At `Tl^(+) |Tl` couple was prepared by saturating 0.1 M KBr with TlBr and allowing the `Tl^(+)` from the relatively insoluble bromide to equilibrate. This couple was observed to have a potential of `-0443`V with respect to `Pb6(2+)|Pb` couple in which `Pb6(2+)` was 0.1 M. What is `K_(sp)` of TlBr?
(Report answer in multiplication of `10^(-8)`)
`(E_(Pb^(2+)//Pb)^(@) = -0.126`, `(E_(Tl^(+)//Tl)^(@) = -0.336V)`
(Take antilog(0.5509) = 3.55, (2.303RT)/(F) = 0.059)
[Hint: Take Pb as anode]

Answer

355

Explanation

`Pb^(2+)+2e^(-)toPb` ltbr `E_(Pb^(2+)//Pb)^(0)=E_(Pb^(2+)//Pb)^(0)-(0.0591)/(2)log((1)/(0.1))=-0.1555"volt"`
`E_(cell)=E_(Pb^(2+)//Pb)-E_(Tl^(+)//Tl)`
`E_(Tl^(+)//Tl)-0.5985V`
`Tl^(+)+e^(-)toTl`
`E_(Tl^(+)//Tl)=E_(Tl^(+)//Tl)^(0)-0.059log((1)/((Tl^(+))))`
`-0.5985=-0.336+0.059log((Tl^(+))`
`Tl^(+)=3.55xx10^(-5)M`
`KsP=[Tl^(+)][Br^(-)]=3.55xx10^(-5)xx0.1=3.55xx10^(-6)M^(2)=355xx10^(-8)M^(2)`