The standard potential of the following cell is 0.23 V at `15^@C and 0.21V` at `35^@C`
`(Pt) H_2(g) |HCl(aq)|AgCl(s)|Ag(s)`
Calculate the solubility of AgCl in water at `25^@C` Given:
`E_(Ag^+,Ag)^@=0.80V` at `25^@C`

Explanation

In the given cell
`E_(cell)^@=E_(AgCl,Cl^-)^@-E_(H^+,H_2)^@`
`0.22=E_(AgCl,Cl^-)-0` …..(Average value of `E_(cell)^@` is considered )
or `E_(AgCl,Cl^-)=0.22`
Let us now consider the following cell to calculate `K_(sp) (AgCl)`
`Ag=Ag^+ +e, E^@=-0.8` (given)
`AgCl+e=Ag+Cl^(-),E^@=0.22` (calculated)
`therefore` for the cell reaction `AgCl=Ag^+ +Cl^-`
`E_(cell)=(E_(AgCl,Cl^-)^@-E_(Ag^+,Ag)^@)-0.0591/1 log[Ag^+][Cl^-]`
At equilibrium `E_(cell)=0 and R.Q=K_(sp)`
`therefore E_(AgCl,Cl^-)^@-E_(Ag^+,Ag)^@=0.0591 log K_(sp)`
`0.22-0.8=0.0591 log K_(sp)`
`therefore K_(sp)=1.535 times 10^-10`
`therefore` solubility of `AgCl=sqrt(K_(sp))=sqrt(1.535 times 10^-10)`
`=1.239 times 10^-5 mol e//L`