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Secondary Cells

H2g and O2g...

Question

$H_{2} (g)$ and $O_{2} (g)$ , can be produced by the electrolysis of water. What total volume (in $L$ ) of $O$ and $H_{2}$ are produced at $S T P$ when a current of $30 A$ is passed through a $K_{2} S O_{4} (a q)$ solution for $193 m i n .$ ?

20.16

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40.32

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60.48

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80.64

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Solution

The correct option is B 40.32
SOLUTION:
When 30 amperes is passed for 193 minutes, the quantity of current passed =193×60×30=347400 coulombs.
convering into faraday we get
$\to \frac{347400}{96500} = 3.6 F a r a d a y .$
$H_{2} O e l e c t r o l y s i s - ------ \to k_{2} s o_{4} (a q .) \frac{1}{2} O_{2} + H_{2}$
On passing 3.6 Faradays of electricity 3.6 moles of Hydrogen and (3.6×0.5)=1.8 moles of Oxygen will be formed.
Volume of oxygen produced at STP =1.8×22.4=40.32 liters.
hence the correct opt: B

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