386 amperes of current is passed through acidulated water for one minute and forty seconds. The volume of gas liberated at cathode will be (at STP conditions)

Cathode : 2H2O+2e−→H2+2OH−×2Anode : 2H2O→4H ++O2+4e−Net Reaction: 2H2O→4F2H2Cathode+O2AnodeQ = itQ = 386 x 100 = 38600 C4 x 96500 C will liberate 2 x 22400 cc of H2 at STP38600 C will liberate ‘X’  of H2 at STPX=2×22400×386004×96500=4480 cc=4.48 l