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JEE Advanced 2008Paper 2
Exam
JEE Advanced
Year
2008
Paper
Paper 2
Subject
Chemistry
Chapter
Electrochemistry

JEE Advanced 2008 · Paper 2 · Chemistry · Electrochemistry

Electrolysis of dilute aq. NaCl solution was carried out by passing...

Electrolysis of dilute aq. NaCl\mathrm{NaCl} solution was carried out by passing 10 mA10 \mathrm{~mA} current. The time required to liberate 0.010.01 mole of H2\mathrm{H}_2 gas at the cathode is (1 Faraday =96500Cmol−1=96500 \mathrm{C} \mathrm{mol}^{-1} )

Options

  1. A9.65×104 s9.65 \times 10^4 \mathrm{~s}
  2. B19.3×104 s19.3 \times 10^4 \mathrm{~s}
  3. C28.95×104 s28.95 \times 10^4 \mathrm{~s}
  4. D38.6×104 s38.6 \times 10^4 \mathrm{~s}

Correct answer

B. 19.3×104 s19.3 \times 10^4 \mathrm{~s}

Step-by-step solution

2H2O+2e−⟶H2+2OH−For2 \mathrm{H}_2 \mathrm{O}+2 e^{-} \longrightarrow \mathrm{H}_2+2 \mathrm{OH}^{-} For0.01moleofH2,mole of \mathrm{H}_2, 0.02moleofelectronsareconsumedChargerequiredmole of electrons are consumed Charge required=0.02 \times 96500 C=\mathrm{C}=i \times t  Time required =0.02×9650010×10−3=19.3×104s\text{ Time required }=\frac{0.02 \times 96500}{10 \times 10^{-3}}=19.3 \times 10^4 \mathrm{ s}
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