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Question #44144

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The Question

Consider an aqueous solution containing dissolved $\ce{Cl2(aq)}$ , $\ce{Cl-(aq)}$ , and $\ce{ClO3-(aq)}$ , all maintained at an identical concentration $C$ . The solution is buffered at $pH = 1.0$ at $298\text{ K}$ .

Given the standard reduction potentials: $E^\circ_{\ce{Cl2}/\ce{Cl-}} = 1.36\text{ V}$ $E^\circ_{\ce{ClO3-}/\ce{Cl2}} = 1.47\text{ V}$

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

Which of the following statements correctly describes the thermodynamic stability of $\ce{Cl2(aq)}$ against disproportionation into $\ce{Cl-}$ and $\ce{ClO3-}$ ?

Options

A.
$\ce{Cl2}$ is stable at $C = 1.0\text{ M}$ but undergoes disproportionation at $C = 10^{-2}\text{ M}$ .
B.
$\ce{Cl2}$ undergoes disproportionation at $C = 1.0\text{ M}$ but is stable at $C = 10^{-2}\text{ M}$ .
C.
$\ce{Cl2}$ is stable at both $C = 1.0\text{ M}$ and $C = 10^{-2}\text{ M}$ because $E^\circ_{\text{cell}} < 0$ .
D.
$\ce{Cl2}$ undergoes disproportionation at both concentrations because the acidic medium ( $pH = 1.0$ ) drives the forward reaction.

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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

Disproportionation Constraints|Nernstian Redox Equilibrium|Electrode Potential Dependencies

Why This Question Is Tricky

Common Mistakes

This question tests the 'Concentration-Dependent Thermodynamic Stability' trap.

Students often calculate $E^\circ_{\text{cell}} = -0.11\text{ V}$ and falsely conclude that the species is unconditionally stable against disproportionation (Option C).

The hidden variable is the reaction quotient $Q$ .

Because the disproportionation produces more moles of aqueous species than it consumes (6 moles of products vs 3 moles of reactants, excluding water and buffered $H^+$ ), dilution significantly decreases $Q$ .

According to Le Chatelier's principle (quantified by the Nernst equation), dilution shifts the equilibrium forward, increasing $E_{\text{cell}}$ until it crosses zero, making the previously stable species unstable at lower concentrations.

Quick Facts About This Question

Question Information

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