One at a time, folks, I only have two arms.
I’m not sure what you find incomplete here. You mention H+, but perhaps you fail to realize that reduction in an aqueous solution is assumed to involve H+ when looking at the final product as H2O. If you look at OH- as the final product, then you assume H2O is involved. Now, H+ can be reduced to H2, but not by iron in anything but an acidic solution.
You guys must truly hate me, because you have forced me to pull out my old Chem manual, which I promised I would never look at again for fear that it might turn me to stone.
To summarize, the reduction potential for Iron is -0.44, the reduction potential for H+ is 0, and the potential for O2 is 0.40. This means that Iron, in theory, will always reduce oxygen over hydrogen ions when both are present (you’ll note that H+ is only going to be in abundance in an acidic solution). In reality, though, iron will not oxidize to rust by either oxygen or H+ unless both are present. Under these condition, O2 is reduced. I quote:
“The rusting of iron is known to require oxygen; iron does not rust in water unless O2 is present. Rusting also requires water; iron does not rust in oil, even if it contains O2, unless H2O is also present.”
The reaction proceeds as:
"Cathode: O2(g) + 4H+(aq) + 4e- --> 2H2O(l)
[Energy of reduction] =1.23V
Anode: Fe(s) --> Fe++(aq) + 2e-
[Energy of reduction] =-0.44V
The Fe++ formed at the anode is eventually oxidized further to Fe+++, which forms the hydrated iron (III) oxide known as rust:
4Fe++(aq) + O2(g) + 4H2O(l) + 2xH2O(l) —>
2Fe2O3xH2O(s) + 8H+(aq)"
-Brown, et al. Chemistry: The Central Science
You’ll note that the anode step must be doubled to balance the sum of the half reactions, and that gaseous oxygen is involved in both steps.
Additionally, I don’t know where you got the idea that there is no oxygen at the ocean floor. You’ll note that many aerobic organisms do just fine down there.
It seems that there is some contention as to what rust is. Fe(OH)2 and Fe(OH)3 are red-brown, kinda icky looking, and both precipitate when formed. Fe2O3 is red, kinda icky looking, and precipitates when formed.
The first oxidation of iron would also tend to alkalize the solution, possibly facilitating Fe(OH)2 formation (see my previous mention of NH3). You’ll note that the second oxidation step in the electrochemical model would seem to work with only a small adjustment, if Fe(OH)2 is substituted (and let’s ignore the hydration for simplicity’s sake).
4Fe(OH)2(s) + O2(g) --> 2Fe2O3(s) + 4H2O
This, of course, assumes that Fe(OH)2 precipitate (a solid, not aqueous) will react.
If the iron is in contact with another oxidizing agent, then it may form Fe+++ seperate from iron oxide. This could then form Fe(OH)3, but it would not technically be rust. However, it would be corrosion which is synonymous enough with rusting for most purposes.
In short, iron will corrode in an acidic environment, or when it is in contact with a suitable oxidizing agent (say nickle or tin or lead). Iron will rust in the presence of oxygen and an aqueous solution with a pH at or below 9. Iron will not rust due simply to exposure to an acid, to water of any pH, to oxygen, or to any other oxidizing agent. Iron will not corrode at all without oxygen unless in an acidic environment or in contact with a suitable oxidizer. Oxygen and water must both be present for real rust to form. Anything else is just an imitator, apparently.