Question

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

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

Answer 1

Answer: (B)

$0.01\, mol$ of $H_2 = 0.02 g$ equivalent
$\Rightarrow $ Coulombs required $= 0.02 \times 96500=1930\, C$
$\Rightarrow Q=It= 1930\, C$
$\Rightarrow t= \frac{1930}{10 \times 10^{-3}}=19.3 \times 10^4 s$