Q21.153CP - Chemistry: Molecular Nature Of Matter And Change

Question

Two concentration cells are prepared, both with   $\mathbf{90}\mathbf{.}\mathbf{0}$mL of $\mathbf{0}\mathbf{.}\mathbf{0100}$M  $\mathbf{Cu}\mathbf{(}\mathbf{N}{\mathbf{H}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}$ and a Cu bar in each half-cell. 

(a) In the first concentration cell,   $\mathbf{10}\mathbf{.}\mathbf{0}$mL of  $\mathbf{0}\mathbf{.}\mathbf{500}$M  is added to one half-cell; the complex ion  $\mathbf{Cu}\mathbf{(}\mathbf{N}{\mathbf{H}}_{\mathbf{3}}{{\mathbf{)}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{+}}$ forms, and ${\mathbf{E}}_{\mathbf{cell}}$  is   $\mathbf{0}\mathbf{.}\mathbf{129}$V. Calculate ${\mathbf{K}}_{\mathbf{f}}$  for the formation of the complex ion. 

(b) Calculate  ${\mathbf{E}}_{\mathbf{cell}}$ when an additional   $\mathbf{10}\mathbf{.}\mathbf{0}$mL of  $\mathbf{0}\mathbf{.}\mathbf{500}$M ${\mathbf{NH}}_{\mathbf{3}}$ is added. 

(c) In the second concentration cell,   $\mathbf{10}\mathbf{.}\mathbf{0}$mL of  $\mathbf{0}\mathbf{.}\mathbf{500}$M NaOH is added to one half-cell; the precipitate $\mathbf{Cu}\mathbf{(}\mathbf{OH}{\mathbf{)}}_{\mathbf{2}}$ forms (${\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}$ ). Calculate ${{\mathbf{E}}^{\mathbf{o}}}_{\mathbf{cell}}$ . 

(d) What would the molarity of NaOH have to be for the addition of $\mathbf{10}\mathbf{.}\mathbf{0}$ mL to result in an ${{\mathbf{E}}^{\mathbf{o}}}_{\mathbf{cell}}$  of  $\mathbf{0}\mathbf{.}\mathbf{340}$ V?

Step-by-Step Solution

Answer

(a) The value of ${\mathrm{K}}_{\mathrm{f}}$  is obtained as: ${\mathrm{K}}_{\mathrm{f}}=2.09\times {10}^{14}$ .

(b)  The value of  ${\mathrm{E}}_{\mathrm{cell}}$ is obtained as: ${\mathrm{E}}_{\mathrm{cell}}=0.156\mathrm{V}$.

(c)  The value of ${\mathrm{E}}_{\mathrm{cell}}$  is obtained as:  ${\mathrm{E}}_{\mathrm{cell}}=0.3548\mathrm{V}$.

(d)  The molarity of NaOH is obtained as:  $[\mathrm{O}{\mathrm{H}}^{-}]=8.84\times {10}^{-4}$.

Step 1: Concentration cell

Concentration cell is the one that has both the electrodes are same and are separated by electrolyte solutions with different concentrations.

Step 2: Calculate the value of K f

(a) When both cells have the same concentration of $\mathrm{Cu}(\mathrm{N}{\mathrm{O}}_3{)}_2$ , the net voltage is zero.

When ${\mathrm{NH}}_3$  is introduced to the first cell, however, part of the  $\mathrm{Cu}(\mathrm{N}{\mathrm{O}}_3{)}_2$ interacts with the ${\mathrm{NH}}_3$ and forms a complex. As a result, the concentration of ${\mathrm{Cu}}^{+2}$  ions in the first cell decreases. ${\mathrm{E}}_{\mathrm{cell}}$  is  $0.129$ V at this stage. The first cell, which has a lower Cu content, will attempt to oxidise Cu to ${\mathrm{Cu}}^{+2}$ and serve as an anode. The cathode will be the second cell.

The half-cell response looks like this:

$\begin{array}{c}\mathrm{Anode}:-\mathrm{Cu}\to {\mathrm{Cu}}^{+2}+2{\mathrm{e}}^{-}\\ \mathrm{Cathode}:-{\mathrm{Cu}}^{+2}+2{\mathrm{e}}^{-}\to \mathrm{Cu}\\ {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}={\mathrm{E}}_{\mathrm{cell}}-\frac{0.0591}{\mathrm{n}}\log \frac{\left[\mathrm{product}\right]}{\left[\mathrm{reactant}\right]}\\ [\mathrm{n}=\mathrm{numberoftransferringelectroninthereactionis}2]\\ {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}={\mathrm{E}}_{\mathrm{cell}}-\frac{0.0591}{\mathrm{n}}\log \frac{{\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}}{{\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{cathode}}}\\ {\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{cathode}}=0.09\mathrm{L}\times 0.01 \mathrm{mol}/\mathrm{L}\\ =9\times {10}^{-4}\mathrm{moles}\\ 0.129=0-\frac{0.0591}{2}\log \frac{{\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}}{9\times {10}^{-4}\mathrm{moles}}\\ {\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}=3.8\times {10}^{-8}\mathrm{moles}\end{array}$

The following is the reaction of  ${\mathrm{NH}}_3$ with ${\mathrm{Cu}}^{+2}$ :

