Question

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)]`

Answer 1

We know,
`E_(cell) = E_(OP_(Ag//Ag^(+))) + E_(RP_(Ag^(+)//Ag))`
L.H.S. R.H.S.
`= 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)) = 2.8 xx 10^(-10)`
`:. [Ag^(+)][Cl^(-)] = 2.8 xx 10^(10)`
`:. [Ag^(+)] = (2.8 xx 10^(-10))/([Cl^(-)]) = (2.8 xx 10^(-10))/(0.2) = 1.4 xx 10^(-9) M`
For `R.H.S. K_(SP_(AgBr)) = 3.3 xx 10^(-13)`
`[Ag^(+)][Br^(-)] = (3.3 xx 10^(-13))/([Br^(-)]) = (3.3 xx 10^(-13))/(0.001)`
`= 3.3 xx 10^(-10)M`
`:.` By Eq. `(1)`,
`E_(cell) = (0.059)/(1)log_(10).(3.3 xx 10^(-19))/(1.4 xx 10^(-9)) = -0.037V`
Thus, to get `E_(cell)` positive, polarity of cells should be reversed.
.e., Cell is `Ag|underset(0.001M)(AgBr_(s)KBr)||underset(0.2M)(AgCl, KCl)|Ag`
and `E = +0.037 V`