How much $O_{2}$ gas will be collected at the anode at $300$ K temperature and $1$ bas pressure if $2.5$ ampere electric current is passed for one hour in electrolysis of aqueous solution of $N a_{2} S O_{4}$ . (F $= 96500$ Coulomb) [ $1$ mole gas volume is $22.4$ litre at STP].

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Solution

No of coulombs = current in amperes $\times$ time in second
no of coulombs = $0.50 \times 30 \times 900$
the first release of oxygen from water molecule
$2 H_{2} O (l) \to O_{2} + 4 H^{+} (a q) + 4 e^{-}$
the alternative way of release of oxygen from water molecule
$4 O H^{-} (a q) \to 2 H_{2} O (l) + O_{2} (g) + 4 e^{-}$
release of one mole of $O_{2}$ involve 4 mole of electron
$4 \times 96500$ coulomb gives $22.4 d m^{3}$ O_2 $at STP
volume of oxygen $= 900 \times 22.4 / 4 \times 96500$
$= 0.052$

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How much O2 gas will be collected at the anode at 300K temperature and 1 bas pressure if 2.5 ampere electric current is passed for one hour in electrolysis of aqueous solution of Na2SO4. (F=96500 Coulomb) [1 mole gas volume is 22.4 litre at STP].

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