What is the cell potential for the following reaction at room temperature?

\({\bf{Al(s)}}\left| {{\bf{A}}{{\bf{l}}^{{\bf{3 + }}}}{\bf{(aq,0}}{\bf{.15M)}} | | {\bf{C}}{{\bf{u}}^{{\bf{2 + }}}}{\bf{(aq,0}}{\bf{.025M)}}} \right|{\bf{Cu(s)}}\)

What are the values of \(n\) and \(Q\) for the overall reaction? Is the reaction spontaneous under these conditions?

In electrochemistry, Galvanic cellis a kind of electrochemical cell in which the current is produced using a redox reaction. Redox reactions involve oxidation as well as reduction. Galvanic cell consists of two half cells. In one half cell, oxidation occurs. This half-cell acts as the anode. In the other half cell, reduction occurs, This half-cell is termed as the cathode. These two half cells work together and constitute an electrochemical cell.

\({E^\circ }\) cell \( = {E^\circ }\) red ( cathode\() - {E^\circ }\) red ( anode )were,\({E^\circ }\) cell = standard emf of cell .

\({E^\circ }\) red = standard reduction potential. And \({E^\circ }\) red ( cathode\() > {E^\circ }\) red( anode )

At anode half-cell.

\(Al(s) \to A{l^{3 + }} + 3{e^ - } \ldots \ldots \ldots .1\)

At cathode half-cell.

\(C{u^{2 + }} + 2{e^ - } \to Cu(s) \ldots \ldots ..2\)

Multiplying equation 1 by 2 and equation 2 by 3 and adding them.

\(\begin{array}{}2{\rm{Al}}(s) \to 2{\rm{Al}}{l^{3 + }} + 6{e^ - }\\3{\rm{C}}{{\rm{u}}^{2 + }} + 6e \to 3{\rm{Cu}}(s) - \ldots - \ldots - 3\end{array}\)

\((overall\)\({\rm{ reaction }}) \Rightarrow 2Al(s) + 3C{u^{2 + }} \to 2A{l^{3 + }} + 3Cu(s)\)

The value of \({\bf{n}} = {\bf{6}}{{\bf{e}}^ - }(\)answer)

Now calculating \(Q = {K_{eq}} = \frac{{{{\left[ {A{l^{ + 3}}} \right]}^2}{{[Cu]}^3}}}{{{{[Al]}^2}{{\left[ {{\rm{C}}{{\rm{u}}^{2 + }}} \right]}^3}}}\)

Here \([{\rm{Al}}]\) solid and \([{\rm{Cu}}]\) solid is taken as 1

\(\begin{array}{l}Q = \frac{{{{\left[ {A{l^{3 + }}} \right]}^2}}}{{{{\left[ {C{u^{2 + }}} \right]}^3}}}\\Q = \frac{{{{(0.15)}^2}}}{{{{(0.025)}^3}}}\\Q = 1442.30({\rm{ answer }})\end{array}\)

To predict the spontaneity of the above reaction, we use the following equations,

\(\begin{array}{}\Delta {G^0} = \frac{{{\bf{0}}.{\bf{0591}}}}{{\bf{n}}}\log Q \ldots \ldots ..3\\\Delta {G^0} = - \frac{{0.0591}}{6}\log (1442.30)\\\Delta {G^0} = - 0.00985 \times 3.159\\\Delta {G^0} = - 0.031J\end{array}\)

Above reaction is spontaneous as \(\Delta {G^0} < 0\).