Question

Calculate the potential of a `Zn -An^(2+)` electrode in which the molarity of `Zn ^(2+)` is `0.001M.` Given that `E_(Zn^(2+)//Zn)^(0)=-0.76V`
`R=0.314JK^(-1) mol ^(-1), F= 96500C mol ^(-1).`

Answer 1

Given electrode `Zn|Zn^(+2)(0.001m)E_(Zn^(+2)//Zn)^(0) =-0.76V`
Nernst equation is
`E=E^(0) +(2.303RT)/(nF) log [M^(n+)]`
Given `R=8.314J//K.`mole
`F = 96500c//`mole
`E= E^(0) +(0.059)/(2)log C`
`=-0.76+(0.059)/(2)log 0.001`
`=-0.76 -(0.059)/(2)xx3`
`= -0.76-0.0295xx3`
`=-0.76-0.0885 =-0.8485V`