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

a. Hydrogen gets oxidised and silver gets reduced.
`(1)/(2)H_(2)(g)+AgCI(s) hArr H^(o+)(aq) +Ag(s) +CI^(Theta)(aq)`
b. `DeltaG^(Theta) = nFE_((288K))^(Theta)`
`=- 1 xx 96500 xx 0.23`
`=- 22195 J`
`DeltaG_((308K))^(Theta) = - nFE_((308K))^(Theta)`
`= -1 xx 96500 xx 0.21`
`=- 20265 J`
`DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)`
`-22195 = DeltaH^(Theta) - 288 DeltaS^(Theta) ...(i)`
`-20265 = DeltaH^(Theta) - 308 DeltaS^(Theta) ...(ii)`
Subtracting (ii) from (i), we get
`DeltaS^(Theta) =- 96.50 J`
`-20265 = DeltaH^(Theta) - 308 DeltaS^(Theta)`
`= DeltaH^(Theta) - 308 xx (-96.50)`
or `DeltaH^(Theta) =- 49987 J`
c. `E_(25^(@)C)^(Theta)` of cell
`E_((15^(@)C))^(Theta) = 0.23 V`
`E_((35^(@)C))^(Theta) = 0.21 V`
`(DeltaE^(Theta))/(DeltaT) = ((0.21 -0.23))/(20) =- 0.01`
`DeltaE^(Theta) for 10^(@)C = 0.01 xx 10 =- 0.1`
`E_((25^(@)C))^(Theta) =E_((15^(@)C))^(Theta) +(-0.1)`
`= 0.23 - 0.1`
`=0.22 V`
`E_(cell)^(Theta) =E_("cathode")^(Theta) -E_("anode")^(Theta)`
or `0.22 = E_(CI^(Theta)//AgCI//Ag)^(Theta) - E_(2H^(o+)//H_(2))^(Theta)`
`= E_(CI^(Theta)//AgCI//Ag)^(Theta)-0`
or `E_(CI^(Theta)//AgCI//Ag)^(Theta) = 0.22V`
`E_(A^(o+)//Ag)^(Theta) = 0.80 V`
`E_(CI^(Theta)//AgCI//Ag)^(Theta) = E_(Ag^(o+)//Ag)^(@) - (0.0591)/(1) log K_(sp) (AgCI)`
or `0.22 = 0.58 - (0.0591)/(1) log K_(sp) (AgCI)`
or `K_(sp) (AgCI) = 1.47 xx 10^(10)`
Also
`K_(sp)(AgCI) =[Ag^(o+)] xx [CI^(Theta)]`
or `1.47 xx 10^(10) = [Ag^(o+)]^(2)`
or `[Ag^(o+)] = 1.21 xx 10^(-5)`