use the nerst equation for electrochemistry. it relates electrode potential to concentration of reduced and oxidised species.
The nernst equation can be used for any redox process to calculate potential for reaction.
E = E(0) + (RT/nF)ln([oxidized]/[reduced])
For standard temp, RT/F is a constant, and converting ln to log yields:
E = E(0) + (.0591/n)log([oxidized]/[reduced])
The standard reduction half rxn (from tables) is:
4H+ + O2(g) + 4e -> 2H2O E(0) = 1.23 V
The oxidized part [oxidized] of the equation is:
([H+]^4 [pO2]
The [reduced] part is [H2O]^2 but it is not included because it is the solvent. n=4 (# of e)
E = 1.23 + (.0591/4)log([H+]^4 [pO2])
where [H+] = 10^-6 because ph=6
and pO2(g) = 0.21 atm because that's the partial pressure of oxygen since it's 21% by volume.
then
E = 1.23 + (.0591/4)log([10^-6]^4 [0.21])
E= 1.23 - .36 = .87 V
PE = -log(E) = -log(.87) = .06