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InterviewSolution
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8.2 times 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 times 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. |
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Answer» Solution :Hydrolysis of water At anode: `2H_(2)O rarr 4H^(+)+O_(2)+4e^(-) "...(1)"` At cathode: `2H_(2)O+2e^(-) rarr H_(2)+2OH^(-)` Overall REACTION `6H_(2)O rarr 4H^(+)+4OH^(-)+2H_(2)+O_(2)` `""`(or) Equation (1) + `(2) times 2 rArr 2H_(2)O rarr 2H_(2)+O_(2)` `therefore` According to Faradays Law of electrolysis, to electrolyse two mole of Water `(36gcong36mL " of "H_(2)O)`, 4F charge is required alternatively, when 36 mL of water is electrolysed, the charge generated `=4 times 96500 C`. `therefore` When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to `""=(4 times 96500C)/(36mL) times 9 times 10^(12)L` `""=(4 times 96500 times 9 times 10^(12))/(36 times 10^(-3))C` `""=96500 times 10^(15)C` `therefore` Given that in 1 second, `2 times 10^(6)` C is generated therefore, the time required to generate `96500 times 10^(15)C " is " =(1S)/(2 times 10^(6) times C) times 96500 times 10^(15)C` `""=48250 times 10^(9)S` `therefore` NUMBER of years `=(48250 times 10^(9))/(365 times 24 times 60 times 60)` `""=1.5299 times 10^(6) years` 1 year = 365 days `""=365 times 24` hours `""=365 times 24 times 60` min `""=365 times 24 times 60 times 60` sec. |
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