JEE Challenger
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ChemMedAdv 2007 • Paper 1

Calculate Total Charge Required for Complete Electrolysis of Aqueous NaCl Solution

Comprehension Passage

Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023×10236.023 \times 10^{23}) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept.

A 4.04.0 molar aqueous solution of NaCl\text{NaCl} is prepared and 500 mL500\text{ mL} of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass : Na=23,Hg=200;1 Faraday=96500 coulombs\text{Na} = 23, \text{Hg} = 200; 1\text{ Faraday} = 96500\text{ coulombs}).

The total charge (coulombs) required for complete electrolysis is

Options

A

24125

B

48250

C

96500

D

193000

Correct

Topics & Concepts

ElectrochemistryElectrolysisMole Concept

Step-by-Step Solution

To find the total charge required for the complete electrolysis of the given aqueous NaCl\text{NaCl} solution, we follow these steps:

  1. Calculate the number of moles of NaCl\text{NaCl} in the solution:
    The molarity (MM) of the aqueous solution is 4.0 M4.0\text{ M} and the volume (VV) is 500 mL=0.5 L500\text{ mL} = 0.5\text{ L}.
    Moles of NaCl=Molarity×Volume in liters\text{Moles of } \text{NaCl} = \text{Molarity} \times \text{Volume in liters}
    nNaCl=4.0 mol/L×0.5 L=2.0 molesn_{\text{NaCl}} = 4.0\text{ mol/L} \times 0.5\text{ L} = 2.0\text{ moles}

  2. Determine the stoichiometry of the electrochemical reaction:
    In aqueous NaCl\text{NaCl} solution, NaCl\text{NaCl} dissociates completely into Na+\text{Na}^+ and Cl−\text{Cl}^- ions:
    NaCl(aq)→Na+(aq)+Cl−(aq)\text{NaCl(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(aq)}

    Thus, 2.0 moles2.0\text{ moles} of NaCl\text{NaCl} yields 2.0 moles2.0\text{ moles} of Cl−\text{Cl}^- ions.

    During electrolysis, chloride ions undergo oxidation at the anode according to the half-reaction:
    2Cl−(aq)→Cl2(g)+2e−2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2e^-

    This shows that oxidation of 1 mole1\text{ mole} of Cl−\text{Cl}^- requires 1 mole1\text{ mole} of electrons (1 Faraday1\text{ Faraday} of charge).

  3. Calculate the total charge required:
    The total number of moles of electrons required (nen_e) for 2.0 moles2.0\text{ moles} of Cl−\text{Cl}^- is:
    ne=2.0 moles of e−n_e = 2.0\text{ moles of } e^-

    Given that 1 Faraday (F)=96500 coulombs1\text{ Faraday } (F) = 96500\text{ coulombs}, the total charge (QQ) is given by:
    Q=ne×FQ = n_e \times F
    Q=2.0×96500 C=193000 CQ = 2.0 \times 96500\text{ C} = 193000\text{ C}

Therefore, the correct option is D.

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