Four moles of electrons were transferred from anode to cathode in an experiment on electrolysis of water. The total volume. of the two gases (dry and at STP) produced will be approximately: (in litres) (A) 22.4 (B) 44.8 (C) 67.2 (D) 11.2

 Four moles of electrons were transferred from anode to cathode in an experiment on electrolysis of water. The total volume. of the two gases (dry and at STP) produced will be approximately: (in litres)
(A) 22.4
(B) 44.8
(C) 67.2
(D) 11.2

Solution=
According to the first law of Faraday,
Wt/Ewt=Q/F
QF= faraday No.
∴ charge of 1 mol of electrons is equal to 1 faraday.
Wt/Ewt=ne/F∴Q=ne and n=4
mole×V.f.=4F/F
mole×2=4
mols of H2O=2
Hence the two moles of H2O will be decomposed into its gaseous components
2H2O⟶2H2+O2
2moles of H2(at the cathode) and 1mol of O2(at the anode) will produce.
moles=v/22.4
So volume of H2 and O2 will be = 44.8+22.4 = 67.2