1.

The K_(sp) for AgCl at 298 K is 1.0xx10^(-10). Calculate E for Ag^(+)//Ag electrode immersed in 1.0 M KCl solution. Given : E^(@)Ag^(+)//Ag=0.799" V".

Answer»

SOLUTION :`AgCl(s)HARR Ag^(+)+Cl^(-)`
`K_(sp)=[Ag^(+)][Cl^(-)]`
`[Cl^(-)]=1.0M`
`[Ag^(+)]=(k_(sp))/([Cl^(-)])=(1xx10^(-10))/(1)=1xx10^(-10)M`
`"Now,"Ag^(+)+e^(-)rarrAg(s)`
`E=E^(theta)-(0.059)/(1)log.(1)/([Ag^(+)])`
`=0.80-(0.059)/(1)log.(1)/(10^(-10))`
`=0.80-0.059xx10=0.21V`


Discussion

No Comment Found

Related InterviewSolutions