In the electrolysis of an aqueous solution of `NaOH`, `2.8` litres of `O_(2)(g)` is liberated at the anode at `NTP`. Volume of hydrogen gas liberated at the cathode at `NTP` will be

Options

A. `2.8` litres
B. `5.6` litres
C. `1.4` litres
D. `11.2` litres

Answer

B

Explanation

Number of equivalents of `O_(2)(g)` liberated at anode = Number of equivalents of `H_(2)(g)` liberated at cathode
eq. of `O_(2)(g)` liberated
`= ("Vol. of" O_(2)(g) "liberated at NTP")/("Vol. occupied by 1 eq. of" O_(2)(g) "at NTP")`
`= (2.8 L)/(5.6 L eqq.^(-1))`
`(1)/(2)eq.`
Thus no of equivalent of `H_(2)(g)` liberated at the cathode is also `(1)/(2)`.
Volume of `H_(2)(g)` liberated at `NTP`
= No. of equivalents of `H_(2)(g)` liberated `xx` Vol. occupied by `1` eq. of `H_(2)(g)` at `NTP`
`= (1)/(2) xx 11.2 L`
`= 5.6` litres
Short cut method:
Overall reaction during electrolysis of water is
`underset(2 mol)(2H_(2)O(l)) rarr underset(2 mol)(2H_(2)(g))+underset(2 mol)(O_(2)(g)`
At constant `T` and `p`
According to Avogadro's law, at the same `T` and `P`, ratio of moles is the ratio of volumes. Thus, volumes of `H_(2)(g)` liberated are in the `2 : 1` ratio i.e. vol of `H_(2)(g)` is twice the vloume of `O_(2)(g)`