Question

Electrolysis of dilute aqueous $NaCl$ solution was carried out by passing $10$ mili ampere current. The time required to liberate $0.01mole$ of $H_2$ gas at the cathode is:
$(1faraday=96500Cmo{l}^{-1})$.

• A. $9.65\times {10}^{4}sec$
• B. $19.3\times {10}^{4}sec$
• C. $28.95\times {10}^{4}sec$
• D. $38.6\times {10}^{4}sec$

Answer 1

The correct option is B $19.3\times {10}^{4}sec$

Mass of $0.01mole{H}_2=0.02g$
$W=\frac{ItE}{96500}$
$0.02=\frac{10\times {10}^{-3}\times t\times 1}{96500}$
$t=19.3\times {10}^{4}sec$.