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

F e^{2 +} | F e

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O_{2} | H_{2} O

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F e^{3 +} | F e^{2 +}

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F e | F e^{3 +}

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Solution

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 $H_{2} C O_{3} a n d H_{2} O$ . 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.
$4 H + O_{2} \to 2 H_{2} O$ …(ii)
Multiplying Eq. (i) with 4 and adding to equation (ii), the complete reduction reaction may be written as
$O_{2} + 4 H^{\oplus} + 4 e^{-} \to 2 H_{2} O (E_{r e d}^{⊖} = 1.23 V) (i i i)$
The dissolved oxygen may take up electrons directly to form $O H^{⊖}$ ions as follows:
$O_{2} + 2 H_{2} O + 4 e^{-} \to 4 O H^{⊖}$
The sites where the above reactions take place act as cathodes.
So, at cathode, reduction takes place as:
$O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O (E_{r e d}^{⊖} = 1.23 V) \Rightarrow O_{2} | H_{2} O$

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Similar questions

F e | F e^{2 +} | | F e^{3 +} | F e

F e | F e^{2 +} | | F e^{3 +}, F e^{2 +} | P t

F e | F e^{3 +} | | F e^{3 +}, F e^{2 +} | P t

2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V,

F e^{2 +} + 2 e^{-} \to F e (s); E^{\circ} = - 0.44 V .

Δ G^{\circ} (i n k J)

2 H^{+} + \frac{1}{2} O_{2} + 2 e^{-} \to H_{2} O; E^{o} = 1.23 V

F e^{2 +} + 2 e^{-} \to F e; E^{o} = - 0.44 V

Δ G^{o}

F e^{2 +} ⇌ F e; E^{\circ} = - 0.44 V

F e^{3 +} ⇌ F e^{2 +}; E^{\circ} = + 0.77 V

F e^{2 +}, F e^{3 +}

F e

(E^{o})

F e^{3 +} \to F e

F e^{2 +} + 2 e 6 - \to F e; E_{F}^{o} e_{F e^{2 +} / F e} = - 0.47 V

F e^{3 +} + e^{-} \to F e^{2 +}; E_{F e^{3 +} / F e^{2 +}}^{o} = + 0.77 V

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