Question

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?

Answer 1

Correct Answer - `[Br]: [CI] = 1:200`
`Ag|AgBr_((s)) |AgCI_((s)) |CI^(-)|Ag`
`E_(cell) =(0.0591)/(1)log. ([Ag^(+)]_(R.H.S))/([Ag^(+)]_(L.H.S))`
For `L.H.S`.
`K_(sp) of AgBr = 5 xx 10^(-13)`
`[Ag^(+)] [Br^(-)] = 5 xx 10^(-13)` ltbr? `[Ag^(+)]_(L.H.S) = (5 xx 10^(-13))/([Br^(-)])`
One for `R.H.S`
`[Ag^(+)] [CI^(-)] = K_(sp) of AgCI`
`[Ag^(+)]_(R.H.S) = (10^(-10))/([CI^(-1)])`
`10^(@) = (|Br^(-)|xx10^(10))/(|CI^(-)|xx5 xx 10^(-13))`
`(|Br^(-)|)/(|CI^(-)|) =(5)/(10^(3)) = (1)/(200)`