Question

In the rusting of Iron, which of the following reaction occurs at cathode?

• A. $F{e}^{2+}|Fe$
• B. $O_2|H_{2}O$
• C. $F{e}^{3+}|F{e}^{2+}$
• D. $Fe|F{e}^{3+}$

Answer 1

The correct option is B $O_2|H_{2}O$

For this, first we need to know the spot which behaves as cathode. During rusting of Iron, The electrons lost by iron are taken up by the $H^{\oplus }$ ions present on the surface of the metal which were produced by the dissociation of $H_{2}C{O}_{3}and{H}_{2}O$. Thus, $H^{\oplus }$ ions are converted into H atom.
$H^{\oplus }+e^{-}\to H$ …(i)
These H atoms either react with dissolved oxygen or oxygen from the air to form water.
$4H+O_2\to 2H_{2}O$ …(ii)
Multiplying Eq. (i) with 4 and adding to equation (ii), the complete reduction reaction may be written as
$O_2+4H^{\oplus }+4e^{-}\to 2H_{2}O(E_{red}^{\ominus }=1.23V)\left(iii\right)$
The dissolved oxygen may take up electrons directly to form $O{H}^{\ominus }$ ions as follows:
$O_2+2H_{2}O+4e^{-}\to 4O{H}^{\ominus }$
The sites where the above reactions take place act as cathodes.
So, at cathode, reduction takes place as:
$O_2+4H^{+}+4e^{-}\to 2H_{2}O(E_{red}^{\ominus }=1.23V)\Rightarrow O_2|H_{2}O$