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Question #44752

Difficulty: 8/10 | 🟢 Free

The Question

A 1.00 L1.00 \text{ L} aqueous solution initially containing 0.10 M0.10 \text{ M} \ce{AgNO3} is electrolyzed using inert Platinum electrodes at 298 K298 \text{ K}. A constant current of 9.65 A9.65 \text{ A} is passed through the solution for 510 s510 \text{ s}. After the electrolysis is complete, 0.10 moles0.10 \text{ moles} of solid \ce{NaCH3COO} is added to the solution. Assume no change in volume occurs upon the addition of the solid and during electrolysis.

Given at 298 K298 \text{ K}: Ksp(\ceAgCH3COO)=2.5×10−3K_{sp}(\ce{AgCH3COO}) = 2.5 \times 10^{-3} Ka(\ceCH3COOH)=2.0×10−5K_a(\ce{CH3COOH}) = 2.0 \times 10^{-5} Faraday's constant F=96500 C mol−1F = 96500 \text{ C mol}^{-1} Molar mass of \ceAgCH3COO=167 g mol−1\ce{AgCH3COO} = 167 \text{ g mol}^{-1} Molar mass of \ceAg=108 g mol−1\ce{Ag} = 108 \text{ g mol}^{-1}

Calculate the mass of the \ce{AgCH3COO} precipitate formed in the solution.

Options

  • A.

    0.00 g0.00 \text{ g}

  • B.

    3.07 g3.07 \text{ g}

  • C.

    5.51 g5.51 \text{ g}

  • D.

    8.35 g8.35 \text{ g}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Faraday's First and Second Laws|Solubility of Solids in Liquids

Why This Question Is Tricky

Common Mistakes

The problem tests the integration of Faraday's laws of electrolysis with ionic equilibrium (solubility and buffer/weak acid).

The 'hidden variable' is the generation of \ceH+\ce{H+} ions at the anode during the electrolysis of water (since \ceNO3−\ce{NO3-} is inert).

Students often focus only on the metal ion (\ceAg+\ce{Ag+}) reduction at the cathode and forget that the anodic oxidation of water produces \ceH+\ce{H+}, which lowers the pH.

When the weak base anion (\ceCH3COO−\ce{CH3COO-}) is added, it is protonated by the electrogenerated \ceH+\ce{H+}, significantly reducing the free [\ceCH3COO−][\ce{CH3COO-}] available for precipitation.

Failing to account for this anodic \ceH+\ce{H+} generation leads to the false conclusion that \ceAgCH3COO\ce{AgCH3COO} will precipitate (Option B).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterElectrochemistry - Class 12
ConceptFaraday's First and Second Laws
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree