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

`a.` At cathode`: 2Cl^(c-) rarr Cl_(2)+2e^(-)`
At anode `: 2H_(2)O+2e^(-) rarr 2 overset(c-)(O)H +H_(2)`
`b. [2e^(-)+2F=2xx96500C-=1 mol `of `Cl_(2)=71g]`
or `1F=96500C=1Eq` or `Cl_(2)=35.5g`
Since current efficiency is `50%`
`:.` Current `=96.5xx(50)/(100)=(96.5)/(2)`
Use direct relation,
`W_(Cl_(2))=(Ew_((Cl_(2)))xxIxxt)/(96500C)`
`35.5g=(35.5gxx96.5//2xxt)/(96500C)`
`:.t=(96500xx35.5xx2)/(35.5xx96.5)=2000s`
`2000s=(2000)/(3600)=0.55h`
`c.` `Eq` of `overset(c-)(O)H` formed`=Eq` of `Cl_(2)`
`=(35.5g)/(35.5g)=1Eq`
`[overset(c-)(O)H]=(Equival ent)/(V_(1))=(1)/(100L)=10^(-2)N`
`d.` `pOH=-log [10^(-2)]=2impliespH=14-2=12`