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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

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}

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

^{-13+10}= 5 x 10

^{-3}= $\frac{5}{1000}=\frac{1}{200}$

Regards

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