The cell designed as `Pt_(H_(2))|HCl_(aq)||Hg_(2)Cl_(2),0.01 N KCl|Hg` has emf of `0.271 V` at `298 K` and `0.2669` at `308 K`. The `E_(H_(2)^(2+)//Hg)` is `0.260 V`
The change in free energy `(DeltaG)` during cell reaction is:

Options

A. `-52.3 kJ`
B. `+52.3 kJ`
C. `-26.15 kJ`
D. `+26.15 kJ`

Explanation

`H_(2)to2H^(+)+2e` and `Hg_(2)^(2+)+2etoHg`
`E_("cell")=E_(OP_(H))^(@)+E_(RP_(Hg_(2)Cl_(2)))+0.0591/2+log([Hg_(2)^(2+)]PH_(2))/([H^(+)]^(2))...(1)`
`:.` At `298 K : -DeltaG=nEF`
`=2xx0.271xx96500`
`=-DeltaG=52.3xx10^(3) J=52.3 k J`
`((deltaE)/(deltaT))_(P)=(E_(2)-E_(1))/(T_(2)-T_(1))=(0.2669-0.271)/10`
`=-4.1xx10^(-4)`
`DeltaH=nF[T((deltaE)/(deltaT))_(P)-E]`
`=2xx96500[298xx(-4.1xx10^(-4))-0.271]`
`DeltaH=-75.9xx10^(3) J=-75.9 kJ`
Also `DeltaG=DeltaH-TDeltaS`
`=-52.3=-75.9-298xxDeltaS`
`:. DeltaS=-0.07992 k J =-79.2 J`
Also be eq. (1)
`0.271=0+0.260+0.0592/2 log([Hg_(2)^(2+)]P_(H_(2)))/([H^(+)]^(2))`