IIT-JEE 2008 · Paper 2 · Chemistry Q51

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ChemistrySingle correct+3 / -1Question 51 of 66

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of $\ce{H2}$ gas at the cathode is (1 Faraday = 96500 C mol⁻¹)

9.65\times10^4

sec

19.3\times10^4

sec

28.95\times10^4

sec

38.6\times10^4

sec

Answer and worked solution

Official answer: B

Charge. $\ce{2H2O + 2e- -> H2 + 2OH-}$ : 0.01 mol $\ce{H2}$ needs 0.02 mol electrons, $0.02\times96500=1930$ C.

Time. $t=\dfrac{1930}{0.010}=1.93\times10^5$ s $=19.3\times10^4$ s.

(B)

Topics:Electrolysis and Faraday's laws

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IIT-JEE 2008 · Paper 2, Chemistry question 51