8.2 xx 10^(12) 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 xx 10^6 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.

8.2 xx 10^(12) litres of water is available in a lake. A power reactor

1.	8.2 xx 10^(12) 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 xx 10^6 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» Solution :Hydrolysis of water: At anode: `2H_2O to 4H^+ + O_2 + 4E^(-) "" …………..(1)` At cathode: `2H_2O + 2e^(-) to H_2 + 2OH^(-)` Overall reaction : `6H_2O to 4H^+ + 4OH^(-) + 2H_2 + O_2` (or) Equation `(1) + (2) xx 2 implies 2H_2O to 2H_2 + O_2` `:.` ACCORDING to Faraday.s Law of electrolysis, to electrolyse two mole of Water `(36g ~= 36 mL " of " H_2O)`, 4F charge is required alternatively, when 36 mL of water is electrolysed,the charge generated = `4 xx 96500 C`. `:.` When the whole water which is avialable on the lake is completely electrolysed the amount of charge generated is equal to `(4 xx 96500 C)/(36 mL) xx 9 xx 10^(12)L = (4 xx 96500 xx 9 xx 10^(12))/(36 xx 10^(-3)) C = 96500 xx 10^(15) C` `:.` Given that in 1 second, `2 xx 10^(6) C` is generated therefore, the TIME required to generate `96500 xx 10^(15) C is = (1S)/(2 xx 10^6C) xx 96500 xx 10^(15) C = 48250 xx 10^9S` `:. ` Number of years = `(48250 xx 10^9)/(365 xx 24 xx 60 xx 60) "1 year= 365 years "` `= 1.5299 xx 10^(6) "years" = 365 xx 24` hours `"" = 365 xx 24 xx 60` min `"" = 365 xx 24 xx 60 xx 60 `sec.