For the cell
`Mg_((s))abs(Mg^(2+)""_((aq)))abs(Ag^(+)""_((aq)))Ag_((s)),`
calculate the equilibrium constant at `25^(@)C` and maximum work that can be obtained during operation of cell. Given :
`E_(Mg^(2+)|Mg)^(@)=-237V " and " E_(Ag^(2+)|Ag)^(@)=0.80V`

Explanation

oxidation at anode
`Mg rarr mg^(2+)+2e^(-)" ...(1)"`
`" "(E_("ox")^(@))=2.37V`
Reduction at cathode
`Ag^(+)+e^(-) rarr Ag " ...(2)"`
`" "(E_("red")^(2))=0.80V`
`therefore E_("cell")^(@)=(E_("ox")^(@))_("anode")+(E_("red")^(@))_("cathode")`
`" "=2.37+0.80`
`" "=3.17V`
Overall reaction
Equation (1) + `2 times " equation (2)" rArr`
`Mg+2Ag^(2+) rarr Mg^(2+)+2Ag`
`" "DeltaG^(@)=-nfE^(@)`
`" "=-2 times 96500 times 3.17`
`" "=611.810J`
`DeltaG^(@)=-6.12 times 10^(5)J`
`W=6.12 times 10^(5)J`
`DeltaG^(@)=-2.803" RT log " K_(c)`
`rArr logK_(c)=(6.12 times 10^(5))/(2.803 times 8.314 times 298)`
`" "K_(c)=` Antilog of (107.2).