Hello Ashish,

 

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 !!!