The reduction potential of a hydrogen electrode at `pH 10` at `298K` is : `(p =1` atm)

Options

A. 0.51 V
B. 0.000 V
C. 0.59 V
D. 0.059 V

Answer

C

Explanation

`E = E^(@) - (0.059)/(n) " log " _(10) [H^(+)]`
Given `pH = 10 , n =1 therefore [H^(+)] = 10^(-10)` ,
`E_(SHE)^(@) = 0 , 10 = - "log"_(10) [ H^(+)]`
`E = E^(@) - 0.059 xx log 10^(-10) = 0-0.059 xx (-10)`
`= 0 + 0.59 = 0.59` V