Calculate `Delta_(r)G^(@)` of the following reaction:
`Ag^(+)(aq)+Cl^(-)(aq)rarr AgCl(s)`
Given: `Delta_(r)G^(@)(AgCl)=-"109kJ.mol"^(-1)`,
`Delta_(r)G^(@)(Cl^(-))=-"129kJ.mol"^(-1), Delta_(r)G^(@)(Ag^(+))="77.kJ.mol"^(-1)`. Represent the reaction in the form of `E^(@)` of the cell. Find `log_(10)K_(sp)" of "AgCl`.

Explanation

Half - cell reactions are -
`{:(" "Ag^(+)(aq)+e" "rarr" "Ag(s)),(" "Ag(s)+Cl^(-)(aq)" "rarr" "AgCl(s)+e),(bar("Cell reaction: "Ag^(+)(aq)+Cl^(-)(aq)rarrAgCl(s))):}`
The cell is : `Ag|AgCl(s)|Cl^(-)(aq)||Ag^(+)(aq)|Ag`
`Ag^(+)(aq)+Cl^(-)(aq)rarrAgCl(s)`
`therefore DeltaG^(@)=DeltaG_(r)^(@)(AgCl)-DeltaG_(r)^(@)(Ag^(+))-DeltaG_(r)^(@)(Cl^(-))`
`"or, "DeltaG^(@)=[-109-77-(-129)]"kJ.mol"^(-1)`
`"or, "DeltaG^(@)=-"57 kJ.mol"^(-1)`
But, `DeltaG^(@)=-nFE^(@)" or, "-57000=-1xx96500xxE^(@)`
`"or, "E^(@)=(57000)/(96500) or, E^(@)=0.59V`
The solubility equilibrium for `AgCl` is
`AgCl(s)hArrAg^(+)(aq)+Cl^(-)(aq)`
For this reaction `E_("cell")^(@)=-0.59V`
`therefore log_(10)K_(sp)=(nFE^(@))/(RT)=-(0.59)/(0.059)=-10`