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)`
(i)
`{:((1)/(2)H^(2)rarrH^(+)+e("Anode")),(AgCl+erarrAg+Cl^(-)("Cathode")):}/((1)/(2)H_(2)+AgClrarrH^(+)+Ag+Cl^(-))`
(ii) `-DeltaG^(@) = nE^(@)F = 1 xx 0.23 xx 96500 xx = 22195 J` (at `15^(@)C`)
`DeltaG^(@) = nE^(@)F = 1 xx 0.21 xx 96500 xx = 20265 J` (at `15^(@)C`)
(ii) `-DeltaG^(@) =nE^(@)F = 1 xx 0.23 xx 96500 = 22195J (at 15^(@)C)`
Also, `{:(DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)),(-22195=DeltaH^(@)-288xxDeltaS^(@)),(-20265=DeltaH^(@)-308xxDeltaS^(@)),(" + - +"):}/(DeltaS^(@)=-96.50 J)`
Also, `-22195 = Delta H^(@) -288 xx (-96.5) = 49987 J`
`:. Delta H^(@) = 49.987 kJ`
(iii) Consider the follwoing reaction at `AgCl_(s)//Ag` electrodes
`{:(AgrarrAg^(+)+e,,E_(OP)^(@)=-0.8V),(AgCl_((s))+erarrAg+Cl^(-),,E_(RP_(Cl^(-)//AgCl//Ag))^(@)=?):}/(AgClrarrAg^(+)+Cl^(-))`
`:. E_(cell) = E_(OP_(Ag//Ag^(+)))^(@)-(0.059)/(1)log[Ag^(+)]+E_(RP_(Cl^(-)//AgCl//Ag))^(@)`
`+ (0.059)/(1)log.(1)/([Cl_(1)])`
`E_(cell) = 0` at equililbrium,
Also, `E_(OP_(H))^(@)+E_(RP_(Cl^(-)//AgCl//Ag))^(@) = 0.22V` at `25^(@)C`
and `E_(OP_(H))^(@) = 0`
`:. E_(OP_(Ag//Ag^(+)))^(@)+E_(RP_(Cl^(-)//AgCl//Ag))^(@)=(0.059)/(1)log[Ag^(+)][Cl^(-)]`
`= 0.059logK_(SP)AgCl`
or `-0.8 + 0.22 = 0.059logK_(SP)AgCl`
`:. K_(SP)AgCl = 1.47 xx 10^(10)`
`:.` Solubility of `AgCl = sqrt(K_(SP)) = sqrt(1.47 xx 10^(-10))`
`= 1.21 xx 10^(-5) "mol litre"^(-1)`