When an electric current is passed through acidified water, `112ml` of `H_(2)` gas at `NTP` is collected at the cathode is `965` seconds. The current passed in amperes is

Options

A. `2.0`
B. `0.1`
C. `0.5`
D. `1.0`

Answer

D

Explanation

Reduction at cathode :
`2 e^(-) + 2 H_(2) O + H_(2) + 2OH^(-) " "` (valence factor) `H_(2) = 2` at NTP 22400 mL of `H_(2)` = 1 mole of `H_(2)`
112 mL of `H_(2) = (1)/(22400) xx 112 = 0.005` mole of `H_(2)`
Moles of `H_(2)` produced `= (I xx t)/(96500) xx ` mole ratio
`0.005 = (1 xx 965)/(96500) xx (1 "mole of " H_(2))/(2 "mole of " e^(-))`
`I =1.0 A `