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

(i) Electrode process :
`{:(1/2 H_(2) rarr H^(+) + e^(-)" (Anode)"),(AgCl+e^(-) rarr Ag+Cl^(-)" (Cathode)"),(ulbar(1/2 H_(2)+AgCl hArr H^(+)+Ag+Cl^(-))):}`
(ii) We know that, `DeltaG^(@)=DeltaH^(@)-T DeltaS^(@)`
`-22195=DeltaH^(@)-288xxDeltaS^(@)`
`-20265=DeltaH^(@)-308xxDeltaS^(@)`
On solving, `DeltaS^(@)=-96.5 J, DeltaH^(@)=49.987 kJ`
(iii) `E=E^(@)-0.0591/n log_(10) Q`
At equilibrium, `E=0, Q=K=[Ag^(+)][Cl^(-)]`
`0=(0.8-0.22)+0.0591/1 log K_(sp)`
`((-0.8+0.22))/0.0591=log K_(sp)`
`K_(sp)=1.47xx10^(-10)`
Solubility, `S=sqrt(K_(sp))`
`=sqrt(1.47xx10^(-10))=1.21xx10^(-5) M`