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Consider the cell Ag| AgBr Br^(-) ||CI|A...

Consider the cell `Ag| AgBr Br^(-) ||CI|AgCl Ag at 25^(@)` C the solubility product of AgCl and AgBr are `1xx10^(@)` and `5xx10^(-13)` respectively. At which ratio of concentration of Br and Clions would the emf of cell be zero?

A

`100:1`

B

`1:100`

C

`200:1`

D

`1:200`

Text Solution

Verified by Experts

The correct Answer is:
C
`Ag|Ag^(+)||Ag^(+)Ag`
`E=E^(@)-(0.59)/(1)log (Ag^(+))/(Ag^(+))` therefore `(Ag^(+))/(Ag^(+))=1`
`(Br^(-))/(CI^(-))=(5xx10^(-13))/(1xx10^(-10))=1/200`
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