Q. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10mA current. The time required to liberate 0.01 mole of $H_{2}$ gas at the cathode is ($1F = 96500\ C\ {mol}^{- 1}$ )
A) $9.65\ \times 10^{4}\ s\$
B) $19.3\ \times \ 10^{4}\ s$
C) $28.95\ \times \ 10^{4}\ s$
D) $38.6\ \times \ 10^{4}\ s$
Explanation
(b)
${2H}_{2}O + \ {2e}^{-}\ \rightarrow \ H_{2} + \ {2OH}^{-}$
For 0.01 mole $H_{2},\$ 0.02 mole of electrons are consumed charge required
$= \ 0.02\ \times \ 96500\ C\ = \ i\ \times \ t$
Time required $= \ \frac{0.02\ \times 96500}{10\ \times \ 10^{- 3}} = 19.3\ \times \ 10^{4}s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.