The `EMF` of the cell `:`
`Ag|Ag_(2)CrO_(4)(s),K_(2)CrO_(4)(0.01 M)||AgNO_(3)(0.01M)||Ag` is `206.5mV.` Calculate the solubility of `Ag_(2)CrO_(4)` in `1M Na_(2)CrO_(4)` soluion.

Explanation

It is a concentration cell, therefore, `E^(c-)._(cell)=0`.
`Ag_((Anode))rarr AG^(o+)._((Anode))+e^(-)`
`Ag^(o+)._((cathode))+e^(c-) rarrAg_((Cathode))`
`ulbar(Ag^(o+)._((Cathode))rarr Ag^(o+)._((Anode)))`
`E=E^(c-)._(cell)log[(Ag^(o+)._((Anode))("let it is x"))/(Ag^(o+)._((cathode))("which is 0.01 M"))]`
`0.2065=-(0.059)/(1)log.(x)/(0.01M)`
`-(02065)/(0.059)=-log x-log10^(-2)`
`log x =-2-3.5=-5.5=bar(6).5`
Taking Antilog `implies x=3.163 xx 10^(-6)M`

`K_(sp)=(x^(2))(0.01)=(3.163xx10^(-6))(0.01)=10^(-13)`
`K_(sp)-[Ag^(o+)][CrO_(4)^(2-)]`
`K_(sp)=S^(2)xx(1M)`
`[Ag^(o+)]=S=sqrt(K_(sp))`
`[Ag^(o+)]` in `1 M Na_(2)CrO_(4)=sqrt(K_(sp))`
`=sqrt(10^(-13))=sqrt(10xx10^(-14))`
`=3.16xx10^(-7)M`