For the reaction,
`Ag_((aq.))^(+) + Cl_((aq.))^(-) hArr AgCl_((s))`
the `Deltag^(@)` Values for `Ag_(aq.)^(@), Cl_(aq.)` and `AgCl_(s)` are `+77, -129` and `-109 kj mol^(-1)`, Write the cell representation of above reaction and calculate `E^(@)` at `298 K`, also calculate the `log_(10) K_(sp)` of `AgCl` at `298K`.
(b) If `6.539 xx 10^(-2)g` of metallic zinc is added to `100 mL` saturated solution of `AgCl`, find the value of `log_(10).([Zn^(2+)])/([Ag^(+)]^(2))`. How many moles of `Ag` will be precipitated in this reaction ?
Given, `E_(Zn^(2+)//Zn)^(@) = -0.76 V`.

Explanation

`Ag_((s)) + (1)/(2)Cl_(2_(g))rarr AgCl_((s)), DeltaG^(@) = -109kJ` ....(1)
`Ag_((s)) rarr Ag_(kJ)^(+) + e, DeltaG^(@) = +77 kJ` ....(2)
`(1)/(2)Cl_(2(g))^(-) + e rarr Cl_((aq.))^(-), DeltaG^(@) = -129 kJ` ....(3)
By eq. `(1)` - `(2)` - `(3)`
`Ag_((aq.))^(+) + Cl_((aq.))^(-) rarr AgCl_((s)), DeltaG^(@) = -109 - 77 + 129 = -57 kJ`
`:' -DeltaG^(@) = nE^(@)F`
`57 xx 10^(3) = 1 xx E^(@)F`
`:. E^(@) = 0.59 V`
The cell is `Ag|AgCl_((s))|Cl_((aq.))^(-)||Ag_((aq.))^(+)|Ag`
`E_(cell) = E_(OP_(Ag//AgCl//Cl^(-))) -0.059log.(1)/([Cl^(-)])+E_(RP_(Ag^(+)//Ag)) + 0.059log [Ag^(+)]`
at equilibrium
`E_(cell) = 0`, thus
`0 = E_(Ag//AgCl//Cl^(-))^(@) + E_(RP_(Ag//AgCl//Cl^(-)))^(@) + 0.059 log [Ag^(+)][Cl^(-)]`
or `0 = E_(cell)^(@) + 0.059 log K_(SP)`
or `E_(cell)^(@) = -0.059 log K_(SP)`
or `logK_(SP) = -(E_(cell)^(@))/(0.059) = - (0.59)/(0.059) = -10`
`:. K_(SP)AgCl = 1 xx 10^(-10) M^(2)`
(b) Let solubility of `AgCl` be `S`, then
`S = sqrt(K_(SP)) = sqrt(10^(-10)) = 10^(-5) M`
Total milli-mole of `AgCl` in its saturated solution of `10^(-5)M` in `100 mL = 10^(-5) xx 10^(2) = 10^(-3)`
`:.` Mole of `AgCl` in `100 mL` solution
`= 10^(-3) xx 10^(-3) = 10^(-6)`
Also mole of `Zn` added `= (6.539 xx 10^(-2))/(65.39) = 10^(-3)`
`DeltaG^(@)` for `Ag rarr Ag^(+) + e` is `77 kJ`
`DeltaG^(@) = nE^(@)F`
`:. E_(Ag//Ag^(+))^(@) = (-77 xx 10^(3))/(1 xx 96500) = - 0.80 V`
For the redox change on addition of `Zn`
`{:(ZnrarrZn^(2+)+2e,E^(@)=+0.76),(2Ag^(+)+2erarr2Ag,E^(@)=+0.80):}/(Zn+2Ag^(+)rarrZn^(2+)+2Ag,E_(cell)^(@)=0.76+0.80=1.56V)`
`E_(cell) = E_(OP_(Zn))^(@) + E_(RP_(Ag))^(@) + (0.059)/(2)log.([Ag^(+)]^(2))/([Zn^(2+)])`
At equilibrium, `E_(cell) = 0`
`:.= E_(cell)^(@) +(0.059)/(2)log.([Ag^(+)]^(2))/([Zn^(2+)])`
`:. log.([Zn^(2+)])/([Ag^(+)]^(@)) = (E_(cell)^(@) xx 2)/(0.059) = (1.56 xx 2)/(0.059) = 52.88`
`K_(c) =([Zn^(2+)])/([Ag^(+)]^(2)) = 7.6 xx 10^(52)`
Since `K_(c)` is appreciably high and thus nearly whole of the `Ag^(+)` is converted to`Ag`. Thus moles of `Ag` formed `=` moles of `Ag^(+)` in `100 mL` saturated solution
`= (10^(-3))/(10^(3)) = 10^(-6)`