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Blazing goIITian

Joined:	16 Dec 2006
Post:	1362

6 Feb 2007 22:45:59 IST

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563

electrochemistry

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What volume of O₂ would be liberated from an aquous sol. of NaOH by a current of 2A passed for 1.5hrs. (Temp. = 27⁰ and pressure = 1 atm )

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Aditya Arora

Blazing goIITian

Joined: 3 Dec 2006

Posts: 1043

6 Feb 2007 23:51:00 IST

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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 !!!

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