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. `9.65xx10^(4)S`
B. `19.3xx10^(4)`
C. `28.95xx10^(4)`
D. `38.6xx10^(4)s`

Answer

B

Explanation

Electrolysis of dil. NaCl produces `H_(2)` gas at cathode because reduction of water is more favourable than reduction of `Na^(+)`.
`2H_2O + 2e^(-) rightarrow H_(2) + 2O^(-)H`
`1mol H_(2) = 2F`
`0.01mol H_(2) equiv 2F xx0.01C = 0.02F`
So, `Q=it Rightarrow 2xx96500 xx0.01 = 10xx10^(-3)xxt` `Rightarrow t= 19.3xx10^(4) sec`