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.

Explanation

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)`