The standard oxidation potential of `Zn` referred to SHE is `0.76V` and that of `Cu` is `-0.34V` at `25^(@)C`. When excess of `Zn` is added to `CuSO_(4),Zn` diplaces `Cu^(2+)` till equilibrium is reached. What is the ratio of `Zn^(2+)` to `Cu^(2+)` ions at equilibrium?

Answer

`[Zn^(2+)]//[Cu^(2+)] =1.941 xx 10^(37)`

Explanation

`E_(cell)^(@) = (0.0591)/(2)log.([Zn^(2+)])/([Cu^(2+)])`
`1.1 =(0.0591)/(2)log.([Zn^(2+)])/([Cu^(2+)])`
`([Zn^(2+)])/([Cu^(2+)]) =1.941 xx 10^(37)`