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

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 × 104 sec (B) 19.3 × 104 sec (C) 28.95 × 104 sec (D) 38.6 × 104 sec

Tardigrade Admin · Accepted Answer

Q = i × t Q =10 × 10-3 × 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 × 96500 C ∴ 0.02 × 96500=10-2 × t t =19.30 × 104 sec

Q. Electrolysis of dilute aqueous $N a Cl$ solution was carried out by passing $10$ milli ampere current. The time required to liberate $0.01 m o l$ of $H_{2}$ gas at the cathode is (1 Faraday $= 96500 C m o l^{- 1}$

$9.65 \times 1 0^{4} sec$

$19.3 \times 1 0^{4} sec$

$28.95 \times 1 0^{4} sec$

$38.6 \times 1 0^{4} sec$

$Q = i \times t$
$Q = 10 \times 1 0^{- 3} \times t$
$2 H_{2} O + 2 e^{-} ⟶ H_{2} + 2 O H^{-}$
To liberate $0.01$ mole of $H_{2}, 0.02$ Faraday charge is required
$Q = 0.02 \times 96500 C$
$∴ 0.02 \times 96500 = 1 0^{- 2} \times t$
$t = 19.30 \times 1 0^{4} sec$

Q. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01mol of H2​ gas at the cathode is (1 Faraday =96500Cmol−1

9.65×104sec

19.3×104sec

28.95×104sec

38.6×104sec