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

Answer

`48.71 hour, [OH^(-)] = 1.41M`

Explanation

`2NaCI hArr 2Na^(+) +2CI^(-)`
`2H_(2)O hArr 2H^(+) +2OH^(-)`
At anode:
`2CI^(-) rarr CI_(2) +2e^(-)`
At anode:
`2H^(+) +2e^(-) rarr H_(2)`
Not reaction
`2NaCI + 2H_(2)O hArr 2OH^(-) +H_(2) + CI_(2) +2Na^(+)` ltbr. `100 = (35.5)/(96500) xx 25.62 xx t`
`t = 175374.83 sec = 48.74 hr`.
moles of `OH^(-) = (1000)/(71) xx2`
volume `= 20L`
`[OH^(-)] = (2000)/(71 xx 20) = 1.408M`