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. `9.65 xx 10^(4) sec`
B. `19.3 xx 10^(4) sec`
C. `28.95 xx 10^(4) sec`
D. `38.6 xx 10^(4) sec`

Answer

B

Explanation

Q= it or Q=`10xx10^(-3)xx t " "…(i)`
`therefore t=(Q)/(10xx10^(-3))`sec
`2H_(2)O + 2e^(-) to H_(2) + 2OH^(-)`
0.1 mole of `H_(2)` is liberated by 0.02 Faraday of charge i.e. Q=0.02 `xx96500 " "…(ii)`
From (i) and (ii) , `10xx10^(-3)xxt=0.02xx96500`
`therefore t=(0.02xx96500)/(10xx10^(-3))=19.3xx10^(4)` sec