Question

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

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

Answer 1

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

Equivalent of $H_2$ = $2\times$ moles of $H_2$
$0.01mol$ of $H_2$ = $0.02g$ equivalent
Coulombs required = equivalent of $H_2\times 96500$
⇒ $0.02\times 96500=1930C$
⇒ $Q=I\times t=1930C$
= $t=\frac{1930}{10\times {10}^{-3}}=19.3\times {10}^{4}s$