Oddsnends
Registered
Wracking my brain, took my inorganic chemistry a few years ago...
If I recall correctly,
Iron + O2 gives Fe(II) species
Iron + O2 + H2O gives you Fe(III) species
The bridge example above is referring to an Fe(II) species: a tightly adhering iron oxide coating. The nightmare scenario referenced in the DAN fatality refers to Fe(III) species (aka. rust).
The key is this:
O2 + 4 e- + 2 H2O → 4 OH-
Anyone with chemistry knowledge knows this is an "Oh %#$" situation. You now have to worry about acid/base reactions.
You get redox reactions like: 2 Fe2+ + 0.5 O2 → 2 Fe3+ + O2− in presence of H2O
And equilibriums that favor: Fe(OH)3O2 and other Fe(III) species
The Key here is that rust is not passive like iron oxides are. It does not form a protective layer. It degrades continuously causing pitting. So yes you could produce enough rust to consume most oxygen in the tank.
The key here is you need BOTH water and oxygen. IF you add some electolytes (read salt water) the reactions will be catalyzed at an even faster rate.
If I recall correctly,
Iron + O2 gives Fe(II) species
Iron + O2 + H2O gives you Fe(III) species
The bridge example above is referring to an Fe(II) species: a tightly adhering iron oxide coating. The nightmare scenario referenced in the DAN fatality refers to Fe(III) species (aka. rust).
The key is this:
O2 + 4 e- + 2 H2O → 4 OH-
Anyone with chemistry knowledge knows this is an "Oh %#$" situation. You now have to worry about acid/base reactions.
You get redox reactions like: 2 Fe2+ + 0.5 O2 → 2 Fe3+ + O2− in presence of H2O
And equilibriums that favor: Fe(OH)3O2 and other Fe(III) species
The Key here is that rust is not passive like iron oxides are. It does not form a protective layer. It degrades continuously causing pitting. So yes you could produce enough rust to consume most oxygen in the tank.
The key here is you need BOTH water and oxygen. IF you add some electolytes (read salt water) the reactions will be catalyzed at an even faster rate.