The standard reduction potential for the half reaction below at pH 6?
O2+2H2O+4e --> 4OH- E=+0.40V
My Attempt:
pH=6 pOH=8
[OH]=10^-8
E=E^nod + 0.059/4 lg 1/(10^-8)^4
=0.4+0.48
=0.88
But the actual answer is 0.64 which is (0.4+0.24), can anyone points out the mistake for me?