How many coulombs of electricity are required for the oxidation of 1 mole of $H_{2} O$ to $O_{2}$ ?

9.65 \times 10^{4} C

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4.825 \times 10^{4} C

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1.93 \times 10^{5} C

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3.86 \times 10^{5} C

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Solution

The correct option is C $1.93 \times 10^{5} C$
Oxidation of $H_{2} O$ to $O_{2}$ is:
$2 H_{2} O \to 4 H^{+} + O_{2} + 4 e^{-}$
Oxidation of 2 moles of water requires $96500 \times 4 C$
$∴$ oxidation of 1 mole of water requires $= \frac{96500 \times 4}{2} = 1.93 \times 10^{5} C$

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