Question

Explain the mechanism of rusting of iron with reactions.

Answer 1

Hint: Rust is an iron oxide, usually red oxide formed by the redox reaction of iron and oxygen in the presence of water or air moisture. It consists of hydrated iron (III) oxides, $\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}.\text{n}{{\text{H}}_{2}}\text{O}$ and iron (III) oxide hydroxide$\\{\text{FeO(OH), Fe(OH}{{\text{)}}_{3}}\\}$.

Complete step by step solution:
Rusting is the common term for corrosion of iron and its alloys, such as steel. Many other metals undergo equivalent corrosion, but the resulting oxides are not commonly called rust.

The overall rusting involves the following steps:
Oxidation occurs at the anodes of each electrochemical cell. Therefore, at each anode neutral iron atoms are oxidized to ferrous ions.

1- At anode:
Thus, the metal atoms in the lattice pass into the solution as ions, leaving electrons on the metal itself. These electrons move towards the cathode region through the metal.
$\text{Fe(s)}\to \text{F}{{\text{e}}^{2+}}(\text{aq)+2}{{\text{e}}^{-}}$

2- At cathode:
The electrons are taken up by hydrogen ions (reduction takes place). The ${{\text{H}}^{+}}$ ions are obtained either from water or from acidic substances in water.
${{\text{H}}_{2}}\text{O}\to{{\text{H}}^{+}}+\text{O}{{\text{H}}^{-}}$ $\text{C}{{\text{O}}_{2}}+{{\text{H}}_{2}}\text{O}\to{{\text{H}}^{+}}+\text{HC}{{\text{O}}_{3}}^{-}$
The hydrogen atoms on the iron surface reduce dissolved oxygen.
$\text{4}{{\text{H}}^{+}}+{{\text{O}}_{2}}\to 2{{\text{H}}_{2}}\text{O}$
Therefore, the overall reaction at cathode of different electrochemical cells may be written as,
$\text{4}{{\text{H}}^{+}}+{{\text{O}}_{2}}+4{{e}^{-}}\to 2{{\text{H}}_{2}}\text{O}$

The overall redox reaction may be written by multiplying reaction at anode by 2 and adding reaction at cathode to equalize number of electrons lost and gained-
Oxidation half reaction:
$\text{Fe(s)}\to \text{F}{{\text{e}}^{2+}}(aq)+2e] \times 2$
Reduction half reaction:
$\text{4}{{\text{H}}^{+}}+{{\text{O}}_{2}}+4{{e}^{-}}\to 2{{\text{H}}_{2}}\text{O}$
Overall cell reaction:
$\text{2Fe(s)+4}{{\text{H}}^{+}}+{{\text{O}}_{2}}\to 2\text{F}{{\text{e}}^{2+}}(aq)+2{{\text{H}}_{2}}\text{O}$
The ferrous ions are oxidized further by atmospheric oxygen to form rust.
$4\text{F}{{\text{e}}^{2+}}(aq)+{{\text{O}}_{2}}(g)+4{{\text{H}}_{2}}\text{O}\to \text{2F}{{\text{e}}_{2}}{{\text{O}}_{3}}+8{{\text{H}}^{+}}$
$\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}+x{{\text{H}}_{2}}\text{O}\to \text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}.x{{\text{H}}_{2}}\text{O}$
Therefore, this is the mechanism of rust with the chemical formula $\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}.x{{\text{H}}_{2}}\text{O}$.

Note: It may be noted that salt water accelerates corrosion. This is mainly due to the fact that salt water increases the electrical conduction of electrolyte solution formed on the metal surface. Therefore, rusting becomes a more serious problem where salt water is present.