The standart potential of the following cell is 0.23 at `15^(@)C and 0.21 V at 35^(@)C`
`Pt|H_(2)(g)|HCI(aq)|AgCI(s)|Ag(s)`
(i) Write the cell reaction
(ii) Calculate `DeltaH^(@) and DeltaS^(@)` for the cell reaction by assuming that these quantites remain unchanged in the range `15^(2) C to 35^(@) C`
[Given SRP of `Ag^(+)(aq)|Ag(s) is 0.80 V at 25^(@)C]`

Explanation

At anode `1/2H_2toH^(+)+e^(-), E^(@)=0`
At cathode `underline(AgCl(s)+e^(-)to Ag +Cl^(-),E^(@)=?)`
(i) Cell reaction : `1/2H_2+AgCl(s)toAg+H^(+)+Cl^(-)`
(ii) `DeltaG^(@)=-nE^(@)F=DeltaH^(@)-T DeltaS^(@)`
At `15^(@)c` :`-0.23xx96500=DeltaH^(@)-288Deltas^(@)`......(i)
At `35^(@) C` : `-0.21xx96500=DeltaH^(@)-308DeltaS^(@)`....(ii)
`implies 96500(0.23-0.21)=-20 DeltaS^(@) implies DeltaS^(@)=-(96500xx0.02)/(20)=-96.5J`
Substituting value of `Deltas^(@)` in (i)
`DeltaH^(@)=288xx(-96.5)-0.23xx96500=-49.987 kJ`
(iii)At `25^(@)C`
`-E^(@)xx96500=-49987-298(-96.5) E^(@)=0.22V`
`implies AgCl(s) +e^(-) to Ag+Cl^(-),E^(@)=0.22V`
`Ag to A g^(+)+e^(-),E^(@)=-0.80V`
Adding: `AgCl(s)to Ag^(+)+Cl^(-), E^(@)=-0.8V`
`implies e^(@)=0.0592 log K_(sp) implies log K_(sp)=(-0.58)/(0.0592)=-9.79`
`K_(sp)=1.6xx10^(-10)`