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

Answer

a

Explanation

Cell reaction `:`
`Ag(s) +Ag^(o+)(` from `AgCl) rarr Ag(s)+Ag^(o+)(` from `AgBr)`
`E_(cell)=E^(c-)._(cell)-(0.059)/(1)log .([Ag^(o+)]_(AgBr))/([Ag^(o+)]_(AgCl))`
`=0-0.059log .((K_(sp)(AgBr))/([Br^(c-)])xx([Cl^(c-)])/(K_(sp(AgCl))))`
For `E_(cell)=0`
`(K_(sp)(AgBr)xx[Cl^(c-)])/(K_(sp)(AgCl)xx[Br^(c-)])=1`
`([Cl^(c-)])/([Br^(c-)])xx(5xx10^(-13))/(1xx10^(-10))implies[(Cl^(c-))/[Br^(c-))]=(1)/(200)`
`[(Br^(c-))/(Cl^(c-))]=200:1`