How do you explain the electrochemical theory of corrosion?

Aug 28, 2017

Well, see this old answer.

Explanation:

Erect a structure, and invariably it will be an iron and steel structure, and from the moment the steel leaves the foundry, you fight a losing battle against, corrosion, against oxidation......This is especially true of bridges over bodies of salt water, where the electrolyte in the solvent facilitates corrosion...

Iron loses electrons as it corrodes......or as it oxidizes....

$F e + {H}_{2} O \rightarrow F e O + 2 {H}^{+} + 2 {e}^{-}$ $\left(i\right)$...else

$2 F e + 3 {H}_{2} O \rightarrow F {e}_{2} {O}_{3} + 6 {H}^{+} + 6 {e}^{-}$...$\left(i i\right)$

And air-borne oxygen is the electron source.....

$\frac{1}{2} {O}_{2} + 2 {H}^{+} + 2 {e}^{-} \rightarrow {H}_{2} O$ $\left(i i i\right)$

And we add $\left(i\right) + \left(i i\right) + 4 \times \left(i i i\right)$ to give a reaction that represents the oxidation......

$3 F e + \cancel{4 {H}_{2} O} + 2 {O}_{2} + \cancel{8 {H}^{+} + 8 {e}^{-}} \rightarrow F e O + F {e}_{2} {O}_{3} + \cancel{8 {H}^{+} + 8 {e}^{-}} + \cancel{4 {H}_{2} O}$

to give finally.....

$3 F e + 2 {O}_{2} \rightarrow F e O + F {e}_{2} {O}_{3}$

And of course the oxides are non-structural materials, and will tend to accumulate in the junctions between different metals, and in those places where the structure bears a load.

I have written before here, that we have all seen pictures of monumental bridges all over the world, in Sydney Harbour, in San Francisco Bay, in New York harbour; all of these have approx. 300-400 year life spans if inspections, and rust control were to cease tomorrow.

I hope that helps!