The standard potential of the following cell is `0.23V` at `15^(@)C` and `0.21V` at `35^(@)C:`
`Pt|H_(2(g)|HCl(aq)|AgCl(s)|Ag(s)`
`a.` Write the cell reaction.
`b.` Calculate `DeltaH^(c-)` and `DeltaS^(c-)` for the cell reaction by assuming that these quantities remain unchanged in the range `15^(@)C` to `35^(@)C`
`c.` Calculate the solubility of `AgCl` in water at `25^(@)C`.
Given `:` The standard reduction potential of `Ag^(o+)(aq)|Ag(s)` is `0.80V` at `25^(@)C`.

Explanation

`Pt_(H_(2(G)))|HCl_((aq.))"||"AgCl_((S))|Ag_((S))`
1) `{:((1)/(2)H_(2)rarrH^(+)+e" (Anode)"),(AgCl+erarrAg+Cl^(-)"(Cathode)"),(bar((1)/(2)H_(2)+AgClrarrH^(+)+Ag+Cl^(-))):}`
ii) `DeltaG^(0)=nE^(@)F=1xx0.23xx96500="22195J (at 10"^(@)C)`
`DeltaG^(0)=nE^(@)F=1xx0.12xx96500="2265J (at "35^(@)C)`
Also, `DeltaG^(0)=DeltaH^(0)-TDeltaS^(0)`
`{:(therefore-22195=DeltaH^(@)-288xxDeltaS^(@)),(" "-22195=DeltaH^(@)-308xxDeltaS^(@)),(" + " -" +"),(bar(therefore" "S^(@)=-96."50J ")):}`
Also, `-22195=DeltaH^(0)-288xx(-96.5)=49987J`
`therefore" "DeltaH^(0)=49.987kJ`
iii) Consider the following reaction at `AgCl_((S))//Ag` electrodes
`{:(" Ag"rarrAg^(+)+e," "E_(OP)^(0)=-0.8V),(AgCl_((S))+erarrAg+Cl^(-)," "E_(RP_(Cl^(-)//AgCl//Ag))^(0)=?),(bar(" "AgClrarrAg^(+)Cl^(-)" ")):}`
`thereforeE_(cell)=E_(OP_(Ag//Ag^(+)))^(0)-(0.059)/(1)log[Ag^(+)]+E_(RP_(Cl^(-)//AgCl//Ag))^(0)+(0.059)/(1)log.(1)/([Cl^(-)])`
`E_(cell)=0` at equilibrium, ltBrgt Also, `E_(OP_(H))^(0)+E_(RP_(Cl^(-)//AgCl//Ag))^(0)=0.22V" at"25^(@)C`
and `E_(OP_(H))^(0)=0`
`thereforeE_(OP_(Ag//Ag^(+)))^(0)+E_(RP_(Cl^(-)//AgCl//Ag))^(0)=(0.059)/(1)log[Ag^(+)][Cl^(-)]`
`=0.059"log K"_(SP)AgCl`
`or" "-0.8+0.22=0.59"kog K"_(SP)AgCl`
`therefore" "K_(SP)AgCl=1.47xx10^(-10)`
`therefore"Solubility of AgCl"=sqrt(L_(SP))=sqrt(1.47xx10^(-10))`
`=1.21xx10^(-5)"mol litre"^(-1)`