Nernst equation for `Zn^(2+)|Zn` electrode is given as -
`E_(Zn^(2+)|Zn)=E_(Zn^(2+)|Zn)^(@)-(0.059)/(2)log.([Zn(s)])/([Zn^(2+)])`
What will be the plot of `E_(Zn^(2+)|Zn)" vs log "[Zn^(2+)]` look like?

Explanation

`E_(Zn^(2+)|Zn)=E_(Zn^(2+)|Zn)^(0)-(0.059)/(2)log([Zn(s)])/([Zn^(2+)])`
As `[Zn(s)]=1`, we have
`E_(Zn^(2+)|Zn)=E_(Zn^(2+)|Zn)^(0)+(0.059)/(2)log[Zn^(2+)]`
This relation represents an equation of a straight line of the type `y=mx+c`.
So, plotting `E_(Zn^(2+)|Zn) " vs log "[Zn^(2+)]` will give a straight line with a slope of `(0.059)/(2)` and an intercept of `E_(Zn^(2+)|Zn).`