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Question #45030

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The Question

Consider the Group 11 monovalent cations \ce{Cu+}, \ce{Ag+}, and \ce{Au+} in aqueous solution at 298 K. The standard reduction potentials are given as: $E^\circ_{\ce{Cu+}/\ce{Cu}} = +0.52\text{ V}, \quad E^\circ_{\ce{Cu^{2+}}/\ce{Cu+}} = +0.16\text{ V}$ $E^\circ_{\ce{Ag+}/\ce{Ag}} = +0.80\text{ V}, \quad E^\circ_{\ce{Ag^{2+}}/\ce{Ag+}} = +1.98\text{ V}$ $E^\circ_{\ce{Au+}/\ce{Au}} = +1.69\text{ V}, \quad E^\circ_{\ce{Au^{3+}}/\ce{Au+}} = +1.41\text{ V}$

Based on thermodynamic feasibility, identify the ion(s) that undergo spontaneous disproportionation in aqueous solution. For the 5d-block cation that disproportionates, calculate its equivalent weight ( $E$ ) in terms of its molar mass ( $M$ ). Furthermore, identify the primary thermodynamic driver for the anomalous stability of its higher oxidation state despite its high ionization energies.

Options

A.
\ce{Cu+} and \ce{Au+}; $E_{\ce{Au+}} = 1.5 M$ ; High hydration enthalpy of \ce{Au^{3+}} compensates for high ionization energies caused by the lanthanide contraction.
B.
\ce{Cu+} and \ce{Au+}; $E_{\ce{Au+}} = 0.67 M$ ; High hydration enthalpy of \ce{Au^{3+}} compensates for high ionization energies caused by the lanthanide contraction.
C.
Only \ce{Au+}; $E_{\ce{Au+}} = 1.5 M$ ; Relativistic expansion of 5d orbitals lowers the third ionization energy.
D.
\ce{Cu+}, \ce{Ag+}, and \ce{Au+}; $E_{\ce{Au+}} = 3.0 M$ ; Lanthanide contraction increases the effective nuclear charge, directly stabilizing the +3 state.

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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

Equivalent Concept in Redox|Disproportionation Constraints|Post-Lanthanoid Radii Equality

Why This Question Is Tricky

Common Mistakes

The question tests the thermodynamic feasibility of disproportionation reactions using standard reduction potentials, specifically targeting the anomalous behavior of Group 11 elements.

Students often assume a uniform periodic trend (e.g., lower oxidation states becoming more stable down the group due to the inert pair effect), but the 5d elements (like Au) exhibit stable higher oxidation states due to exceptionally high hydration/lattice enthalpies overcoming the high ionization energies caused by the lanthanide contraction.

Furthermore, the problem traps students in the equivalent weight calculation of a disproportionating species with asymmetric electron transfer ( $n_{\text{ox}} \neq n_{\text{red}}$ ), where the common error is to invert the n-factor or use a simple average, leading to incorrect values like $0.67 M$ .

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Question Information

Subject	Chemistry
Chapter	Redox Reactions - Class 11 Part 2
Concept	Equivalent Concept in Redox
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free