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Question #44232

Difficulty: 8/10 | 🟢 Free

The Question

Consider the standard Gibbs free energy changes (ΔG∘\Delta G^\circ) for the following processes at 298 K:

For a transition metal MM: Atomization: M(s)→M(g)M(s) \rightarrow M(g), ΔG∘=330 kJ mol−1\Delta G^\circ = 330 \text{ kJ mol}^{-1} Ionization: M(g)→M2+(g)+2e−M(g) \rightarrow M^{2+}(g) + 2e^-, ΔG∘=2700 kJ mol−1\Delta G^\circ = 2700 \text{ kJ mol}^{-1} Hydration: M2+(g)→M2+(aq)M^{2+}(g) \rightarrow M^{2+}(aq), ΔG∘=−1998.8 kJ mol−1\Delta G^\circ = -1998.8 \text{ kJ mol}^{-1}

For a transition metal NN: Atomization: N(s)→N(g)N(s) \rightarrow N(g), ΔG∘=130 kJ mol−1\Delta G^\circ = 130 \text{ kJ mol}^{-1} Ionization: N(g)→N2+(g)+2e−N(g) \rightarrow N^{2+}(g) + 2e^-, ΔG∘=2640 kJ mol−1\Delta G^\circ = 2640 \text{ kJ mol}^{-1} Hydration: N2+(g)→N2+(aq)N^{2+}(g) \rightarrow N^{2+}(aq), ΔG∘=−1950 kJ mol−1\Delta G^\circ = -1950 \text{ kJ mol}^{-1}

The absolute standard Gibbs free energy change for the oxidation half-reaction H2(g)→2H+(aq)+2e−H_2(g) \rightarrow 2H^+(aq) + 2e^- is 965.6 kJ mol−1965.6 \text{ kJ mol}^{-1}.

An aqueous solution containing 1.0 M1.0 \text{ M} of M2+M^{2+} and 1.0 M1.0 \text{ M} of N2+N^{2+} at pH = 0 is electrolyzed using inert electrodes at 298 K and 1 bar pressure. The overpotential for H2H_2 evolution at the cathode is 0.40 V0.40 \text{ V}. Given F=96500 C mol−1F = 96500 \text{ C mol}^{-1}.

What are the standard reduction potentials EM2+/M∘E^\circ_{M^{2+}/M} and EN2+/N∘E^\circ_{N^{2+}/N} respectively, and which species is discharged first at the cathode?

Options

  • A.

    +0.34 V,−0.75 V,M(s)+0.34 \text{ V}, -0.75 \text{ V}, M(s)

  • B.

    −0.34 V,+0.75 V,N(s)-0.34 \text{ V}, +0.75 \text{ V}, N(s)

  • C.

    +0.34 V,−0.75 V,H2(g)+0.34 \text{ V}, -0.75 \text{ V}, H_2(g)

  • D.

    +0.68 V,−1.51 V,M(s)+0.68 \text{ V}, -1.51 \text{ V}, M(s)

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Correct Answer

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Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Hess's Law Application|Electrolysis Product Prediction|E° (M²⁺/M) Anomalies

Why This Question Is Tricky

Common Mistakes

This problem tests the thermodynamic origin of standard reduction potentials via the Born-Haber cycle.

Students often memorize that d-block metals have negative E∘E^\circ values, missing the anomaly of metals like Copper (represented by MM) where high atomization energy and lower hydration energy result in a positive E∘E^\circ.

It also tests the correct application of overpotential.

A common trap is adding the overpotential to the reduction potential instead of subtracting it, which would incorrectly make H2H_2 evolution appear more favorable (+0.40 V+0.40 \text{ V}) than MM reduction (+0.34 V+0.34 \text{ V}).

Another trap is reversing the SHE reference subtraction, flipping the signs of the potentials.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterThermodynamics - Class 11 Part 1
ConceptHess's Law Application
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree