A galvanic cell consists of a Mg electrode in \({\bf{1M}}\)\({\bf{Mg}}{\left( {{\bf{N}}{{\bf{O}}_{\bf{3}}}} \right)_{\bf{2}}}\)solution and a Ag electrode in 1M AgNO solution. Calculate the standard cell potential at \({25^\circ }{\rm{C}}\).

In electrochemistry, a 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. A galvanic cell consists of two half cells. In the 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 the cell.

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

- Given-

Electrodes

\({\rm{Mg}}\)in \(1{\rm{M Mg}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\) solution

\({\rm{Ag}}\)in \(1{\rm{M AgN}}{{\rm{O}}_3}\) solution

Temperature \( = {25^\circ }{\rm{C}}\)

- Reduction of half-reactions-

\(\begin{aligned}{}{\rm{M}}{{\rm{g}}^{2 + }}({\rm{aq}}) + 2{{\rm{e}}^ - } \to {\rm{Mg}}({\rm{s}})\quad \\{{\rm{E}}^\circ }{\rm{ red }} = - 2.372\;{\rm{V}}\\{\rm{Ag}}_{({\rm{aq}})}^ + + 1{{\rm{e}}^ - } \to {\rm{Ag}}({\rm{s}})\quad \\{{\rm{E}}^\circ }{\rm{ red }} = + 0.7996\;{\rm{V}} - 2.372\;{\rm{V}} < + 0.7996\;{\rm{V}}\end{aligned}\)

Therefore, Mg is the anode electrode and Ag is the cathode electrode.

\(\begin{aligned}{}{E^\circ }{\rm{ cell }} &= + 0.7996\;{\rm{V}} - ( - 2.372\;{\rm{V}})\\{E^\circ }{\rm{ cell }} &= + 3.1716\;{\rm{V}}\end{aligned}\)

\({\rm{3}}{\rm{.17 V}}\) is the standard cell potential of the cell at \({25^\circ }{\rm{C}}\)