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

Explanation

In corrsoion , a metal is oxidised by the loss of electrons to oxygen with the formation of oxides. So, an electrochemical cell is set-up,
e.g. rusting of iron involves the following steps `:`
(i) The water layer present on the surface of iron dissolves acidic oxides from air like `CO_(2)` and forms acid to produce `H^(+)` ions.
`H_(2) O + CO_(2) rarr H_(2) CO_(3) hArr 2H^(+) + CO_(3)^(2-) `
(ii) In the present of `H^(+)` ions, iron starts losing electrons at some spot to form ferrous ions. The spot behaves as anode.
`Fe(s) overset( "Oxidation")(rarr) Fe^(2+) (aq) + 2e^(-) , ( E_(Fe^(2+) //Fe)^(@)= 0.44V )`
(iii) The electrons released at anode move to another spot where `H^(+)` ions and the dissolved oxygen gain these electrons. This sopt becomes a cathode.
`O_(2)(g) + 4H^(+)(aq) + 4e^(-)overset( "Reduction")(rarr) 2H_(2) O(l), ( E_(H^(+) //O_(2) //H_(2) O )^(@) = 1.23V)`
(iv) Overall reaction, i.e., redox reaction is
`2Fe(s) + O_(2)(g) + 4H^(+) (aq) rarr 2Fe^(2+) (aq) + 2H_(2) O (L), E_("cell")^(@) = 1.67V`
(v) Ferrous ions are further oxidised by the atmospheric oxygen to ferric ions which combine with water molecules to form hydrated ferric oxides.
Oxidation `Fe(s) rarr Fe^(2+) (aq) + 2e^(-)`
Reduction `O_(2)(g) + 4H^(+) (aq) + 4e^(-) rarr 2H_(2) O(l)`
Amospheric oxidation
`2Fe^(2+) (aq)+ 2H_(2)O(l) + ( 1)/(2) O_(2)(g) rarr Fe_(2) O_(3)(s) + 4H^(+) (aq)`