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Electrolysis and Electrolytes

On passing 4 ...

Question

On passing 4 amperes of current for 10min, during electrolysis of acidulated water at STP, what quantity of oxygen in $d m^{3}$ will be liberated?

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Solution

$H_{2} O \to O_{2} + 2 H^{+} + 4 e^{-}$
$∴$ Faraday $= \frac{c h a n g e}{96500} = \frac{4 \times 10 \times 60}{96500} = 0.0248 F$
4 mol F $\to 22.4 {d m}^{3}$ of $O_{2}$
$0.0298 F = \frac{1}{4} \times 22.4 \times 0.0248 = 0.14 {d m}^{3}$ of oxygen

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