For the cell `Zn|ZnCl_(2)(m)|AgCl,E` is `1.24V` at `25^(@)C` and `1.260V` at `35^(@)C` of `m=10^(-3)`. Write down the cell reaction and calculate `DeltaG, DeltaH,` and `DeltaS` at `25^(@)C`.

Explanation

`Zn rarr Zn^(2+|+2e^(-)`

`ulbar(Zn(s)+2AgCl(s)rarrZn^(2+)(aq)+2Ag(s)+2Cl^(c-)(aq))`
`:. (DeltaE)/(DeltaT)=(1.26)/(308)=(1.26-1.24)/(308-298)=2xx10^(-3)V degree^(-1)`
`DeltaG^(c-)(at 298K)=-nFE^(c-)=-2xx96500xx1.24`
`=-239320J`
`=-239.32kJ`
`DeltaS^(c-)=nF((delE)/(delT))=2xx96500xx2xx10^(-3)=386J`
`DeltaG=Delta H-TDeltaS`
`-239320=DeltaH-298xx386`
`:. DeltaH=-124292J`