Electrolysis of dilute aqueous `NaCl` solution was carried out by passing `10mA` current. The time required to liberate `0.01mol` of `H_(2)` gas at the cathode is `(1F=96500C mol ^(-1))`

Options

A. `9.65 xx 10^(4)`sec
B. `19.3 xx 10^(4)` sec
C. `28.95 xx 10^(4)`sec
D. `38.6 xx 10^(4)`sec

Answer

B

Explanation

No. of Faraday of electricity ot liberate `0.01`
Mole `= 0.01 xx 2F = 0.02F`
`=2 xx 10^(-2) xx 9600 = 1930` coulomb
`A = 10` miliampere `= 0.01` ampere
`Q = 1 xx t`
`t = (Q)/(I) = (1930)/(0.01) = 1.93 xx 10^(5) = 19.3 xx 10^(4)` sec