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Question #44124

Difficulty: 8/10 | 🟢 Free

The Question

Gold(I) ions are thermodynamically unstable in aqueous solution and undergo disproportionation to Gold(III) and solid gold. Given the standard reduction potentials at $298 \text{ K}$ :

$E^\circ_{\ce{Au^{3+}/Au}} = +1.36 \text{ V}$ $E^\circ_{\ce{Au^{+}/Au}} = +1.68 \text{ V}$

If a hypothetical aqueous solution is prepared with an initial concentration of $0.03 \text{ M}$ $\ce{Au^+_{(aq)}}$ , what will be the equilibrium concentration of $\ce{Au^+_{(aq)}}$ remaining in the solution?

(Assume $\frac{2.303RT}{F} = 0.06 \text{ V}$ at $298 \text{ K}$ )

Options

A.
$1.0 \times 10^{-6} \text{ M}$
B.
$1.0 \times 10^{-18} \text{ M}$
C.
$3.0 \times 10^{-2} \text{ M}$
D.
$4.6 \times 10^{-4} \text{ M}$

Check Your Answer

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

Stock Notation Logic|Comproportionation Reversibility|Electrode Potential Dependencies

Why This Question Is Tricky

Common Mistakes

This problem tests the thermodynamic stability of metal ions in specific Stock notations (Gold(I) vs Gold(III)).

The core trap lies in the disproportionation stoichiometry and the calculation of the intermediate reduction potential.

Students often incorrectly assume $n=1$ for the disproportionation of a $+1$ ion, forgetting that forming $\ce{Au^{3+}}$ requires a 2-electron oxidation per atom, making the balanced reaction $3\ce{Au^+} \rightleftharpoons \ce{Au^{3+}} + 2\ce{Au}$ with $n=2$ .

Furthermore, failing to cube the $[\ce{Au^+}]$ term in the equilibrium expression leads to the $10^{-18} \text{ M}$ trap.

Quick Facts About This Question

Question Information

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