Question

If a water electrolysis cell operates at a current of 7.6 AA , how long will it take to generate 27.0 LL of hydrogen gas at a pressure of 27.0 atm and a temperature of 26 ∘C∘C?

R = 0.08206 L atm mol^-1 K^-1

Answer 1

Reduction of H2O to give H2 is as follows

  

21,0 + 2e + H2 + 2OH

thus 2 mole of electron is used up in formation of one mole of hydrogen gas.
here, mole of hydrogen can be calculated using PV=nRT

Thus, n = 29.711 ( P=27 atm, V= 27 litre and T = 299 K)

PI n = RT

from above we had obtained 1 mole H2 is liberated from 2 mole electron thus to obtain 29.711 mole H2 we need 59.422 mole of electron

Charge on one mole electron =
thus charge on 59.422 mole = 5728280.8 coulomb

1.6 x 10-19 x 6.022 x 1023 = 96400coulomb(approx))

now we know

Q=IxTinsec)

thus T = 753721.15 sec ( I = 7.6 A)

= 209.36 hours