JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Redox Reactions and Electrochemistry

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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) $9.65 \times 1 0^{4}$ sec
(B) $19.3 \times 1 0^{4}$ sec✓ Correct
(C) $28.95 \times 1 0^{4}$ sec
(D) $38.6 \times 1 0^{4}$ sec

Answer: (B)

Solution

At the cathode:

2 H_{2} O + 2 e^{-} \to H_{2} + 2 OH^{-}

To liberate 1 mole of

H_{2}

, 2 Faraday of charge is required.
For 0.01 mole of

H_{2}

, charge required = 0.02 × 96500 = 1930 C

Using

Q = i \times t

1930 = 10 \times 1 0^{- 3} \times t

t = \frac{1930}{0.01} = 193000

sec =

19.3 \times 1 0^{4}

sec

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option

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