Electrolysis of dilute aq. $\mathrm{NaCl}$ solution was carried out by passing $10 \mathrm{~mA}$ current.…
Electrolysis of dilute aq. $\mathrm{NaCl}$ solution was carried out by passing $10 \mathrm{~mA}$ current. The time required to liberate $0.01$ mole of $\mathrm{H}_2$ gas at the cathode is (1 Faraday $=96500 \mathrm{C} \mathrm{mol}^{-1}$ )
$9.65 \times 10^4 \mathrm{~s}$
$19.3 \times 10^4 \mathrm{~s}$
$28.95 \times 10^4 \mathrm{~s}$
$38.6 \times 10^4 \mathrm{~s}$
Solution
$2 \mathrm{H}_2 \mathrm{O}+2 e^{-} \longrightarrow \mathrm{H}_2+2 \mathrm{OH}^{-}$
For $0.01$ mole of $\mathrm{H}_2, 0.02$ mole of electrons are consumed Charge required $=0.02 \times 96500 \mathrm{C}=i \times t$
$
\text { Time required }=\frac{0.02 \times 96500}{10 \times 10^{-3}}=19.3 \times 10^4 \mathrm{~s}
$