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Question #44868

Difficulty: 8/10 | 🟢 Free

The Question

Consider the standard reduction potentials for the \ceM(aq)3+/M(aq)2+\ce{M^{3+}_{(aq)} / M^{2+}_{(aq)}} redox couple of the Group 5 transition metals Niobium (Nb) and Tantalum (Ta) at 298 K. Due to the lanthanoid contraction, their corresponding ionic radii are virtually identical (r\ceNbn+≈r\ceTan+r_{\ce{Nb^{n+}}} \approx r_{\ce{Ta^{n+}}}). Assuming the entropy change (ΔS∘\Delta S^\circ) for the reduction half-reaction is identical for both metals, which of the following expressions correctly approximates the difference in their standard reduction potentials, ΔE∘=E∘(\ceNb3+/Nb2+)−E∘(\ceTa3+/Ta2+)\Delta E^\circ = E^\circ(\ce{Nb^{3+}/Nb^{2+}}) - E^\circ(\ce{Ta^{3+}/Ta^{2+}})?

Options

  • A.

    1F[IE3(\ceNb)−IE3(\ceTa)]\frac{1}{F} [IE_3(\ce{Nb}) - IE_3(\ce{Ta})]

  • B.

    0 V0 \text{ V}

  • C.

    1F[ΔHsub(\ceNb)−ΔHsub(\ceTa)]\frac{1}{F} [\Delta H_{sub}(\ce{Nb}) - \Delta H_{sub}(\ce{Ta})]

  • D.

    13F[ΔHhyd(\ceNb3+)−ΔHhyd(\ceTa3+)]\frac{1}{3F} [\Delta H_{hyd}(\ce{Nb^{3+}}) - \Delta H_{hyd}(\ce{Ta^{3+}})]

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Post-Lanthanoid Radii Equality|E° (M³⁺/M²⁺) Redox Potential|Standard Electrode Potential Sign Conventions

Why This Question Is Tricky

Common Mistakes

This question targets the 'Correlation Fallacy' regarding the Lanthanoid Contraction.

Students frequently assume that because 4d and 5d elements have identical radii, their standard electrode potentials must also be identical (Option B trap).

Furthermore, students often blindly include the enthalpy of sublimation (ΔHsub\Delta H_{sub}) in all E∘E^\circ calculations, forgetting that an ion-to-ion reduction (\ceM3+−>M2+\ce{M^{3+} -> M^{2+}}) does not involve the solid metal phase (Option C trap).

The question forces the student to derive the exact thermodynamic cycle and recognize that the divergence in E∘E^\circ is driven purely by the divergence in the third ionization energies (IE3IE_3), which do not scale linearly with radius due to relativistic effects and poor shielding of 4f electrons.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterD and F Block Elements - Class 12
ConceptPost-Lanthanoid Radii Equality
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree