Question

A voltaic cell consists of a standard hydrogen electrode in one half-cell and a ${\mathbf{\text{Cu/Cu}}}^{\mathbf{\text{2+}}}$ half-cell. Calculate$\mathbf{\left[}{\mathbf{\text{Cu}}}^{\mathbf{\text{2+}}}\mathbf{\right]}$ when${\mathbf{\text{E}}}_{\mathbf{\text{cell}}}^{\mathbf{\text{°}}}$ is $\mathbf{0}\mathbf{.}\mathbf{22}$V

Short answer

Hence, the concentration of ${\text{Cu}}^{\text{+2}}$ ions is $8.9\times {10}^{-5}\text{\hspace{0.17em}M}$.

Step-by-step solution

Calculating Ecello

The electrode potential of ${\text{Cu/Cu}}^{\text{+2}}$ is greater than 0. Thus, it can oxidize hydrogen to ${\text{H}}^{\text{+}}$ ions. The redox reaction is given below.

${\text{Cu}}^{\text{2+}}{\text{(aq)+H}}_{\text{2}}\text{(g)}\to {\text{Cu(s)+2H}}^{\text{+}}\text{(aq)}$

The half-cell cathode and anode reaction is given below.

Anode:${\text{H}}_2(\text{g})\to 2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{e}}^{-}{{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{\text{°}}}_{{\text{H}}^{\text{+}}{\text{/H}}_{\text{2}}}\text{=0.00V}$

Cathode: ${\text{Cu}}^{\text{+2}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Cu}\left(\text{s}\right){{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}E}}^{°}}_{{\text{Cu}}^{2+}/\text{Cu}}=0.34\text{V}$

${\mathbf{E}}_{\text{cell}}^{°}={\mathbf{E}}_{\text{cathode}}^{°}-{\mathbf{E}}_{\text{anode}}^{°}$

$\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{=E}}_{{\text{Cu}}^{\text{2+}}\text{/Cu}}^{\text{°}}{\text{-E}}_{{\text{H}}^{\text{+}}{\text{/H}}_{\text{2}}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{o}}\text{=0.34V-0.00}\\ {\text{E}}_{\text{cell}}^{\text{o}}\text{=0.34V}\end{array}$

Calculating [Cu+2]

The Nernst equation relates the cell potential to the concentration of the chemical species in the solution through the following expression.

$\begin{array}{l}{\text{E}}_{\text{cell}}{\text{=\hspace{0.17em}E}}_{\text{cell}}^{\text{°}}\text{-}\frac{\text{RT}}{\text{nF}}\text{lnQ}\\ {\text{E}}_{\text{cell}}{\text{=\hspace{0.17em}E}}_{\text{cell}}^{\text{°}}\text{-}\frac{\text{8.314\hspace{0.17em}J/Kmol×298\hspace{0.17em}K}\times \text{2.303}}{\text{n×96500C/mol}}\text{logQ}\\ {\text{E}}_{\text{cell}}{\text{=\hspace{0.17em}E}}_{\text{cell}}^{\text{°}}\text{-}\frac{0.0592\text{\hspace{0.17em}\hspace{0.17em}V}}{\text{n}}\text{logQ}\end{array}$

$\begin{array}{c}\text{0.22=0.34-}\frac{\text{0.0592}}{\text{2}}\text{log}\frac{\left[{\text{H}}^{\text{†}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}\\ \text{0.22-0.34=\hspace{0.17em}-0.0296}\times \text{log}\frac{\text{1M}}{\left[{\text{Cu}}^{\text{2+}}\right]}\\ \text{-0.12=-0.0296}\times \text{log}\frac{\text{1M}}{\left[{\text{Cu}}^{\text{2+}}\right]}\\ \text{4.05=ln}\frac{\text{1M}}{\left[{\text{Cu}}^{\text{2+}}\right]}\\ {10}^{\text{4.05}}{\text{=1.12×10}}^{\text{4}}\text{=}\frac{\text{1M}}{\left[{\text{Cu}}^{\text{2+}}\right]}\\ \left[{\text{Cu}}^{\text{2+}}\right]\text{=}\frac{\text{1M}}{{\text{1.12×10}}^{\text{4}}}\\ \left[{\text{Cu}}^{\text{2+}}\right]{\text{=8.9×10}}^{\text{-5}}\text{\hspace{0.17em}\hspace{0.17em}M}\end{array}$

Hence, the concentration of ${\text{Cu}}^{\text{+2}}$ ions is $8.9\times {10}^{-5}\text{\hspace{0.17em}M}$.