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

`E_(cell)=E_("Oxid. Pot. LHS electrode")+E_("Red. Pot. RHS electrode")`
`=E_("Oxid. Pot. "Ag//Ag^(+))^(@)-0.0591 log [Ag^(+)]_(LHS)+E_("Red. Pot. "Ag^(+)//Ag)^(@)+ 0.0591 log [Ag^(+)]_(RHS)`
`=0.0591 "log"([Ag^(+)]_(RHS))/([Ag^(+)]_(LHS))" "["Since, "E_(Ag//Ag^(+))^(@)+E_(Ag^(+)//Ag)^(@)=0]`
`=0.0591" log"((K_("sp AgBr"))/([Br^(-)]))/((K_("sp AgCl"))/([Cl^(-)]))`
`=0.0591"log" (3.3xx10^(-13))/0.001xx0.2/(2.88xx10^(-10))`
`=-0.0371" volt"`
The cell potential is negative, therefore, the cell reaction is non-spontaneous. For spontaneous reaction emf should be positive. therefore, the correct cell reaction is
`Ag|underset("Anode")(AgBr), KBr||KCl, underset("Cathode")(AgCl)|Ag`