Question

Question: An electrochemical cell consists of a silver metal electrode immersed in a solution with $\left[{\mathrm{Ag}}^{+}\right]=\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{M}$separated by a porous disk from a compartment with a copper metal electrode immersed in a solution of role="math" localid="1663918603238" $\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{M}\mathbf{}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}$that also contains $\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{M}$. The equilibrium between ${\mathbf{Cu}}^{\mathbf{+}\mathbf{2}}$and ${{\mathbf{NH}}_{\mathbf{3}}}^{\mathbf{+}}$is:

data-custom-editor="chemistry" ${\mathbf{Cu}}^{\mathbf{+}\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{NH}}_{\mathbf{3}}\stackrel{}{\mathbf{\rightleftharpoons }}\mathbf{Cu}{{\left({\mathrm{NH}}_3\right)}_{\mathbf{4}}}^{\mathbf{+}\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}}$

and the two cell half-reactions are:

$$\begin{gathered} {\mathbf{Ag}}^{\mathbf{+}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{Ag}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{E}}^{\mathbf{\circ }}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{\hspace{0.17em}}\mathbf{V} \\ {\mathbf{Cu}}^{\mathbf{+}\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{Cu}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{E}}^{\mathbf{\circ }}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{34}\mathbf{\hspace{0.17em}}\mathbf{V} \end{gathered}$$

Assuming Ag⁺ is reduced, what is the cell potential at 25°C?

Short answer

The cell potential is $1.05\mathrm{V}$.

Step-by-step solution

Analyzing the given data

$$\begin{gathered} {\mathrm{Cu}}^{+2}\left(\mathrm{aq}\right)+4{\mathrm{NH}}_3\stackrel{}{\rightleftharpoons }\mathrm{Cu}{{\left({\mathrm{NH}}_3\right)}_4}^{+2}\left(\mathrm{aq}\right) \\ \mathrm{K}=1.0\times {10}^{13} \\ \mathrm{K}=\frac{\left[\mathrm{Cu}{{\left({\mathrm{NH}}_3\right)}_4}^{+2}\right]}{\left[{\mathrm{Cu}}^{+2}\right]{\left[{\mathrm{NH}}_3\right]}^4} \end{gathered}$$

Concentration of$\left[{\mathrm{Cu}}^{+2}\right]=2.40\times {10}^{-20}\mathrm{M}$

Calculating the cell potential  

Writing the half-reactions,

$$\begin{gathered} {\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Ag}{\mathrm{E}}^{\circ }=0.80\hspace{0.17em}\mathrm{V} \\ {\mathrm{Cu}}^{+2}+2{\mathrm{e}}^{-}\to \mathrm{Cu}{\mathrm{E}}^{\circ }=0.39\hspace{0.17em}\mathrm{V} \end{gathered}$$

Reorganizing the equations to flip the second equationlocalid="1663920536184" $\mathrm{Cu}\to {\mathrm{Cu}}^{+2}+2{\mathrm{e}}^{-}{\mathrm{E}}^{\circ }=-0.39\hspace{0.17em}\mathrm{V}$

The final equation;

localid="1663920579054" $$\begin{gathered} 2{\mathrm{Ag}}^{+}+\mathrm{Cu}\to 2\mathrm{Ag}+{\mathrm{Cu}}^{+2}{\mathrm{E}}^{\circ }=0.46\mathrm{V} \\ \mathrm{E}={\mathrm{E}}^{\circ }-\frac{0.0591}{\mathrm{n}}\mathrm{logQ} \end{gathered}$$

Where Q is the cell potential, localid="1663920600240" $\frac{\left[{\mathrm{Cu}}^{+2}\right]}{{\left[{\mathrm{Ag}}^{+}\right]}^2}$.

Hence, the cell potential will be localid="1663920622161" $1.05\mathrm{V}$