For the galvanic cell `:`
`Ag|AgCl(s)),KCl(0.2M)||KBr(0.001M),AgBr(s)|Ag,` calculate the `EMF` generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at `25^(@)C`.
`[K_(sp)(AgCl)=2.8xx10^(-10),K_(sp)(AgBr)=3.3xx10^(-13)]`

Explanation

We know,
`E_(cell)=E_(OPunderset(L.H.S)(Ag//Ag^(+)))+E_(RPunderset(R.H.S)(Ag^(+)//Ag))`
`=E_(OP_(Ag//Ag^(+)))-(0.059)/(1)log_(10)[Ag^(+)]_(L.H.S)+E_(RP_(Ag^(+)//Ag))+(0.059)/(1)log_(10)[Ag^(+)]_(R.H.S)`
`E_(cell)=(0.059)/(1)log_(10).([Ag^(+)]_(R.H.S))/([Ag^(+)]_(L.H.S))`.....(1)
Now for L.H.S. `K_(SP_(AgCl))=2.8xx10^(-10)`
`therefore" "[Ag^(+)][Cl^(-)]=2.8xx10^(-10)`
`therefore[Ag^(+)]=(2.8xx10^(-10))/([Cl^(-)])=(2.8xx10^(-10))/(0.2)=1.4xx10^(-9)M`
For For R.H.S. `K_(SP_(AgBr))=3.3xx10^(-13)`
`[Ag^(+)][Br^(-)]=(3.3xx10^(13))/([Br^(-)])=(3.3xx10^(13))/([Br^(-)])=(3.3xx10^(-13))/(0.001)`
`=3.3xx10^(-10)M`
`therefore`by Eq. (1),
`E_(cell)=(0.059)/(1)log_(10).(3.3xx10^(-10))/(1.4xx10^(-9))=-0.037V`
Thus, to get `E_(cell)` positive, polarity of cell should be reversed.
`"i.e., cell is Ag"|underset(0.001M)(AgBr_((S))KBr)||underset(0.23M)(AgCl,KCl)|Ag`
and `E=+0.037V`