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)^(0) = - 237 V " and "E_(Ag^(2+)//Ag)^(0) = 0.80 V`.

Explanation

Oxidation at anode :
`Mg to Mg^(2+)+2e^(-) ` ....(1)
`(E_("ox")^(0)) = 2.37 V`
Reduction at cathode: `Ag^(+) + e^(-) to Ag `...(2)
`(E_("red")^(0)) = 0.80 V`
`E_("cell")^(0) = (E_("ox")^(0))_("anode") + (E_("rod")^(0))_("cathode")`
` = 2.37 + 0.80`
`= 3.17V`
Overall reaction :
Equation (1) `+ 2 xx " equation (2) " rArr`
`Mg + 2Ag^(2+) to Mg^(2+) + 2Ag`
` DeltaG^(0) = - nFE^(0)`
` = - 2 xx 96500 xx 3.17`
` = 611.810 J`
`DeltaG^(0) = - 6.12 xx 10^(5)J`
`W = 6.12 xx 10^(5)J`
` Delta G^(0) = - 2.803 " RT log " K_(C) `
` rArr log K_(C) = (6.12 xx 10^(5))/(2.803 xx 8.314 xx 298)`
`K_(C) = "Antilog of " (107.2)`