Question

Question:You have a concentration cell with Cu electrodes and $\left[{\mathrm{Cu}}^{+2}\right]\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\hspace{0.17em}}\mathbf{M}$(right side) and $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\hspace{0.17em}}\mathbf{M}$(left side).

a. Calculate the potential for this cell at ${\mathbf{25}}^{\mathbf{\circ }}\mathbf{C}$

b. The ion reacts with NH₃ to form Cu(NH₃)_4,where the stepwise formation constants are ${\mathbf{K}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{,}\mathbf{}{\mathbf{K}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{,}\mathbf{}{\mathbf{K}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{,}$and ${\mathbf{K}}_4\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}$.Calculate the new cell potential after enough NH₃ is added to the left cell compartment such that at equilibrium$\left[{\mathrm{NH}}_3\right]\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{M}$

Short answer

1. $0.118\mathrm{V}$
   
2. $0.384V$

Step-by-step solution

analyzing the given data

The two half reactions are given as

$\begin{array}{l}Cu\stackrel{}{\to }C{u}^{+2}+2e^{-}\left(E^0=-0.34V\right)\\ C{u}^{+2}+2e^{-}\stackrel{}{\to }Cu\left(E^0=-0.34V\right)\end{array}$

Determining the cell potential at 25∘C

$$\begin{gathered} E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{n}\log Q \\ {E}_{cell}=0-\frac{0.0591}{2}\log \frac{2}{1\times {10}^3} \\ {E}_{cell}=0.118V \end{gathered}$$

Hence, the cell potential at ${25}^{\circ }C$is $0.118\mathrm{V}$.

Calculating the new cell potential  

$$\begin{gathered} E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{n}\log K \\ {E}_{cell}=0-\frac{0.0591}{2}\log \left(K_1\times K_2\times K_3\times K_4\right) \\ {E}_{cell}=-\frac{0.0591}{2}\log \left(1\times {10}^{13}\right) \\ {E}_{cell}=0.384\mathrm{V} \end{gathered}$$

Hence, the new cell potential is$0.384\mathrm{V}$.