Question

Calculate the equilibrium constant for the reaction at 298 K
`Zn(s)+Cu^(2+)(aq)harr Zn^(2+)(aq)+Cu(s)`
Given` " " E_(Zn^(2+)//Zn)^(@)=-0.76 V` and `E_(Cu^(2+)//Cu)^(@)=+0.34 V`

Answer 1

We know that, `logK_(c)=(nE_(cell)^(@))/(0.0591)`
`E_(cell)=[E_(cathode)^(@)-E_(cathode)^(@)]=[(+0.34 V)-(-0.76 V)]=1.10 V,n=2,`
`:. " "logK_(c)=(2xx(1.10 V))/(0.0591 V)=37.22,K_(c)="Antilog " 37.22=1.66xx10^(37)`