How many coulomb are required for the oxidation of $1 m o l e$ of $H_{2} O_{2}$ to $O_{2}$ ?

9.65 \times 10^{4} C

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93000 C

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1.93 \times 10^{5} C

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19.3 \times 10^{2} C

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Solution

The correct option is C $1.93 \times 10^{5} C$
$H_{2} O_{2} \to O_{2} + 2 H^{+} + 2 e^{-}$
$1 m o l H_{2} O_{2} \equiv 2 m o l e^{-}$
$\equiv 2 \times 96500 C$
$\equiv 193000 C$
$= 1.93 \times 10^{5} C$ .

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