Question

In the electrolysis of water, one faraday of electrical energy would liberate
(a) one mole of oxygen
(b) one gram atom of oxygen
(c) 8 g oxygen
(d) 22.4 lit. of oxygen

Answer 1

C IS CORRECT ANSWER.......

Answer 2

In the electrolysis of water, one faraday of electrical energy would liberate will be 8 g of oxygen (option c)

• We are using Faraday's second law of electrolysis which states that "when the same quantity of electricity is passed through an electrolytic solution, a number of substances liberated are proportional to their chemical equivalent weight."
• We have a formula derived from Faraday's second law, weight is directly proportional to its equivalent weights.
• We are writing the chemical reaction of electrolysis,                        $2H_2O$ → $4H^+$ + $2O^{2-}$
• At anode: $2O^{2-}$ → $O_2$ + $4e^-$
• At cathode: $4H^+$ + $4e^-$ → $2H_2$
• From the reactions mentioned above, we get to know that, 1 Faraday's charge can liberate 1 mole of oxygen. Therefore the weight of 1 mole of oxygen is 32 g.
• So, 1 F = 32/4 = 8 g of oxygen is liberated. We are dividing the weight of oxygen by 4 because ions are released during the chemical process.
• Therefore option c is the correct answer.