For the half cell

At `pH=2`, the electrode potential is

Options

A. `1.36V`
B. `1.30V`
C. `1.42V`
D. `1.20V`

Answer

c

Explanation

In this electrode,
`[A]=[B]` in quinhydrone electrode
Hence, `Q=[H^(o+)]^(2)`

`E=E^(c-)-(0.0591)/(2)log[H^(o+)]^(2)`
`=E^(c-)+0.0591log [H^(o+)]`
`=E^(c-)+0.0591pH`
`1.30+0.059xx2`
`=1.42V`