How much electric charge is required to oxidise:
(a) $1 m o l e$ of $H_{2} O$ to $O_{2}$ and
(b) $1 m o l e$ of $F e O$ to $F e_{2} O_{3}$

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Solution

(a) The oxidation reaction is:
$H_{2} O 1 m o l e \to \frac{1}{2} O_{2} + 2 H^{+} + 2 e^{-} 2 m o l e$
$Q = 2 \times F$
$= 2 \times 96500 = 193000 c o u l o m b$
(b) The oxidation reaction is:
$F e O + \frac{1}{2} H_{2} O \to \frac{1}{2} F e_{2} O_{3} + H^{+} + e^{-}$
$Q = F = 96500 c o u l o m b$ .

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