An aqueous solution of `NaCl` on electrolysis given `H_(2_(g), Cl_(2)(g),` and `NaOH` according to the reaction `:`
`2Cl^(c-)+2H_(2)O rarr2overset(c-)(O)H+H_(2)(g)+Cl_(2)(g)`
A direct current of `96.5 A` with current efficiency of `50%` is passed through `100L` solution of `NaCl(20%` by weight `)`.
`a.` Write down the reactions taking place at the electrodes.
`b.` How long will it take to produce `35.5g` of `Cl_(2)?`
`c.` What will be the molartiy of the solution with respect to `overset(c-)(O)H` ion ?
`d.` What will be the final `pH` of the solution ?
Assume no loss in volume due to evaporation.

Answer

`b.t=48.71h`,`c.1.408M`

Explanation

`A.` Following reaction takes place at anode and cathode
At anode `: 2Cl^(c-) rarr Cl_(2)+2e^(-)`
At cathode `: 2H_(2)O+2e^(-) rarr 2overset(c-)(O)H +H_(2)`
`b.` Weight of `Cl_(2)=10^(3)g`
`Z` for `Cl_(2)=(35.5)/(96500)( :' ("Equivalent weight")/(96500)=Z)`
Current efficiency `=62%=0.62`
Current passe `(I)=25xx0.62A`
According to first law of Faraday,
`m=ZxxIt`
`10^(3)=(35.5)/(96500)xx25xx0.62xxt`
`t=175374.83s`
or `t=48.71h`
`c.` Equivalent of `overset(c-)(O)H` formed `=` Equivalent of `Cl_(2)` formed
`=(10^(3))/(35.5)=28.17`
Equivalent weight and molecular weight of `overset(c-)(O)H` ions are equal .
So, moles of `overset(c-)OH` formed `=28.17`
`:. [overset(c-)(O)H]=(Mol es)/(Volume(i n litre))=(28.17)/(20)=1.408M`