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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 entropy during cell reaction is :A. `+79.2 k J`B. `-75.9 kJ`C. `+75.9 kJ`D. `-79.2 kJ` |
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Answer» `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))` |
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