(a) Calculate the standard free energy change and maximum work obtainable for the reaction. `Zn(s)+Cu^(2+)(aq)hArrCu(s)+Zn^(2+)(aq)` [Given `E_(Zn^(2+)//Zn)^(@)=-0.76V,E_(Cu^(2+)//Cu)^(@)=+0.34V,F=96500" C "mol^(-1)`] (b) also calculate the equilibrium constant for the reaction.

(a) Calculate the standard free energy change and maximum work obtaina

1.	(a) Calculate the standard free energy change and maximum work obtainable for the reaction. `Zn(s)+Cu^(2+)(aq)hArrCu(s)+Zn^(2+)(aq)` [Given `E_(Zn^(2+)//Zn)^(@)=-0.76V,E_(Cu^(2+)//Cu)^(@)=+0.34V,F=96500" C "mol^(-1)`] (b) also calculate the equilibrium constant for the reaction.
Answer» (a) The cell may be represented as `Zn(s)\|Zn^(2+)(1M)\|\|Cu^(2+)(1M)\|Cu(s)` `E_(cell)^(@)=E_(RHS)^(@)-E_(LHS)^(@)=0.34-(-0.76)=1.10`volt For the given cell reaction, n=2 `DeltaG^(@)=-nFE_(cell)^(@)=-2xx96500" C "mol^(-1)xx1.10V=-212,300CV" "mol^(-1)` Thus, the maximum work that can be obtained from the Daniell cell=212.3 kJ. (b) `DeltaG^(@)=-RT" ln "K_(c)=-2.303RT" log "K_(c)` `therefore-212300=-2.303xx8.314xx298xxlog" "K_(c)" or "logK_(c)=(212300)/(2.303xx8.314xx298)=37.2074` `thereforeK_(c)="Antilog "37.2074=1.6xx10^(37)`

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What are elementary reactions ?

(a) Calculate the standard free energy change and maximum work obtainable for the reaction. `Zn(s)+Cu^(2+)(aq)hArrCu(s)+Zn^(2+)(aq)` [Given `E_(Zn^(2+)//Zn)^(@)=-0.76V,E_(Cu^(2+)//Cu)^(@)=+0.34V,F=96500" C "mol^(-1)`] (b) also calculate the equilibrium constant for the reaction.

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