91Ó°ÊÓ

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.

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

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

Anode:${\text{H}}_2(\text{g})→2{\text{H}}^{+}\left(\text{aq}\right)+2{\text{e}}^{−}{{\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰E}}^{\text{°}}}_{{\text{H}}^{\text{+}}{\text{/H}}_{\text{2}}}\text{=0.00V}$

Cathode: ${\text{Cu}}^{\text{+2}}\left(\text{aq}\right)+2{\text{e}}^{−}→\text{Cu}\left(\text{s}\right){{\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰E}}^{°}}_{{\text{Cu}}^{2+}/\text{Cu}}=0.34\text{V}$

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

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

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

$\begin{array}{l}{\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{°}}\text{-}\frac{\text{RT}}{\text{nF}}\text{lnQ}\\ {\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{°}}\text{-}\frac{\text{8.314 J/°­³¾´Ç±ô×298‿é}×\text{2.303}}{\text{²Ô×96500°ä/³¾´Ç±ô}}\text{logQ}\\ {\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{°}}\text{-}\frac{0.0592\text{ â¶Ä‰V}}{\text{n}}\text{logQ}\end{array}$

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

Hence, the concentration of ${\text{Cu}}^{\text{+2}}$ ions is $8.9×{10}^{−5}\text{ M}$.