How many Coulombs of electricity are required for the oxidation of one mol of water to dioxygen?

$9.65 \times 10^{4} C$

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$1.93 \times 10^{4} C$

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$1.93 \times 10^{5} C$

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$19.3 \times 10^{5} C$

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Solution

The correct option is C
$1.93 \times 10^{5} C$

$H_{2} O \to H_{2} + \frac{1}{2} O_{2}$
$O^{2 -} \to O_{2}^{\circ}$
$2 e^{-}$
2F
$2 \times 96500 = 1.93 \times 10^{5} C$

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