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Consider the cell AG|AgBr(s)|Br^(-)||AgC...

Consider the cell `AG|AgBr(s)|Br^(-)||AgCI(s)|CI^(-)|Ag` at `25^(@)C`. The solubility product constants of `AgBr & AgCI` are respectively `5 xx 10^(-13) & 1 xx 10^(-10)`. For what ratio of the concentration of `Br^(-) & CI^(-)` ions would the emf of the cell be zero?

A

200

B

`(1)/(200)`

C

`(1)/(300)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
let x and y are the concentration of `Br^(-)` and `Cl^(-)` at equilibrium when `E_(cell)=0`
`therefore[Ag^(+)]_(LHS)=(K_(sp)(ArBr))/([Br^(-)])=(5xx10^(-13))/(x)`
`[Ag^(+)]_(RHS)=(K_(sp)(AgCl))/([Cl^(-)])=(1xx10^(-10))/(y)`
Also `E_(cell)=E_(Ag//Ag^(+))^(@)(RHS)+E_(Ag//Ag^(+))^(@)(LHS)+(0.0591)/(1)log(([Ag^(+)]_(RHS))/([Ag^(+)]_(LHS)))`
`0=0+(0.0591)/(1)log((1xx10^(-10)xx x)/(5xx10^(-13)xxy))`
`therefore(x)/(y)=(1)/(200)`
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