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`.

Answer

`DeltaS^(c-)=-96.5J K ^(-1) mol^(-1)`
`DeltaH^(c-)=-49987J mol^(-1)`

Explanation

`a.` The half cell reactions are `:`
`i. ` At anode `:(1)/(2)H_(2)rarr H^(o+)(aq)+e^(-)`
`ii.` At cathode `:AgCl(s)+e^(-) rarr Ag(s)+Cl^(c-)(aq)`

`b.` We know that `DeltaS=nF((dE)/(dT))`
`n=`Number of transferred electrons `=1` ltbr. `F=`Faraday number of `=96500C`
`dE=` Difference of electrode potential at two different temperatures
`=(0.21-0.23)=-0.02V`
`dT=`Difference of two temperatures
`=(35^(@)C-15^(@)C)=20^(@)C`
`:. DeltaS^(c-)=1xx96500xx((-0.02)/(20))=-96.5 J K^(-1)mol^(-1)`
`DeltaG^(c-)=-nE^(c-)F`
`DeltaG^(c-)._(15^(@)C)=-1xx0.23xx96500J ( :.E^(c-)._(15^(@)C)=0.23V)`
`=-22195J mol^(-1)`
`DeltaH^(c-)=Delta G^(c-)-T DeltaS^(c-)`
`=-22195-288xx(-96.5)`
`=-49987Jmol^(-1)`
`c.` `E^(c-)._(25^(@)C)` of cell `=E^(c-)._(15)-(dE)/(dT)=(0.23-(0.02)/(20))V`
`=0.229V`
The corresponding cell is
`Ag(s)|Ag^(o+)(aq)||Cl^(c-)(aq)(AgCl)(s)|Ag(s)`
In the form of oxidized electrode potential
`E^(c-)._(cell)=E^(c-)._(anode )-E^(c-)._(cathode)`
`=E^(c-)._(Ag|Ag^(o+))-E^(c-)._(Ag|AgCl|Cl^(c-))`
`=-0.80-(-0.229)=-0.571V`
`E^(c-)._(cell)=(0.591)/(n)logK_(eq)`

`E^(c-)._(cell)=(0.0591)/(n)log[Ag^(o+)][Cl^(c-)]`
`=(0.0591)/(n)logK_(sp)`
`:.-0.571=(0.0591)/(1)logK_(sp)`
or `logK_(sp)=-9.6615=bar(10).3385`
`K_(sp)=2.18xx10^(-10)`
`K_(sp)` of `AgCl=2.18xx10^(-10)(molL^(-1))^(2)`
Solubility of `AgCl=sqrt(K_(sp))=sqrt(2.18xx10^(-10)) `
`=1.476xx10^(-5)molL^(-1)`