Calculate the equilibrium constant for the following reaction at 298K. `Cu(s)+Cl_(2)(g)toCuCl_(2)(aq)` `R=8.314JK^(-1)mol^(-1),E_(Cu^(2+)//Cu)^(@)=0.34V,E_(1//2" "Cl_(2)//Cl^(-))^(@)=1.36V,F=96500" C "mol^(-1)`

Calculate the equilibrium constant for the following reaction at 298K.

1.	Calculate the equilibrium constant for the following reaction at 298K. `Cu(s)+Cl_(2)(g)toCuCl_(2)(aq)` `R=8.314JK^(-1)mol^(-1),E_(Cu^(2+)//Cu)^(@)=0.34V,E_(1//2" "Cl_(2)//Cl^(-))^(@)=1.36V,F=96500" C "mol^(-1)`
Answer» Correct Answer - `3.295xx10^(34)`. Anode raction: `Cu(s)toCu^(2+)(aq)+2e^(-)` Net cell reaction: `Cu(s)+Cl_(2)(g)toCuCl_(2)(aq)` `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=1.36-0.34=1.02V` `E_(cell)^(@)=(0.0591)/(n)log" "K_(c),i.e., 1.02=(0.0591)/(2)" log "K_(c)` or `log" "K_(c)=34.5178` or `K_(c)=`antilog `34.5178=3.295xx10^(34)`.

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Calculate the equilibrium constant for the following reaction at 298K. `Cu(s)+Cl_(2)(g)toCuCl_(2)(aq)` `R=8.314JK^(-1)mol^(-1),E_(Cu^(2+)//Cu)^(@)=0.34V,E_(1//2" "Cl_(2)//Cl^(-))^(@)=1.36V,F=96500" C "mol^(-1)`

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