InterviewSolution
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InterviewSolution
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An aqueous solution of NaCl on electrolysis gives `H_(2(g))Cl_(2(g))` and NaOH according to reaction : `2Cl_((aq.))^(-)+2H_(2)Orarr 2OH^(-)+H_(2(g))+Cl_(g)` A direct current of 25 ampere with a current efficiency of 62% is passes though 20 litre of NaCl solution (20% by weight). Write drown the reactions taking place at teh electrodes. (b) How long will it take to produce 1kg of `Cl_(2)` ? (c ) What will be the molarity od solution with respect to `OH^(-)` ? Assume no loss in volume due to evaporation. |
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Answer» We know, (a) Anode : `2Cl^(-)rarr Cl_(2)+2e` Cathod : `2e+2H_(2)O rarr 2OH^(-)+H_(2)` (b) `w=(E.i.t)/(96500)` `because W_(Cl_(2))=10^(3)g, E_(Cl_(2))=35.5` `10^(3)=(35.5xx25xx62xxt)/(100xx96500)` `therefore` Current efficiency = 62% `t = 175374.83` sec `therefore i = (25xx62)/(100)` ampere `therefore t=48.71` hr (c ) Eq. of `OH^(-)` formed = Eq. of `Cl_(2)` formed `=(10^(3))/(35.5)=28.17` `therefore` Mole `OH^(-)` formed `=28.17` (`because` monovalent) `therefore [OH^(-)]=("mole")/("valume in litre")=(28.17)/(20)=1.408"mol litre"^(-1)` |
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