Question

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)`

Answer 1

`"For Cu"(OH)_(2),K_(SP)=[Cu^(2+)][OH^(-)]^(2)`
`because[H^(+)]=10^(-14),thus [OH^(-)]=10^(0)=1`
therefore,
`[Cu^(2+)]=(K_(SP))/([OH^(-)]^(2))=(1.0xx10^(-19))/(1)=1.0xx10^(-19)`
Now `E_(RP)` for the couple `Cu^(2+)//Cu` is
`E_(RP)=E_(RP)^(0)+(0.059)/(2)log_(10)[Cu^(2+)]`
`=0.34+(0.059)/(2)log_(10)[1xx10^(-19)]=-0.2205V`