$\begin{array}{c}{\mathrm{Cu}}^{+2}+4{\mathrm{NH}}_3\to {\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{+2}\\ {\mathrm{K}}_{\mathrm{f}}=\frac{{\left[\mathrm{Cu}{\left({\mathrm{N}}_3\right)}_4\right]}^{+2}}{{\mathrm{Cu}}^{+2}\times {\left({\mathrm{NH}}_3\right)}_4}\\ \left[{\left({\mathrm{NH}}_3\right)}_4\right]=0.01 \mathrm{L}\times 0.5 \mathrm{mol}/\mathrm{L}\\ =0.005\mathrm{moles}\\ {\mathrm{K}}_{\mathrm{f}}=\frac{0.005}{3.8\times {10}^{-8}\times (0.005{)}^4}\\ {\mathrm{K}}_{\mathrm{f}}=2.09\times {10}^{14}\end{array}$

Therefore, the value is: ${\mathrm{K}}_{\mathrm{f}}=2.09\times {10}^{14}$ .

Step 3: Evaluate the value of E c e l l

(b) If  $10$mL of  $0.5$M   is added, the total  ${\mathrm{NH}}_3$ concentration is   $20$mL and the   $0.5$M concentration is  $0.01$ moles.

Calculate the ${\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}$  using this value.

$\begin{array}{c}{\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}=\frac{{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{+2}}{{\mathrm{K}}_{\mathrm{f}}\times {\left({\mathrm{NH}}_3\right)}_4}\\ =\frac{0.01}{2.09\times {10}^{14}\times (0.01{)}^4}\\ =4.78\times {10}^{-9}\\ {\mathrm{E}}_{\mathrm{cell}}{=\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.0591}{\mathrm{n}}\log \frac{{\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}}{{\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{cathode}}}\\ =0-\frac{0.0591}{2}\log \frac{4.78\times {10}^{-9}}{9\times {10}^{-4}}\\ {\mathrm{E}}_{\mathrm{cell}}=0.156\mathrm{V}\end{array}$

Therefore, the value is: ${\mathrm{E}}_{\mathrm{cell}}=0.156\mathrm{V}$ .

Step 4: Evaluate the value of E cell

NaOH is added  $10$ml,  $0.5$M =   $0.01$L times  $0.5$M which gives  $0.005$ Moles

$C{u}^{2+}+2NaOH\to Cu{\left(OH\right)}_2+2N{a}^{+}$

$Cu{\left(OH\right)}_2\left(s\right)\rightleftharpoons C{u}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)$

$\begin{array}{c}{\mathrm{K}}_{\mathrm{sp}}=\left[\right\{\mathrm{C}{\mathrm{u}}^{+2}\left]\right[\mathrm{O}{\mathrm{H}}^{-}{]}^2\}\\ 2.2\times {10}^{-20}=[\mathrm{C}{\mathrm{u}}^{+2}\left]\right(0.05{)}^2\\ {\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}=8.8\times {10}^{-16}\end{array}$

The total volume of the solution is:

$90+10=100\mathrm{ml}$

The moles then are:

$\begin{array}{c}\left[{\mathrm{Cu}}^{+2}\right]=100\mathrm{ml}\times 8.8\times {10}^{-16}\\ =8.8\times {10}^{-17}\end{array}$

$\begin{array}{c}{\mathrm{E}}_{\mathrm{cell}}{=\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.0591}{\mathrm{n}}\log \frac{{\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}}{{\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{cathode}}}\\ =0-\frac{0.0591}{2}\log \frac{8.8\times {10}^{-16}}{9\times {10}^{-4}}\\ {\mathrm{E}}_{\mathrm{cell}}=0.3548\mathrm{V}\end{array}$

Therefore, the value is:  ${\mathrm{E}}_{\mathrm{cell}}=0.3548\mathrm{V}$.

Step 5: Molarity

The value of ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=0.34\mathrm{V}$  and then ${\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}$  is evaluated as:

$\begin{array}{c}{\mathrm{E}}_{\mathrm{cell}}{=\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.0591}{\mathrm{n}}\log \frac{{\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}}{{\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{cathode}}}\\ =0-\frac{0.0591}{2}\log \frac{{\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}}{9\times {10}^{-4}}\\ {\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}=2.807\times {10}^{-15}\mathrm{moles}\end{array}$

The molarity then will be:

$\begin{array}{c}{\left[{\mathrm{Cu}}^{+2}\right]}_{\mathrm{anode}}=2.807\times {10}^{-15}\mathrm{moles}/0.1\mathrm{L}\\ =2.807\times {10}^{-14}\mathrm{M}\end{array}$

$\begin{array}{c}{\mathrm{K}}_{\mathrm{sp}}=[\mathrm{C}{\mathrm{u}}^{+2}\left]\right[\mathrm{O}{\mathrm{H}}^{-}]\\ 2.2\times {10}^{-20}=[2.807\times {10}^{-14}\mathrm{M}\left]\right[\mathrm{O}{\mathrm{H}}^{-}{]}^2\\ [\mathrm{O}{\mathrm{H}}^{-}{]}^2=7.8\times {10}^{-7}\end{array}$

Therefore, the value is: $[\mathrm{O}{\mathrm{H}}^{-}]=8.84\times {10}^{-4}$ .