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March 30, 2024, 02:50:54 am

Author Topic: SHIPWRECKS QUESTION  (Read 617 times)  Share 

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preliminary17hsc18

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SHIPWRECKS QUESTION
« on: November 02, 2018, 12:57:39 pm »
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HEY GORLS

Need some help with this absolutely horrendous question for shipwrecks:
How does iron corrode in acidic conditions, because:
Oxidation: Fe --> Fe2+ 2e-
Reduction: O2 + 4H+ + 4e- --> 2H20
So in the end you are left with iron ions and water so how does the iron form the hydroxide that's the actual rust?


infinityandbeyond

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Re: SHIPWRECKS QUESTION
« Reply #1 on: November 04, 2018, 01:01:59 pm »
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The reduction reaction that occurs is actually:

O2 + 2H2O + 4e --> 4OH- (this is on the periodic table, easy to remember)

Once REDOX happens, you are left with iron ions and water that form precipitate Fe(OH)2.
This precipitate further reacts with oxygen to form rust.

4Fe(OH)2 + O2 --> 2H2O + 2Fe2O3.H2O (<--this is rust)