Determine the overall reaction and its standard cell potential at \({25circ} {\rm{C}}\) for this reaction. Is the reaction spontaneous at standard conditions?

\({\rm{Cu}}(s)\left| {{\rm{C}}{{\rm{u}}^{2 + }}(aq) || {\rm{A}}{{\rm{u}}^{3 + }}(aq)} \right|{\rm{Au}}(s)\)

For positive value of standard cell potential, the reaction isspontaneousand it isnon-spontaneousfor negative value of the standard cell potential.

In the given reaction copper is oxidized at anode and Au is reduced at cathode

Anode reaction:

\(Cu(s) \to C{u^{2 + }}(aq) + 2{e^ - }\quad {E^o} = + 0.34V\)

Cathode reaction:

\(A{u^{3 + }}(aq) + 3{e^ - } \to Au(s)\quad {E^o} = + 1.498\;{\rm{V}}\)

The overall or net cell reaction is as follows:

\(\begin{aligned}3\left( {{\rm{Cu}}(s) \to {\rm{C}}{{\rm{u}}^{2 + }}(aq) + 2{e^ - }} \right)\\\frac{{2\left( {A{u^{3 + }}(aq) + 3{e^ - } \to Au(s)} \right)}}{{3Cu(s) + 2A{u^{3 + }}(aq) \to 3C{u^{2 + }}(aq) + 2Au(s)}}\end{aligned}\)

Now calculate the standard cell potential at \({25\circ }{\rm{C}}\), using the following expression:

\(\begin{aligned}E_{{\rm{cell }}}^o &= E_{{\rm{cathode }}}^o - E_{{\rm{anode }}}^o\\E_{{\rm{cell }}}^o &= + 1.498\;{\rm{V}} - ( + 0.34\;{\rm{V}})\\ &= + 1.158\;{\rm{V or }} + 1.16\;V\end{aligned}\)

Here the standard cell potential is positive, and then the reaction is spontaneous.