a. For the reaction
Given : `Ag^(o+)(aq)+Cl^(o+)(aq)rarr AgCl(s)`

Write the cell representation of above reaction and calculat `E_("cell")^(@)` at 298K.
Also find the solubility product of AgCl.
b. If `6.539xx10^(-2)g` of metallic zinc is added to 100ml saturated solution of AgCl. Find the value of
`log_(10).([Zn^(2+)])/([Ag^(+)]^(2))`
How many moles of Ag will be preciptated in the above reaction. Given that
`Ag^(+)+e^(-)rarrAg, E^(@)=0.80V`
`Zn^(2+)+2e^(-)rarr Zn,E^(@)=-0.76V`

Explanation

a From the given details, the reactions can be written as: At anode:
`Ag(s) +Cl^(-)(aq) rarr AgCl (s)+e^(-)`
At anode: `Ag^(+)(aq) +e^(-) rarr Ag(s)`
Complete reaction
`Ag^(+)(aq)+Cl^(-)(aq) rarr AgCl (s)`
Hence cell representation is
`Ag(s)//AgCl//Cl^(-)(aq)//Ag^(+)(aq)//Ag(s)`
`DeltaG^(@) = DeltaG_(f)^(@) (AgCl) -[DeltaG_(f)^(@) (Ag^(+))+DeltaG_(f)^(@) (Cl^(-))]`
`=- 109 - (-129 +77) =- 57 kJ//mol`
We know that `DeltaG^(@) =- nFE_(cell)^(@)`
`-57000 = - 1 xx 96500 xx E_(cell)^(@)`
`( :. n=` electron transferred = 1)
`E_(cell)^(@) = (57000)/(96500) = 0.59` volts
Again `E_(cell)^(@) = (0.0591)/(n)logK_(c)`
or `K_(cell)^(@) = (0.0591)/(1)log.(AgCl)/([Ag^(+)][Cl^(-)])`
`E_(cell)^(@) = (0.0591)/(1)log ((1)/(k_(sp)))`
`:. k_(sp) = 10^(-10)`
b. When Zn is added to 100ml of saturated `AgCl` solution.
`2Ag^(+) + Zn(s) underset(larr)(rarr)2Ag(s) + Zn^(2+)`
`Ag^(+) +e^(-) underset(larr)(rarr)Ag: E^(@) = 0.80V`
`Zn^(2+) +2e^(-) underset(larr)(rarr) Zn, E^(@) =- 0.76V`
`E_(cell)^(@) = E_(Ag^(+)(s))^(@) - E_(Zn^(2+)//ZN(s))^(@) = 1.56V`
`E_(cell)^(@) = (0.059)/(n)log_(10). ([Zn^(2+)])/([Ag^(+)]^(2))`
`rArr 1.56 = (0.059)/(2)log_(10).([Zn^(2+)])/([Ag^(+)]^(2))`
`rArr log_(10).([Zn^(2+)])/([Ag+]^(2)) = 52.9`
As the value of equilibrium constant is very high so the reaction moves in forward direction completely.
`[Ag^(+)]` from `(a) = sqrt(10^(-10)) = 10^(-5)`
`[ : K_(sp) = 10^(-10) = [Ag^(+)] [Cl^(-)]]`
`:. Ag^(+)` in 100ml of solution
`= (10^(-5) xx 100)/(1000) = 10^(-6) mol`