Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of `H_(2)` gas at the cathode is (1 Faraday=96500 C `"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+aqrarrNa++Cl^(-)`
`H_(2)OhArrH^(+)+OH^(-)`
`H_(2)^(+)+e^(-)rarr+1//2H_(2)`
`therefore` 0.5 mole of `H_(2)` is liberated by 1F= 96500C
0.01 mole of `H_(2)` will be liberated by charge
`=(96 500)/(0.5)xx0.01=1930C`
`Q=1xxt` or `t=(Q)/(I)=(1930)/(10xx10^(-3))A`
`=193000"sec" =19.3xx10^(4)` sec.