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)sec`
B. `19.3xx10^(4)sec`
C. `28.95xx10^(4)sec`
D. `38.6xx10^(4)sec`

Answer

B

Explanation

`NaCl+aq to Na^(+)+Cl^(-)`
`H_(2)OhArrH^(+)+OH^(-)`
`H^(+)+e^(-)to(1)/(2)H_(2)`
Thus, 0.5 mole of `H_(2)` is liberated by 1F=96500C
`therefore0.01V` mole of `H_(2)` will be liberated by charge
`=(96500)/(0.5)xx0.01=1930C`
`Q=Ixxt` or `t=(Q)/(I)=(1930C)/(10xx10^(-3)A)`