JEE Challenger
More from Electrochemistry
ChemMedAdv 2008 • Paper 2

Time Required to Liberate Hydrogen Gas in Electrolysis

Electrolysis of dilute aqueous NaCl\text{NaCl} solution was carried out by passing 10 milli10\text{ milli} ampere current. The time required to liberate 0.01 mol0.01\text{ mol} of H2\text{H}_2 gas at the cathode is (1 Faraday=96500 C mol−11\text{ Faraday} = 96500\text{ C mol}^{-1})

Options

A

9.65×104 sec9.65 \times 10^4\text{ sec}

B

19.3×104 sec19.3 \times 10^4\text{ sec}

Correct
C

28.95×104 sec28.95 \times 10^4\text{ sec}

D

38.6×104 sec38.6 \times 10^4\text{ sec}

Topics & Concepts

ElectrochemistryElectrolysis

Step-by-Step Solution

To find the time required to liberate 0.01 mol0.01\text{ mol} of H2\text{H}_2 gas during the electrolysis of a dilute aqueous NaCl\text{NaCl} solution, we first consider the reduction reaction occurring at the cathode:

2H2O(l)+2e−⟶H2(g)+2OH−(aq)2\text{H}_2\text{O}(l) + 2e^- \longrightarrow \text{H}_2(g) + 2\text{OH}^-(aq)

From the reaction stoichiometry, 1 mole1\text{ mole} of H2(g)\text{H}_2(g) requires 2 moles2\text{ moles} of electrons (2 Faradays2\text{ Faradays} of charge).

The total charge QQ required to liberate n=0.01 moln = 0.01\text{ mol} of H2(g)\text{H}_2(g) is calculated as:
Q=n×2×FQ = n \times 2 \times F

Given:

  • n=0.01 moln = 0.01\text{ mol}
  • 1 F=96500 C mol−11\text{ F} = 96500\text{ C mol}^{-1}

Substituting these values into the equation:
Q=0.01 mol×2×96500 C mol−1=1930 CQ = 0.01\text{ mol} \times 2 \times 96500\text{ C mol}^{-1} = 1930\text{ C}

The current passed through the solution is:
I=10 mA=10×10−3 A=10−2 AI = 10\text{ mA} = 10 \times 10^{-3}\text{ A} = 10^{-2}\text{ A}

Using Faraday's law of electrolysis, Q=I×tQ = I \times t, the time tt in seconds is given by:
t=QI=1930 C10−2 A=193000 sec=19.3×104 sect = \frac{Q}{I} = \frac{1930\text{ C}}{10^{-2}\text{ A}} = 193000\text{ sec} = 19.3 \times 10^4\text{ sec}

Therefore, the correct option is B.

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