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)`.

Answer

`-0.037 V`

Explanation

At anode:
`Ag +CI^(-) rarr AgCI +e^(-)`
At anode:
`AgBr +e^(-) rarr Ag +Br^(-)`
`E = E_(Ag//AgCI//CI^(-))^(@) +E_(Br^(-)//AgBr//Ag)^(@) - 0.0591 log. ([Br^(-)])/([CI^(-)])` ..(i)
For reaction:
`Ag +CI^(-) rarr AgCI -e^(-)`
`Ag^(+) +e^(-) rarr Ag`
`E = O = E_(Ag//AgCr//CI^(-))^(@) +E_(Ag^(+)//Ag)^(@) - 0.0591`
`log.(1)/([Ag^(+)][CI^(-)])`
`E_(Ag//AgCI//CI^(-))^(@) = E_(Ag//Ag)^(@) - 0.0591 log K_(sp) (AgCI)` ...(ii)
for reaction
`AgBr +e^(-) rarr Ag +Br^(-)`
`Ag rarr Ag^(+) +e^(-)`
`E = 0 = E_(Br^(-)//AgBr//Ag)^(@) +E_(Ag//Ag^(+))^(@) - 0.0591 log [Ag^(+)] [Br^(-)]`
`E_(Br^(-)//AgBr//Ag)^(@) = E_(Ag^(+)//Ag)^(@) +0.0591 log K_(sp) (AgBr)` ...(iii)
from equation (1),(2) & (3)
`E = 0.0591 log.(K_(sp)(AgBr))/(K_(sp)(AgCI)) - 0.0591 log.([Br^(-)])/([CI^(-)])`
`E = 0.0591 log.(3.3 xx 10^(-13) xx 0.2)/(2.8 xx 10^(10) xx 0.001) =- 0.037 V`