Do the question as follows :
When you write the half-oxidation and half-reduction reactions, you will see ONE EQUATION :
2 ( OH ( -ve) ) --> 1 oxygen molecule + 2 ( Hydrogen ions ) +
4 ( electrons )
[ Sorry, there is some problem with the notation bar ]
i.e there is an exchange of 4 electrons in the reaction.
So, for making 1 mole of oxygen we need 4 mole electrons i.e 4 Farads charge
[ 1 Farad = 96500 C (approx. ) ]
Now, charge provided, Total = 2Ampere * (1.5)(60)(60) sec
= 10,800 C
Now, Charge needed for liberating 1 mole oxygen = 96500 * 4 = 386000 C
I.e moles of oxygen produced = 10800 / 386000 = 0.028 moles
As, conditions provided are of STP, so volume occupied =( 0.028 * 22400 ) ml
= 627.2 ml
This is because at STP one mole of gas occupies 22400 ml volume
Forum experts please check, i am not too sure !!!
Ashish, wait for foru experts to check or other students for reply. I may be wrong !!!