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)`

Options

A. `+ 0.22 V`
B. `- 0.44 V`
C. `- 0.22 V`
D. `+0.44 V`

Answer

c

Explanation

When `pH = 14 [H^(+)][10^(-4) and [OH^(-)] = 1M`
`K_(sp) = [Cu^(2+] [OH^(-)]^(2) = 10^(-19)`
` :. [Cu^(2+)] = (10^(-19))/([OH^(-)]^(2)) = 10^(-10)`
The half cell reaction
`Cu^(2+) + 2e rarr Cu`
`E = E = (0.059)/(2) log. (1)/([Cu^(2+)])`
`= 0.34 - (0.39)/(2) log. (1)/(10^(-19)) = - 0.22V`