[usenet69]

it's a toughie.... lets see who is smart on this board.

What is the redox potential of an acid mine water sample having Fe3+ = 6
x E-03 M and Fe2+ = 5 x E-04 M given that E0 for the Fe3+/Fe2+ couple is 0.77v.

[Demotivator]

But the problem only provides info for the iron half rxn.

Fe3+ + e -> Fe2+  E(0) = 0.77 V
E1 = E(0) + (RT/nF)ln([oxidized]/[reduced])
E1 = 0.77 + (.0591/1)log(.006/.0005)
for the fe3+/fe2+ couple
E1 = 0.77 + .0591(1.08) = 0.83 V

Details on the other half rxn must be assumed based on the described "acid mine water".
Assuming PH = 4 for acidity and an oxygen partial pressure of 0.22 atm:
4H+ + O2  + 4e -> 2H2O   E(0) = 1.23 V
E2 = E(0) + (.0591/n)log([H+]^4 [pO2])
E2 = 1.23 + (.0591/4)log(.0001^4 (0.22))
E2 = 1.23 - .21 = 1.02 V

For the system:
4H+ + O2 + 4Fe2+ -> 4Fe3+  +  2H2O
E solution = E2 - E1 = 1.02 - (0.83) = 0.19 V