Electrochemistry 1 Question 12
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12. Electrolysis of dilute aqueous $\mathrm{NaCl}$ 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 $\left(1 \mathrm{~F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right)$
======= ####12. Electrolysis of dilute aqueous $\mathrm{NaCl}$ 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 $\left(1 \mathrm{~F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right)$
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $9.65 \times 10^{4} \mathrm{~s}$
(b) $19.3 \times 10^{4} \mathrm{~s}$
(c) $28.95 \times 10^{4} \mathrm{~s}$
(d) $38.6 \times 10^{4} \mathrm{~s}$
(2008, 3M)
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Answer:
Correct Answer: 12. (a)
Solution:
- $0.01 \mathrm{~mol}$ of $\mathrm{H}_{2}=0.02 \mathrm{~g}$ equivalent
$\Rightarrow$ Coulombs required $=0.02 \times 96500=1930 \mathrm{C}$
$$ \begin{array}{lc} \Rightarrow & Q=I t=1930 \mathrm{C} \ \Rightarrow & t=\frac{1930}{10 \times 10^{-3}}=19.3 \times 10^{4} \mathrm{~s} \end{array} $$
