JEE Challenger
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ChemMedAdv 2007 • Paper 1

Maximum Weight of Amalgam Formed in Electrolysis with Hg Cathode

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 4.0 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}).

If the cathode is a Hg\text{Hg} electrode, the maximum weight (g) of amalgam formed from this solution is

Options

A

200

B

225

C

400

D

446

Correct

Topics & Concepts

ElectrochemistryElectrolysisMole Concept

Step-by-Step Solution

To find the maximum weight of the amalgam formed during the electrolysis using a mercury (Hg\text{Hg}) cathode, we proceed step-by-step:

  1. Calculate the initial moles of NaCl\text{NaCl} in the solution:
    Given:

    • Molarity of NaCl\text{NaCl} solution (MM) = 4.0 M4.0\text{ M}
    • Volume of solution (VV) = 500 mL=0.5 L500\text{ mL} = 0.5\text{ L}

    Moles of NaCl=Molarity×Volume (in L)\text{Moles of }\text{NaCl} = \text{Molarity} \times \text{Volume (in L)}
    Moles of NaCl=4.0 mol/L×0.5 L=2.0 moles\text{Moles of }\text{NaCl} = 4.0\text{ mol/L} \times 0.5\text{ L} = 2.0\text{ moles}

    Therefore, the maximum number of moles of Na+\text{Na}^+ ions available for reduction is 2.0 moles2.0\text{ moles}.

  2. Cathodic reaction using a mercury electrode:
    When an aqueous solution of NaCl\text{NaCl} is electrolysed using a mercury cathode, sodium ions are discharged at the cathode and dissolve in mercury to form sodium amalgam (NaHg\text{NaHg}):
    Na++e−+Hg→NaHg\text{Na}^+ + e^- + \text{Hg} \rightarrow \text{NaHg}

  3. Calculate the molar mass of sodium amalgam (NaHg\text{NaHg}):
    Given:

    • Atomic mass of Na=23 g/mol\text{Na} = 23\text{ g/mol}
    • Atomic mass of Hg=200 g/mol\text{Hg} = 200\text{ g/mol}

    Molar mass of NaHg=23+200=223 g/mol\text{Molar mass of }\text{NaHg} = 23 + 200 = 223\text{ g/mol}

  4. Calculate the maximum weight of the amalgam formed:
    Since 2.0 moles2.0\text{ moles} of Na+\text{Na}^+ are present, the maximum amount of NaHg\text{NaHg} formed is 2.0 moles2.0\text{ moles}.

    Weight of amalgam=Moles of NaHg×Molar mass of NaHg\text{Weight of amalgam} = \text{Moles of }\text{NaHg} \times \text{Molar mass of }\text{NaHg}
    Weight of amalgam=2.0 mol×223 g/mol=446 g\text{Weight of amalgam} = 2.0\text{ mol} \times 223\text{ g/mol} = 446\text{ g}

Thus, the maximum weight of amalgam formed is 446 g446\text{ g}.

Correct Option: (D)

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