The standard reduction potential for `Cu^(2+)|Cu` is `+0.34V`. Calculate the reduction potential at`pH=14` for the above couple . `K_(sp)` of `Cu(OH)_(2)` is `1.0xx10^(-19)`

Explanation

pH=14
`implies pOH=0 implies [OH^(-)]=1.0M`
`K_(sp)=10^(-19)=[Cu^(2+)][OH^(-)]^(2) implies [Cu^(2+)]=(10^(-19))/([OH^(-)]^(2))=10^(-19)`
For reaction : `Cu^(2+)+2e^(-)to Cu, E^(@)=0.34V`
`E=E^(@)-(0.0592)/2 log. 1/([Cu^(2+)])=0.34-0.0592/2 log 10^(19) =-0.222V`