For the cell Mg (s) `| Mg^(2+) (aq) ||Ag^(+)(aq) |Ag(s),` calculate the equilibrium constant at `25^(@)` C and maximum work that work that can be obtained during operation of cell .Given : `E_(Mg^(2+)|Mg)^(@)=-273V` and `E_(Ag^(2+)|Ag)^(@)=0.80` V

Explanation

oxidation at anode
`Mg rarr mg^(2+) + 2e^(-) ………… (1) " " (E_("ox")^(@) = 2.37 V `
Reduction at cathode
`Ag^(+) + e^(-) rarr Ag ……….. (2) " " (E_("red")^(@)) = 0.80 V `
`:. E_("cell")^(@) = (E_("ox")^(@))_("anode")+ (E_("red")^(@))_("cathode")`
`= 2.37 + 0.80 `
`= 3.17 V `
Overall reaction
Equation (1) + 2 `xx` equation (2) `implies`
`Mg+ 2Ag^(2+) rarr Mg^(2+) +2Ag`
`Delta G^(@) = - nfE^(@) `
` = - 2xx 96500 xx 3.17`
`= 611.810 `J
`DeltaG^(@) = -6.12 xx 10^(5)` J
`W =6.12 xx 10^(5) J`
`DeltaG^(@) = -2.803 ` RT `log^(k)C`
`implies ` log `K_(c ) = (6.12 xx 10^(5))/(2.803xx 8.314 xx 298)`
`K_(c ) ` = Antilog of (107 .2)