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Explain with the help of diagram wet corrosion in neutral medium
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Diagram

Corrosion in neutral media takes place due to absorption of Oxygen.

If electrolyte is neutral or alkaline aqueous solution, corrosion takes place by absorption of $O_2$ rusting of iron in water containing dissolved oxygen occurs by oxygen absorption mechanism.

At anodic area iron will dissolve by oxidation. The surface of iron is usually coated with the thin film of iron oxide. But if this iron oxide film develops some cracks, anodic areas are created on the surface; while the metal (iron) act as cathodes.

Here the anodic areas are small surface while rest of the surface of the metal forms large cathodes.

At anode, $Fe \rightarrow Fe^{2+} + 2e^-$

The electrons flow to cathodic area through and will be accepted by $O_2$.

At cathode, $\frac{1}{2} O_2 + H_2O + 2e^- \rightarrow 2OH^- Fe^{2+} + 2OH \rightarrow Fe(OH)_2$

If enough $O_2$ is present, ferrous hydroxide easily oxidizes to ferric hydroxide

$2Fe(OH)_2 + \frac{1}{2} O_2 + H_2O \rightarrow 2Fe(OH)_3 ----\gt \text{Ferric hydroxide rust}$

This product called yellow rust, which is nothing but $Fe_2O_3.H_2O$

If $O_2$ is limited, the corrosion product will be black unhydrous magnetite, $Fe_3O_4$.

If environment is aqueous solution of $NaCl$ containing dissolved $O_2$,

$NaCl \rightarrow H_2O \rightarrow Na^{+} + Cl^{-}$

At cathode, $Na^+ + OH \rightarrow NaOH$

At anode: $Fe^{2+} + 2Cl \rightarrow FeCl_2 -----\gt Ferrous Chloride$

Both the products $NaOH$, $FeCl_2$ are soluble in water they react with each other and ferrous hydroxide precipitates which further oxidizes to ferric hydroxide $Fe(OH)_3$.