Calculate emf of a cell Ag(s) $/{\mathrm{Ag}}^{+}$(1M) /${\mathrm{Mg}}^{+2}$(1M) / Mg(s) and also predict the spontaneity of cell reaction.$\left({\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^0=+0.8\mathrm{V};{\mathrm{E}}_{{\mathrm{Mg}}^{+2}/\mathrm{Mg}}^{\mathrm{o}}=-2.37\mathrm{V}\right)$

LHE ….Anode

RHE ….Cathode

Both the values given are SRP values

$\therefore {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}={\mathrm{E}}_{\mathrm{ca} \mathrm{thode}}^{\mathrm{o}}-{\mathrm{E}}_{\mathrm{anode}}^{\mathrm{o}}$

            = (–2.37V)–(+0.8V)

           = –3.17V

$\therefore {\mathrm{E}}_{\mathrm{cell}}^0$ is negative, the cell reaction is non-spontaneous