Question

Electrolysis of dilute aqueous solution of NaCl was carried out by passing 10 milliampere current. The time required to liberate 0.01 mole $H_2$ gas at cathode is:

• 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 $19.3\times {10}^{4}sec$

In electrolysis of aq. solution of NaCl,

as reduction potential of sodium ion is lower than that of hydrogen ion so
Cathode : $2H^{+}+2e^{-}\to H_2$

According to faraday's law,

$W=\frac{Ei.t}{96500}$

$0.01\times 2=\frac{10\times {10}^{-3}\times t}{96500}$ (as $E=1$ for hydrogen)

$t=19.3\times {10}^{4}sec$

Option B is correct.