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0.1F of electricity is passed through aqueous solution of copper sulphate using Pt electrodes. If 3.18 grams of copper is deposited at cathode, the volume of oxygen liberated (at STP) at anode will be (Atomic weight of copper = 63.6 ; Atomic weight of oxygen = 16)

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560 ml

224 ml

672 ml

336 ml

answer is A.

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Detailed Solution

Applying Faraday's II law

$\frac{W_{O_{2}}}{W_{Cu}} = \frac{E_{O_{2}}}{E_{Cu}}; E_{Cu} = \frac{63.6}{2}; E_{O_{2}} = \frac{16}{2}; \frac{W_{O_{2}}}{3.18} = \frac{8}{31.8}; W_{O_{2}} = 0.8 grams; V_{O_{2}} at STP = \frac{0.8}{32} \times 22.4 litres = 0.56 litres = 560 ml$

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