Electrolysis of dilute aqueous $\mathrm{NaCl}$ solution was carried out by passing $10 \mathrm{~mA}$ current…

Electrolysis of dilute aqueous $\mathrm{NaCl}$ solution was carried out by passing $10 \mathrm{~mA}$ current. The time required to liberate $0.01 \mathrm{~mol}$ of $\mathrm{H}_{2}$ gas at the cathode is $\left(1 \mathrm{~F}=96500 \mathrm{C} \mathrm{~mol}^{-1}ight)$
  1. $9.65 \times 10^{4} \mathrm{~s}$
  2. $19.3 \times 10^{4} \mathrm{~s}$
  3. $28.95 \times 10^{4} \mathrm{~s}$
  4. $38.6 \times 10^{4} \mathrm{~s}$

Solution

$2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} ightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}$
For $0.01$ mole $\mathrm{H}_{2} ~0.02$ mole of electrons are consumed
charge required $=0.02 \times 96500 \mathrm{~C}=\mathrm{i} \times \mathrm{t}$
Time required $=\frac{0.02 \times 96500}{10 \times 10^{-3}}=19.3 \times 10^{4} \mathrm{~s}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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