Question

For rusting of iron in presence of air and water:

• A. At cathode, reduction of $O_2$ takes place.
• B. $E_{cell}^0>0$
• C. Both (a) and (b)
• D. None of the above

Answer 1

The correct option is C Both (a) and (b)

Anode
At a particular place, on the surface of iron. Oxidation takes place that spot behaves as anodic region
Reaction:
$Fe\left(s\right)\rightleftharpoons F{e}^{2+}\left(aq\right)+2e^{-}$
$E_{F{e}^{2+}/Fe}^0=-0.44V$
Cathode:
Electrons released at anodic spot move through the metal. They go to another spot on the metal and reduce oxygen (of air) in the presence of $H^{+}$. That region behaves as cathodic region.

Reaction:
$O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-}\rightleftharpoons 2H_{2}O\left(l\right)$
$E_{O_2/H_{2}O}^0=1.23V$

Overall reaction:
$4H^{+}\left(aq\right)+O_2\left(g\right)+Fe\left(s\right)\rightleftharpoons 2H_{2}O\left(l\right)+F{e}^{2+}\left(aq\right)$
$E_{\text{cell}}^0=E_{\text{cathode}}^0-E_{\text{anode}}^0{E}_{\text{cell}}^0=1.23-(-0.44)=1.67V$

$E_{\text{cell}}^0>0$
The ferrous ions $\left(F{e}^{2+}\right)$ are
further oxidised by atmospheric oxygen to ferric ions $\left(F{e}^{3+}\right)$
Which comes out as rust in the
form of hydrated ferric oxide i.e. $(F{e}_2{O}_3.x{H}_{2}O)$