Question

Electrolysis of dilute aqueous $NaCl$ solution was carried out by passing $10$ milli ampere current. The time required to liberate $0.01\, mol$ of $H _{2}$ gas at the cathode is (1 Faraday $=96500 \,C\, mol ^{-1}$

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

Answer 1

Answer: (B)

$Q = i \times t $
$Q =10 \times 10^{-3} \times t $
$2 H _{2} O +2 e ^{-} \longrightarrow H _{2}+2 OH ^{-}$
To liberate $0.01$ mole of $H _{2}, 0.02$ Faraday charge is required
$Q =0.02 \times 96500 \,C $
$\therefore 0.02 \times 96500=10^{-2} \times t $
$t =19.30 \times 10^{4}\, sec$