1.

Calculate the potential of a zinc-zinc ion electrode in which the zinc ion activity is 0.001M (E_(Zn^(2+)//Zn)^(@)=-0.76V,R=8.314KJ^(-1)mol^(-1),F=96,500" C "mol^(-1))

Answer»


SOLUTION :`E=E^(@)-2.303(RT)/(nF)"log"(1)/(a_(Zn^(2+)))=-0.76-(2.303xx8.314xx298)/(2xx96500)"log"(1)/(10^(-3))=-0.849V`.


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