For the Galvanic cell, Ag|AgCl(s) |KCl (0.2 M)|| KBr (0.001 M) |AgBr(s) |Ag Calculate the emf generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell at 25^(@) C. Given K_(sp)(AgCl) = 2.8 xx 10^(-10), K_(sp)(AgBr) = 3.3 xx 10^(-13)

For the Galvanic cell, Ag|AgCl(s) |KCl (0.2 M)|| KBr (0.001 M) |AgBr(

1.	For the Galvanic cell, Ag\|AgCl(s) \|KCl (0.2 M)\|\| KBr (0.001 M) \|AgBr(s) \|Ag Calculate the emf generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell at 25^(@) C. Given K_(sp)(AgCl) = 2.8 xx 10^(-10), K_(sp)(AgBr) = 3.3 xx 10^(-13)
Answer» Solution :`AgCl `K_(SP)(AgCl) `therefore [Ag^(+)]_(LHS) = (K_(sp)[AgCl])/[Cl^(-)] = (2.8 xx 10^(-10))/0.2 = 1.4 xx 10^(-9)` M `AGBR `K_(sp)(AgBr) = [Ag^(+)][Br^(-)]` `[Ag^(+)]_(LHS) = (K_(sp)(AgBr))/[Br^(-)] = (3.3 xx 10^(-13))/(0.001) = 3.3 xx 10^(-10)` M `{:("Cell", "Cell Reaction", E^(@)),(LHS,Ag to Ag^(+)+e^(-),E_(Ag//Ag^(+))^(@) =xv),(RHS, Ag^(+) + e^(-) to Ag, E_(Ag^(+)//Ag)=-E_(Ag//Ag^(+))^(@) =-xV):}` `K= ([Ag^(+)]_(LHS))/([Ag^(+)]_(RHS))` `E_("cell") = E_("cell"^(@)) -(0.0591)/N log K = 0-(0.0591)/1 log (1.4 xx 10^(-9))/(3.3 xx 10^(-10)) = -0.0371` V To make `E_("cell")` POSITIVE, LHS cell should be cathode (+ve half cell).