The standard potential of the following cell is 0.23 V at `15^(@)C and 0.21 V" at "35^(@)C:`
`Pt|H_(2)(g)|HCl(aq)|AgCl(s)|Ag(s)`
(3) Calculate the solubility of `AgCl` in water at `25^(@)C`. Given : Standard reduc-tion potential of `Ag^(+)(aq)|Ag(s)` is 0.80 V at `25^(@)C`.

Explanation

Given cell : `Pt|H_(2)(g)|HCl(aq)|AgCl(s)|Ag(s)`
Given `E_((15^(@)C))^(@)=0.23V and E_((35^(@)C))^(@)=0.21V`
`therefore(DeltaE^(@))/(DeltaT)=((0.21-0.23))/(20)=-0.01`
`therefore" "DeltaE^(@)" for "10^(@)C=-0.01xx10=-0.1`
`therefore" "E_((25^(@)C))^(@)=E_((15^(@)C))^(@)+(0.1)=0.23-0=0.22V`
`E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)`
`0.22V=E_(Cl^(-)|AgCl|Ag)^(@)-E_(2H^(+)|H_(2))^(@)=E_(Cl^(-)|AgCl|Ag)^(@)-0`
`therefore E_(Cl^(-)|AgCl|Ag)^(@)=0.22V and E_(Ag^(+)|AG)^(@)=0.80V" (given)"`
`E_(Cl^(-)|AgCl|Ag)^(@)=E_(Ag^(+)|Ag)^(@)-(0.059)/(1)xxlogK_(sp)(AgCl)`
`"or, "0.22V=0.80V-(0.059)/(1)logK_(sp)(AgCl)`
`therefore" "K_(sp)(AgCl)=1.47xx10^(-10)`
and `K_(sp)(AgCl)=[Ag^(+)]xx[Cl^(-)]`
`therefore [Ag^(+)]^(2)=1.47xx10^(-10)`
`or, [Ag^(+)]=sqrt(1.47xx10^(-10))=1.21xx10^(-5)`