(a). Calculate `triangleG_(f)^(@)` of the following raection:
`Ag^(+)(aq)+Cl^(-)toAgCl(s)`
given `triangleG_(f)^(@)(AgCl)=-109kJ//"mole",triangleG_(f)^(@)(Cl^(-))=-129kJ//"mole",triangleG_(f)^(@)(Ag^(+))=77kJ//"mole"`
Represent the above reaction in form of a cell.
Calculate `E^(@)` of the cell. find `log_(10)K_(sp)` of `AgCl` at `25^(@)C`
(b). `6.539xx10^(-2)g` of metallic Zn (atomic mass `=65.39amu`) was added to 100 mL of saturated solution of AgCl.
Calculate `log_(1)([Zn^(2+)])/([Ag^(+)]^(2))` at equilibrium at `25^(@)C` given that
`Ag^(+)+e^(-)toAg" "E^(@)=0.80V`
`Zn^(2+)+2e^(-)toZn" "E^(@)=-0.76V`
Also find how many moles of Ag will be formed (take `(114)/(193)=0.59,(1.56)/(0.059)=26.44)`

Answer

A, B

Explanation

(a). `triangleG_(r)^(@)=-109+129-77=-57kJ//mol`
cell representation `Ag|AgCl||Cl^(-)|Ag^(+)|Ag`.
`-1xx96500xxE^(@)=057xx10^(3)`
`E^(@)=0.59"volt"`
`0=0.59-(0.059)/(1)log((1)/(K_(SP)))`
`logK_(SP)=-10`.
(b). `ZntoZn^(2+)+2e^(-)," "0.76"volt"`
`underline(2Ag^(+)+2e^(-)to2Ag," "0.80"volt")`
`underline(Zn+2Ag^(+)toZn^(2+)+2Ag,) " "E_(cell)^(@)=1.56"volt"`
`n_(2n)=(6.539)/(65.39)=10^(-3)mol" "[Ag^(+)]=sqrt(K_(sp))=10^(-5)M`
`0=1.56-(0.059)/(2)logK" "n_(Ag^(+))=10^(-5)xx0.1=10^(-6)M`.
`n_(Ag)=10^(-6)molimplieslogK=52.8`