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Question #44245

Difficulty: 8/10 | 🟢 Free

The Question

Consider the standard electrode potential E\ceM2+/M∘E^\circ_{\ce{M^{2+}/M}} for a 3d transition metal \ceM\ce{M} at 298 K298\text{ K}. The thermodynamic parameters for the Born-Haber cycle of the half-reaction \ceM(s)−>M2+(aq)+2e−\ce{M(s) -> M^{2+}(aq) + 2e^-} are given below:

ΔHsublimation∘=280 kJ mol−1\Delta H^\circ_{\text{sublimation}} = 280\text{ kJ mol}^{-1} ΔSsublimation∘=100 J K−1 mol−1\Delta S^\circ_{\text{sublimation}} = 100\text{ J K}^{-1}\text{ mol}^{-1} ΔGionization∘(\ceM(g)−>M2+(g)+2e−)=1800 kJ mol−1\Delta G^\circ_{\text{ionization}} (\ce{M(g) -> M^{2+}(g) + 2e^-}) = 1800\text{ kJ mol}^{-1} ΔHhydration∘(\ceM2+(g))=−1900 kJ mol−1\Delta H^\circ_{\text{hydration}} (\ce{M^{2+}(g)}) = -1900\text{ kJ mol}^{-1} ΔShydration∘(\ceM2+(g))=−260 J K−1 mol−1\Delta S^\circ_{\text{hydration}} (\ce{M^{2+}(g)}) = -260\text{ J K}^{-1}\text{ mol}^{-1}

Given the Faraday constant F=96500 C mol−1F = 96500\text{ C mol}^{-1}, calculate the standard reduction potential E\ceM2+/M∘E^\circ_{\ce{M^{2+}/M}} in Volts. (Assume standard state conditions and ideal behavior).

Options

  • A.

    −1.18 V-1.18\text{ V}

  • B.

    −0.93 V-0.93\text{ V}

  • C.

    −1.33 V-1.33\text{ V}

  • D.

    −0.69 V-0.69\text{ V}

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

Entropy of Phase Change|E° (M²⁺/M) Anomalies

Why This Question Is Tricky

Common Mistakes

This question targets the 'Enthalpy-Centric Bias' in evaluating standard electrode potentials.

Students frequently approximate ΔG∘≈ΔH∘\Delta G^\circ \approx \Delta H^\circ for solid-state transitions, ignoring the significant entropy gain during the sublimation of complex metallic lattices (like Manganese) and the entropy loss during hydration due to solvent ordering.

Failing to account for the TΔSsubT\Delta S_{\text{sub}} and TΔShydT\Delta S_{\text{hyd}} terms leads to the incorrect 'enthalpy-only' value of −0.93 V-0.93\text{ V}.

Ignoring just the sublimation entropy yields −1.33 V-1.33\text{ V}.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterThermodynamics - Class 11 Part 1
ConceptEntropy of Phase Change
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree