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

(a) At anode `2Cl^(-) to Cl_(2)+2e`
At cathode `2H_(2)O+2e to 2OH^(-) +H_(2)`
(b) `W_(Cl_(2))=("Eit")/(96500) rArr 10^(3)=(35.5 xx 25 xx 62 xx t)/(100 xx 96500)`
`("Current"=25 xx (62)/(100))`
t=175374.83 sec =48.71 hour
(c) Eq. of `OH^(-)` formed `="Eq. of "Cl_(2)=(10^(3))/(35.5)=28.17`
Mole of `OH^(-)` formed =28.17
`[OH^(-)]=(28.17)/(20)=1.408" mol L"^(-1)`
`[OH^(-)]=(28.17)/(20)=1.408" mol L"^(-1)`
`[[OH^(-)]=("Mole")/(L)]`