For galvanic cell:
`Ag|AgCI(s), KCI(0.2M|| KBr (0.001M), AgBr(s)|Ag` Calculate `EMF` generated and assign correct polarity to each electorde for spontaneous process after taking into accunt the celol reaction at `25^(@)C`.
`K_(sp) AgCI = 2.8 xx 10^(-10), K_(sp)AgBr = 3.3 xx 10^(-3)`.

Explanation

a. `LHS` electrode
`K_(sp) = [Ag^(oplus)] [CI^(Θ)] = [Ag^(oplus)] [0.2]`
`RHS` electrode `K_(sp) = [Ag^(oplus)] [Br^(Θ)] = [Ag^(oplus)] [10^(-3)]`
Calculate `[Ag^(o+)]` for each electrode and than calculate electrode potential and finally the `EMF`.
It is a concentration cell.
`:. E^(Θ) = 0`
`underset("Cathode(c)")(Ag^(oplus))rarrunderset("Anode(a)")(Ag^(oplus))`
`E = E^(Θ) - (0.59)/(n) lo g[Ag^(oplus)]_(a)/([Ag^(o+)]_(c))`
`K_(sp) of [AgCI] = 2.8 xx 10^(-10)`
`[Ag^(o+)]_(a) = (2.8 xx 10^(-10))/(0.2) = 14 xx 10^(-10)`
`K_(sp) [AgBr] = 3.3 xx 10^(-3)`
`[Ag^(oplus)]_(c) = (3.3 xx 10^(-13))/(0.001) = 3.3 xx 10^(-10)`
`E =- (0.059)/(1) lo g (14 xx 10^(-10))/(3.3 xx 10^(-10)) =- 0.059 log[(14)/(3.3)]`
`=- 0.059 xx 0.06276`
`=- 0.037`