Corrosion of iron is essentially an electrochemical phenomenon where the cell reaction are:

F e

is oxidised to

F e^{2 +}

and dissolved oxygen in water is reduced to

O H^{-}

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F e

is oxidised to

F e^{3 +}

and

H_{2} O

is reduced to

O_{2}^{2 -}

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F e

is oxidised to

F e^{2 +}

and

H_{2} O

is reduced to

O_{2}^{-}

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F e

is oxidised to

F e^{2 +}

and

H_{2} O

is reduced to

O_{2}

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Solution

The correct option is A $F e$ is oxidised to $F e^{2 +}$ and dissolved oxygen in water is reduced to $O H^{-}$
$F e \to F e^{2 +} + 2 e$ (anode reaction)

$O_{2} + 2 H_{2} O + 4 e \to 4 O H^{-}$ (cathode reaction)

The overall reaction is:
$2 F e + O_{2} + 2 H_{2} O \to 2 F e (O H)_{2}$

$F e (O H)_{2}$ may be dehydrated to iron oxide $F e O$ , or furtue oxidised to $F e (O H)_{3}$ and then dehydrated to iron rust, $F e_{2} O_{3}$

Suggest Corrections

Similar questions

While $F e^{3 +}$ is stable, $M n^{3 +}$ is not stabe in acid solution because:
$(M n^{3 +} + e^{-} \to M n^{2 +}, E^{\circ} = 1.5 V; O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V; F e^{3 +} + e^{-} \to F e^{2 +}, E^{\circ} = 0.77 V)$

$E^{\circ}$ value of $M g^{2 +} | M g, Z n^{2 +} | Z n$ and $F e^{2 +} | F e$ are – 2.37 V, - 0.76 V and 0.44 V respectively. Which of the following is correct?