Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

Explanation

The water presence on the surface of iron dissolves acidic oxides of air like `CO_(2),SO_(2)`, etc. to form acids which dissociate to give `H^(+)` ions,
`H_(2)O+CO_(2)toH_(2)CO_(3)hArr2H^(+)+CO_(3)^(2-)`
In the presence of `H^(+)`, iron looses `e^(-1)s` to form
`Fe^(3+)`. Hence, this spot acts as anode.
`Fe(s)toFe^(2+)(aq)+2e^(-)`
the `e^(-1)`s released move through the metal to reach another spot where `H^(+)` ions and dissolved oxygen take up these `e^(-1)`s and reduction occurs.
this spot, thus, acts as cathode:
`O_(2)(g)+4H^(+)(aq)+4e^(-)to2H_(2)O(l)`
The overall reaction is
`2Fe(s)+O_(2)(g)+4H^(+)(aq)to2Fe^(2+)(aq)+2H_(2)O(l)`
Thus, an electrochemical cell is set up on the surface. ferrous ions are further oxidised by atmospheric oxygen to ferric ions which combine with water to form hydrated ferric oxide, `Fe_(2)O_(3).xH_(2)O`, which is rust.