Question

8.2 x 1012 litres of water is available in a lake. A power reactor using the electrolysis of water in the lake produces electricity at the rate of 2 x 106 Cs-1 at an appropriate voltage. How many years would it like to completely electrolyse the water in the lake. Assume that there is no loss of water except due to electrolysis.

Answer 1

Hydrolysis of water:

At anode: 2H2O → 4H+ + O2 + 4e-……..(1)

At cathode: 2H2O + 2e- → H2 + 2 OH-

Overall reaction: 6H2O → 4H- + 4OH- +2H2 + O2 (or)

Equation (1) + (2) x 2

= 2H2O → 2H2 + O2

According to Faraday’* Law of electrolysis, to electrolyse two mole of Water (36g ≃ 36 mL. of H2O), 4F charge is required alternatively, when 36 mL of water is electrolysed, the charge generated = 4 x 96500 C.

When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to

Given that in 1 second, 2 x 106 C is generated therefore, the time required to generate

= 1.5299 x 106

= 365 x 24 hours

= 365 x 24 x 60 min

= 365 x 24 x 60 x 60 sec