Question

Electrolysis of dilute aqueous NaCl solution was carried ut be passing 10 milli ampere current. The time required to liberate 0.01 mol of `H_(2)` gas at the cathode is (1Faraday=96500 C `mol^(-1)`)
A. `9.65xx10^(4)sec`
B. `19.3xx10^(4)sec`
C. `28.95xx10^(4)sec`
D. `38.6xx10^(4)sec`

Answer 1

Correct Answer - B
`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)`