Consider the cell `Ag(s)|AgBr(s)Br^(c-)(aq)||AgCl(s),Cl^(c-)(aq)|Ag(s)` at `298 K`. The `K_(sp)` of `AgBr` and `AgCl`, respectively are `5xx10^(-13)` and `1xx10^(-10)` . At what ratio of `[Br^(c-)]` and `[Cl^(c-)]` ions, `EMF_(cell)` would be zero ?

Options

A. `1:200`
B. `1:100`
C. `1:500`
D. `200:1`

Answer

A

Explanation

`E_(Br^(-)//AgBr//Ag)^@=E_(Ag^+//Ag)^@+0.059/1"log"K_(SP)AgBr=E_(Ag^+//Ag)^@-0.7257`
and `E_(BCl^(-)//AgCl//Ag)^@=E_(Ag^+//Ag)^@+0.059/1"log"K_(SP)AgCl=E_(Ag^+//Ag)^@-0.59`
Now cell reaction is
`({:(Ag+Br^(-) toAgBr+1e^(-)),(AgCl+1e^(-)toAg+Cl^(-)):})/ {:(Br^(-)+AgCloverset(1e^(-))toCl^(-)+AgBr):}`
`0=(0.7257-0.59)+0.059/1"log"([Br^(-)])/([Cl^(-)])implies ([Br^(-)])/([Cl^(-)])=0.005`