For the cell `Mg|Mg^(2+)||Ag^(+)|Ag` calculate equilibrium constant at `25^(@)C` and also the maximum work that can be obtained from it.
`E^(@)(Mg^(2+),Mg)=-2*37V and E^(@)(Ag^(+),Ag)=+0*80V`.

Explanation

Step I: To calculate EMF
`Mg|Mg^(2+)||Ag^(+)|Ag`
Half cell reaction are,
Oxidation: `Mg(s) to Mg^(2+)(aq)+2e^(-)`
Reductioni: `[Ag^(+)(aq)+e^(-) to Ag(s)]xx2`
`Mg(s)+2Ag^(+)(aq)toMg^(2+)(aq)+2Ag(s)`
Thus n=2
`EMF^(@)=E^(@)(Ag^(+)//Ag)-E^(@)(Mg^(2+)//Mg)`
`=0*80V-(-2*37V)`
`=(0*80+2*37)V=3*17V`