1.

Calculate the equilibrium constant for the reaction, Zn+Cu^(2+)hArrCu+Zn^(2+) Given: E^(@) for Zn^(2+)//Zn=-0.763V and for Cu^(2+)//Cu=+0.34V,R=8.314JK^(-1)mol^(-1),F=96500" C "mol^(-1)

Answer»


SOLUTION :`E_(CELL)^(@)=E_(Cu^(2+)//Cu)^(@)-E_(ZN^(2+)//Zn)^(@)=0.34-(-0.0763)V=1.103V`.


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