For the cell `Mg(s) |Mg^(2+)(aq)||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)^@ = + 2.37 V and E_(Ag^(2+)|Ag)^@ = 0.80V`

Explanation

`{:("Oxidation at anode", Mg to Mg^(2+)+2e^(-), (E_("ox")^(@)) = 2.37V),("Reduction at cathode ", Ag^+ + e^(-) to Ag, (E_("red")^@) = 0.80 V):}`
`:. E_("cell")^(@) = (E_("ox")^(@))_("anode") + (E_("red")^(@))_("cathode") = 2.37 + 0.80 = 3.17V`
Overall reaction,
`Mg + 2Ag^(+) to Mg^(2+) +2Ag`
`DeltaG^@ = -nFE^@`
`= -2 xx 96400 xx 3.17`
`= -6.118 xx 10^5J`
We know that `W_("max") = DeltaG^@`
`:. W_("max") = +6.118 xx 10^5 J`
Relationship between `DeltaG^@` and `K_(eq)` is
`Delta G = -2.303 RT log K_(eq)`
`DeltaG = -2.303 xx 8.314 xx 298 log K_(eq) " " [ :. 25^@C = 298 K]`
`log K_(eq) = (6.118 xx 10^5)/(2.303 xx 8.314 xx 298) implies (6.118 xx 10^5)/(5705.84)`
`log K_(eq) = 107.223`
`K_(eq) = "Antilog" (107.223)` .