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Question #101481

Difficulty: 5/10 | 🟢 Free

The Question

The Inert Pair Effect profoundly influences the redox chemistry of heavy p-block elements, making \ce{Pb(IV)} and \ce{Tl(III)} powerful oxidizing agents. Consider the redox competition between these states in an aqueous acidic medium at 298 K:

\cePbO2(s)+4H+(aq)+Tl+(aq)<=>Pb2+(aq)+2H2O(l)+Tl3+(aq)\ce{PbO2(s) + 4H+(aq) + Tl+(aq) <=> Pb^{2+}(aq) + 2H2O(l) + Tl^{3+}(aq)}

Given the standard reduction potentials: E\cePbO2/Pb2+∘=+1.46 VE^\circ_{\ce{PbO2/Pb^{2+}}} = +1.46 \text{ V} E\ceTl3+/Tl+∘=+1.25 VE^\circ_{\ce{Tl^{3+}/Tl^+}} = +1.25 \text{ V}

Assuming all aqueous species (except \ce{H+}) are maintained at 1 M1 \text{ M} concentration, determine the standard Gibbs free energy change (ΔG∘\Delta G^\circ) for this reaction and the threshold pH above which the forward reaction becomes thermodynamically non-spontaneous. (Use F=96500 C mol−1F = 96500 \text{ C mol}^{-1} and 2.303RTF=0.06 V\frac{2.303 RT}{F} = 0.06 \text{ V})

Options

  • A.

    −40.53 kJ mol−1,pH=1.75-40.53 \text{ kJ mol}^{-1}, \text{pH} = 1.75

  • B.

    −20.26 kJ mol−1,pH=0.88-20.26 \text{ kJ mol}^{-1}, \text{pH} = 0.88

  • C.

    −40.53 kJ mol−1,pH=7.00-40.53 \text{ kJ mol}^{-1}, \text{pH} = 7.00

  • D.

    −81.06 kJ mol−1,pH=3.50-81.06 \text{ kJ mol}^{-1}, \text{pH} = 3.50

Check Your Answer

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

Inert Pair Effect|Fractional Oxidation State Misconception|Disproportionation Constraints

Why This Question Is Tricky

Examiner Levers

Approximation Bait

Options are extremely close; tests precision vs. early rounding.

Difficulty Boost : +3

Common Mistakes

This question tests the intersection of the Inert Pair Effect and electrochemical thermodynamics.

Students often memorize that Tl(I) and Pb(II) are the most stable states, leading them to incorrectly assume that Tl(I) cannot be oxidized.

However, the relative magnitude of the inert pair effect makes Pb(IV) a stronger oxidant than Tl(III), yielding a spontaneous standard reaction.

The hidden variable trap lies in the stoichiometry of the Nernst equation: students frequently misidentify the number of transferred electrons (n=1n=1 or n=4n=4) or forget to raise the [\ceH+][\ce{H+}] concentration to the 4th power, leading to incorrect pH thresholds.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterClassification of Elements and Periodicity in Properties - Class 11 Part 1
ConceptInert Pair Effect
Difficulty5
Cognitive Load6
Target Solve Time150 sec
AccessFree