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

Answer

B

Explanation

(b) We know that Q=it. In the electrolysis of dilute NaCl solution, water actually participates as
`2H_(2)O+2e^(-) to H_(2)+2OH^(-)`
For 0.01 mole of `H_(2)`, 0.02 mol of `e^(-)` are consumed
Charge (Q) required`=(0.02" mol"xx96500" C "mol^(-1)`)
Charge (i)`=10" mA"=10xx10^(-3)" amp"`.
`t=(Q)/(i)=((0.02xx96500C))/((0.01" amp"))=((0.02+96500" amp"-sec))/((0.01" amp"))`
`=19.3xx10^(4)s`