The quantity of charge liberated during the formation of 3.6 grams of water in hydrogen fuel cell is

Fuel cell reaction:

Net reaction : 2H2 + O2 → 2H2O 

$2\times 18=36$

 Note : Electrolysis of water

2H2O → 2H2 + O2

At Anode : 2H2 → 4H+ + 4e- → In the formation of 2 moles of H2O transfer of 4moles of e-s takes place.

At Cathode :

→ 36gr of H2O → 4f

3.6gr of H2O → ??

$\frac{3.6\times 4}{36}=0.4 \mathrm{F}$