249. For the standard cell, Cu(s) | Cu²+(aq) || Ag⁺ (aq) | Ag(s)
Given: E° _Cu²⁺/Cu = 0.34V, E° = 0.80V
(a) Identify cathode and anode as the current is drawn through the cell.
(b) Write the reactions taking place at the electrodes.
(c) Calculate the standard cell potential.

$$\begin{gathered} \mathrm{a}){\mathrm{E}}_{\raisebox{1ex}{${\mathrm{Cu}}^{2+}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Cu}$}\right.}^0 <{\mathrm{E}}_{\raisebox{1ex}{${\mathrm{Ag}}^{+}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Ag}$}\right.}^0 \\ \mathrm{Therefore} \mathrm{copper} \mathrm{electorode} \mathrm{isthe} \mathrm{anode} \mathrm{and} \mathrm{silver} \\ \mathrm{electrode} \mathrm{act} \mathrm{as} \mathrm{a} \mathrm{cathode}. \mathrm{The} \mathrm{half} \mathrm{reaction} \mathrm{are}: \\ \mathrm{anode}: \mathrm{Cu}(\mathrm{s})\to {\mathrm{Cu}}^{2+} +2{\mathrm{e}}^{-} \\ \mathrm{Cathode}: 2{\mathrm{Ag}}^{+} +2{\mathrm{e}}^{-} \to 2\mathrm{Ag} \\ \mathrm{b}) \mathrm{net} \mathrm{reaction} : \\ \mathrm{Cu}\left(\mathrm{s}\right) +2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right) \to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right) +2\mathrm{Ag}\left(\mathrm{s}\right) \\ \mathrm{c}) \mathrm{The} \mathrm{potential} \mathrm{of} \mathrm{the} \mathrm{cell} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{E} ={\mathrm{E}}_{\mathrm{cell}}^0 -\frac{0.059}{\mathrm{n}}\log \frac{\left[{\mathrm{Cu}}^{2+}\right]}{[{\mathrm{Ag}}^{+}{]}^2} \\ 0 =(+0.80-0.34) -\frac{0.059}{2}\log \frac{0.01}{[{\mathrm{Ag}}^{+}{]}^2} \\ \log \left[\frac{0.01}{({\mathrm{Ag}}^{+}{)}^{2+}}\right] =\frac{ 2 \mathrm{x} 0.46}{0.059} =15.593 \\ \mathrm{taking} \mathrm{antilog} \\ \frac{ \\ 0.01}{[{\mathrm{Ag}}^{+}{]}^2} =\mathrm{antilog} 15.593 =3.919 \mathrm{x} {10}^{15} \\ [{\mathrm{Ag}}^{+}{]}^2 =\frac{0.01}{3.919 \mathrm{x}{10}^{15}} =2.55 \mathrm{x}{10}^{-18} \\ \left[{\mathrm{Ag}}^{+}\right] =1.59 \mathrm{x} {10}^{-9}\mathrm{M} \end{gathered}$$