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Question #44053

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The Question

Given the standard reduction potentials at 298 K298\text{ K}: E\ceCu+/Cu∘=+0.52 VE^\circ_{\ce{Cu^+/Cu}} = +0.52\text{ V} E\ceCu2+/Cu+∘=+0.16 VE^\circ_{\ce{Cu^{2+}/Cu^+}} = +0.16\text{ V} And the overall formation constants (β\beta) for copper-ammonia complexes: β2 for \ce[Cu(NH3)2]+=1011\beta_2 \text{ for } \ce{[Cu(NH_3)_2]^+} = 10^{11} β4 for \ce[Cu(NH3)4]2+=1013\beta_4 \text{ for } \ce{[Cu(NH_3)_4]^{2+}} = 10^{13} Consider the comproportionation reaction in an aqueous medium where the equilibrium concentration of free \ceNH3\ce{NH_3} is maintained at 1.0 M1.0\text{ M}: \ceCu(s)+[Cu(NH3)4]2+(aq)⇌2[Cu(NH3)2]+(aq)\ce{Cu(s) + [Cu(NH_3)_4]^{2+}(aq) \rightleftharpoons 2[Cu(NH_3)_2]^+(aq)} Which of the following correctly identifies the standard potential of this reaction (Erxn∘E^\circ_{\text{rxn}}) under these conditions and the thermodynamic stability of the \ceCu(I)\ce{Cu(I)} state? (Take 2.303RTF=0.06 V\frac{2.303 RT}{F} = 0.06\text{ V})

Options

  • A.

    Erxn∘=+0.18 VE^\circ_{\text{rxn}} = +0.18\text{ V}; \ceCu(I)\ce{Cu(I)} complex is stable against disproportionation.

  • B.

    Erxn∘=−0.36 VE^\circ_{\text{rxn}} = -0.36\text{ V}; \ceCu(I)\ce{Cu(I)} complex disproportionates spontaneously.

  • C.

    Erxn∘=+0.42 VE^\circ_{\text{rxn}} = +0.42\text{ V}; \ceCu(I)\ce{Cu(I)} complex is stable against disproportionation.

  • D.

    Erxn∘=−0.18 VE^\circ_{\text{rxn}} = -0.18\text{ V}; \ceCu(I)\ce{Cu(I)} complex disproportionates spontaneously.

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

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

Concepts Behind This Question

Standard State Conditions|Electrode Potential Sign Conventions|Fuel Cell Efficiency

Why This Question Is Tricky

Common Mistakes

This question tests the 'Concentration Dependency via Nernst Equation' trap.

Students rote-memorize that Cu(I) is unstable in aqueous solutions because its standard disproportionation potential is positive (+0.36 V).

The hidden variable here is differential complexation.

Because the formation constant for the Cu(I) complex (β2\beta_2) is exceptionally high relative to the Cu(II) complex (β4\beta_4), the concentration of free Cu+ is suppressed much more drastically than Cu2+.

This thermodynamic shift reverses the sign of the Gibbs free energy for the reaction, making the otherwise unstable +1 oxidation state highly stable.

Students who only look at the standard uncomplexed E° values will incorrectly choose option B.

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Question Information

SubjectChemistry
ChapterThermodynamics - Class 11 Part 1
ConceptStandard State Conditions
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree