The volume of `H_(2)` gas at NTP obtained by passing 4 amperes through acidified `H_(2)O` for 30 minutes is

Options

A. 0.0836 L
B. 0.0432 L
C. 0.1672 L
D. 0.836 L

Answer

D

Explanation

`H_(2)O (l) to H_(2)(g) + 1/2 O_(2) (g)`
`2H^(+)(aq) + 2e^(-) to H_(2)(g)`
For the production of one mole `H_(2), 2` mole electron. (i.e., 2F charge) are required.
`therefore 2` Faraday = `2 xx 96500` coulomb charge liberate.
`H_(2) = 22400 ml` at NTP
`therefore 4 xx 30 xx 60` coulomb charge will liberate.
`H_(2) = (4 xx 30 xx 60 xx 22400)/(2 xx 96500) = 835.6 mL`
= 0.836 L