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

Options

A. `9.65 xx10^(4)sec`
B. `19.3 XX10^(4)sec`
C. `28. 95xx10^(4)sec`
D. `38.6 xx10^(4)sec`

Answer

B

Explanation

`H_(2)O hArrH^(*)* OH^(*)`
`H^(*) * c ^(*) * * 1/2 H_(2)`
`therefore0.5` mole of `H_(2)` is liberated by 1 `F = 96500C 0.01 ` mole of `H_(2)` will be liberated by
`=(96500)/(0.5)0.01" "=1930C`
`Q=1xx1`
`t * (Q)/(T)* (1930)/(10 * 10^(*3)A)* 19.3* 10^(4)sec`