Question

1.3. How much gram of hydrogen is needed to operate a 50 kW hydrogen/oxygen fuel cell for 3 hours? The potential of the cell is 0.7 V (remember that Power-IV). Molecular weight of H2 is 2 g/mol. The reaction at the anode is H2 2H+ 2e

Answer 1

Sol)

Given
P = 50 K w

We know 1 kilowatt=1000 watt

So here 50 kw is given

i.e.,50 kw×1000= 50000 W

V= 0.7 volts

We know the formula for I is I=P/V equation(1)

  Substitute “p” and :v” value in equation (1)

I=50000/0.7

I=71428.57143 Amp

Here we know 1 Amp = 1 Coulomb/sec

Therefore we need 71428.5714 Coulomb charge per second

From data given

Running time = 3 hours

1 hour=60 minutes

1 minute=60 sec

So here Running time =3×60×60 sec = 10800 sec

So here multiply “I” with “t” we get

Total charge = 71428.5714×10800 = 771428571 c

Here reaction suggests 1 molecular of H ₂ produces 2 molecular of electrons

We learnt that Charge of 1 electron = 1.6x10-19 C

Charge of 2 molecular of electrons = (1.6×10^-19)(2×6.022×10²³)

=1.6×2×6.022×10^-19×10²³

19.2704×10⁴

= 192704 C

1 molecular of H ₂ produces 192704 C charge

So, Moles of H ₂ required = Total charge/charge of 1 molecular

= 771428571/192704

= 4003.17 molecule

We know 1 molecule H ₂ weighs 2 g ms

So, we have to multiply with 2

i.e.,2×4003.17

Hence mass of H 2 required = 8006.34 g ms =8 kg