Question
Discuss electrochemical principle regarding rusting of iron.

Answer

Electrochemical Principle: It can be explained by taking the example of formation of rust. The water vapour present on the surface of metal dissolved $CO _2$ and $O _2$ to form carbonic acid.
$H _2 O + CO _2 \longrightarrow H _2 CO _3, H _2 CO _3$ act as electrolytic
solution and ionise
$H _2 CO _3 \longrightarrow 2 H ^{+}+ CO _3^{2-}$
$\Rightarrow \quad H _2 O \longrightarrow H ^{+}+ OH ^{-}$
Iron in contact with water gets oxidise to $Fe ^{2+}$
$Fe \longrightarrow Fe ^{2+}+2 e^{-}\left( E _{ Fe ^{2+}}^0 Fe =-0.44 V\right)$
The above reaction takes places at a place which act as anode. Electrons lost by iron are taken by $H ^{+}$ions. It react with oxygen to form water.
$4 H ^{+}+ O _2 \longrightarrow 2 H _2 O$
Overall reaction is
$4 H ^{+}+ O _2+4 e^{-} \longrightarrow 2 H _2 O$
The sites where these reaction's take place act on cathodes. The overall cell reaction can we written,
$2 Fe +4 H ^{+}+ O _2 \longrightarrow 2 Fe ^{2+}+2 H _2 O \quad\left[ E _{\text {cell }}=1.67 V\right]$
$Fe ^{2+}$ ions get oxidise by atmospheric oxygen into $Fe ^{3+}$ which react with water to form rust.
$4 Fe ^{2+}+ O _2+4 H _2 O \longrightarrow 2 Fe _2 O _3+8 H ^{+}$
$\Rightarrow \quad Fe _2 O _3+x H _2 O \longrightarrow Fe _2 O _3 . x H _2 O$ rust (hydrated ferric oxide)

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