cell}^o=0V$.

Step 1: Equilibrium

When a chemical reaction does not convert all reactants to products, it is said to be in equilibrium; many reactions attain a state of balance or dynamic equilibrium, in which both reactants and products are present.

Step 2: Subpart (a)

Cadmium will be oxidised because copper has a higher positive reduction potential.

$\begin{array}{l}{\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{Cu};\mathrm{E}}_{\mathrm{half}-\mathrm{cell}}^{\mathrm{o}}=0.34\\ {\mathrm{Cd}}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{Cd};\mathrm{E}}_{\mathrm{half}-\mathrm{cell}}^{\mathrm{o}}=-0.40\end{array}$

$\begin{array}{r}{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}{=\mathrm{E}}_{\mathrm{reduced}}^{\mathrm{o}}{-\mathrm{E}}_{\mathrm{oxidized}}^{\mathrm{o}}\\ =0.34-(-0.40)\\ =0.74\mathrm{V}\\ {\mathrm{\Delta G}}^{\mathrm{o}}{=-\mathrm{nFE}}_{\mathrm{cell}}^{\mathrm{o}}\\ =-2\times 96,500\times 0.74\\ =-1.42\times {10}^5\mathrm{J}\\ {\mathrm{\Delta G}}^{\mathrm{o}}=-\mathrm{RTlnK}\\ \mathrm{lnK}=\frac{\mathrm{\Delta G}}{-\mathrm{RT}}\end{array}$

$\begin{array}{c}\mathrm{K}={\mathrm{e}}^{\frac{{\mathrm{\Delta G}}^{\mathrm{o}}}{-\mathrm{RT}}}\\ ={\mathrm{e}}^{\frac{-1.42\times {10}^5}{-8.314\times 298}}\\ =1.08\times {10}^{25}\end{array}$

Therefore, the values are $1.08\times {10}^{25}$ and $-1.42\times {10}^5\mathrm{J}$.

Step 3: Subpart (b)

As there are equal numbers of $\mathrm{Cd}$ and $\mathrm{Cu}$ ions in solution and their molar ratios are $1:1$, a $0.95\mathrm{M}$ rise in $\mathrm{Cd}$ concentration equals a $0.95\mathrm{M}$ reduction in $\mathrm{Cu}$ concentration.

The Nernst equation can be used to calculate the new ${\mathrm{E}}_{\mathrm{cell}}$,

$\begin{array}{c}{\mathrm{E}}_{\mathrm{cell}}{=\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}\times \mathrm{lnQ}\\ \mathrm{Q}=\frac{\left[{\mathrm{Cd}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}\\ {\mathrm{E}}_{\mathrm{cell}}{=\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}\times \mathrm{lnQ}\\ {=\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}\times \ln \frac{\left[{\mathrm{Cd}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}\\ =0.74-\frac{8.314\times 298}{2\times 96500}\times \ln \frac{[1.95]}{[0.05]}\\ =0.69\mathrm{V}\end{array}$

$$" width="9" height="19" style="max-width: none;" >$\begin{array}{c}\mathrm{\Delta G}=-{\mathrm{nFE}}_{\mathrm{cell}}\\ =-2\times 96,500\times 0.69\\ =-1.34\times {10}^5 \mathrm{J}\end{array}$

Therefore, the values are $-1.34\times {10}^5 \mathrm{J}$, and $$" width="9" height="19" style="max-width: none; vertical-align: -4px;" >$0.69\mathrm{V}$.

Step 4: Subpart (c)

When the cell is at equilibrium, the reaction's free energy is zero, hence $\mathrm{\Delta G}=0$

In the same way, the cell potential is zero.

Based on the equilibrium constant, the equilibrium concentrations of the given vary.

$\begin{array}{c}\mathrm{K}=\frac{\left[{\mathrm{Cd}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}\\ \mathrm{K}=\frac{\left[{\mathrm{Cd}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}\\ 1.08\times {10}^{25}=\frac{\left[{\mathrm{Cd}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}\\ =\frac{1.00+\mathrm{x}}{1.00-\mathrm{x}}\end{array}$

As, $K=1.08\times {10}^{25}$very high, so the change in concentration is very less and hence, at equilibrium the $\left[C{u}^{2+}\right]\cong 1.0 M$.

Therefore, Cu ions have an approximate concentration of 1.0 M.