During electrolysis of aq. NaCl solution for 6 hours, 11.2 litres of Cl₂ gas at STP is evolved at anode. The current in amperes passing through the cell during electrolysis is

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During electrolysis of acidified water, O2 gas is formed at the anode. To produce O2 gas at the anode at the rate of 0.224 mL per second at STP, current passed is:

The correct option is D 3.86AAs we know,At STP, 22400 ml=32 g of O2 gas.0.224 mL O2=32×0.22422400 g=3.2×10−4gAgain,I=WZt=3.2×10−4×96500×432×1=3.86A

During electrolysis of acidified water, $O_{2}$ gas is formed at the anode. To produce $O_{2}$ gas at the anode at the rate of 0.224 mL per second at STP, current passed is:

The correct option is D $3.86 A$
As we know,

At STP, $22400 m l = 32 g$ of $O_{2}$ gas.

$0.224$ mL $O_{2} = \frac{32 \times 0.224}{22400} g = 3.2 \times 10^{- 4} g$

Again,

$I = \frac{W}{Z t} = \frac{3.2 \times 10^{- 4} \times 96500 \times 4}{32 \times 1} = 3.86 A$