Electrochemistry - Result Question 12
12. 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
$\left(1 F=96500 C mol^{-1}\right)$
(a) $9.65 \times 10^{4} s$
(b) $19.3 \times 10^{4} s$
(c) $28.95 \times 10^{4} s$
(d) $38.6 \times 10^{4} s$
(2008, 3M)
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Answer:
Correct Answer: 12. (b)
Solution:
- $0.01 mol$ of $H _2=0.02 g$ equivalent
$\Rightarrow$ Coulombs required $=0.02 \times 96500=1930 C$
$ \begin{array}{lc} \Rightarrow & Q=I t=1930 C \\ \Rightarrow & t=\frac{1930}{10 \times 10^{-3}}=19.3 \times 10^{4} s \end{array} $
BETA
