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The standard potential of the following ...

The standard potential of the following cell is 0.23 V at `15^(@)C and 0.21 V" at "35^(@)C:`
`Pt|H_(2)(g)|HCl(aq)|AgCl(s)|Ag(s)`
(2) Calculate `DeltaH^(@) andDeltaS^(@)` for the cell reaction by assuming that these quantities remain unchanged in the range `15^(@)C" to "35^(@)C`.

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Given cell : `Pt|H_(2)(g)|HCl(aq)|AgCl(s)|Ag(s)`
`DeltaS^(0)=nF((dE^(0))/(dT))`, where n = Number of electrons involved in the cell reaction, F = Faraday = 96500C, `dE^(@)` = Difference of standard electrode potential at two different temperature `=(0.21-0.23)=-0.02V` and dT = difference of two temperatures `=(308-288)K=20K`
`therefore DeltaS^(+@)=1xx96500xx((-0.02)/(20))=-"96.5J.K"^(-1)."mol"^(-1)`
We know, `DeltaG^(@)=-nFE^(@)`
`therefore" "DeltaG_(15^(@)C)^(@)=-1xx96500xx0.23[because E_(15^(@)C)^(@)=0.23V]=-"22195 J. mol"^(-1)`
`therefore DeltaH^(@)=DeltaG^(@)-TDeltaS^(@)=-22195-288xx(-96.5)=-"49987 J.mol"^(-1)`
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