Pls answer this:
At  what ratio of [Br-], [Cl-] ions, emf of cell would be zero. Ksp AgBr = 5x10^​-13 ; Ksp AgCl = 10^-10 
Ag | AgBr_(s) , Br-_(aq) || AgCl_(s) , Cl-_(aq) | Ag

Dear Student

The half cell may be regarded as Ag ​| Ag⁺ on the left and right side of the cell with lower concentration of Ag⁺ ions.

Therefore, E_Ag+/Ag = E^o_Ag+/Ag + $\frac{0.0591}{n}$ log [Ag⁺]
EMF of the cell would be zero, if the electrode potential of both the half-cells remain the same. Therefore, the concentration of [Ag⁺] can be considered in both the cases.
So, ​[Ag⁺] [Br^-] = 5 x 10^-13 ;        ​[Ag⁺] [Cl^-] = 1 x 10^-10

∴ $\frac{\left[B{r}^2\right]}{\left[C{l}^{-}\right]}$ = $\frac{5 x {10}^{-13}}{1 x {10}^{-10}}$ = 5 x 10^-13+10 = 5 x 10^-3 = $\frac{5}{1000} = \frac{1}{200}$

Regards