An aqueous solution of `NaCl` on electrolysis gives `H_(2)(g), Cl_(2)(g),` and `NaOH` accroding to the reaction `:`
`2Cl^(c-)(aq)+2H_(2)Orarr2overset(c-)(O)H(aq)+H_(2)(g)+Cl_(2)(g)`
A direct current of `25A` with a current efficiency of `62%` is passed through `20L` of `NaCl` solution `(20%` by weight`)`. Write down the reactions taking place at the anode and cathode. How long will it take to produce `1 kg ` of `Cl_(2)`? What will be the molarity of the solution with respect to hydroxide ion ? `(` Assume no loss due to evaporation . `)`

Explanation

We know,
(a) Anode: `2Cl^(-) rarr Cl_(2)+2e`
Cathode: `2e + 2H_(2)O rarr 2OH^(-) + H_(2)`
`{:((b) w=(E.i.t)/96500),(10^(3) =(35.5xx25xx62xxt)/(100xx96500)),(t=175374.83 sec),( :. t=48.71 hr):}:|{:( :' W_(Cl_(2))=10^(3)g ","E_(Cl_(2))=35.5),( :. "Current efficiency" =62%),( :. i=(25xx62)/100 "ampere"):}`
(c) Eq. of `OH^(-)` formed = Eq. of `Cl_(2)` formed
`(10^(3))/(35.5) = 28.17`
`:.` Mole of `OH^(-)` formed `= 28.7 (because "monovalent")`
`:. [OH^(-)] = ("mole")/("Volume in litre") = (28.17)/(20)`
`= 1.408 "mol litre"^(-1)`