# 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, EAg+/Ag = EoAg+/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

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

Regards

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