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Question #45036

Difficulty: 8/10 | 🟢 Free

The Question

A redox half-cell is constructed by immersing a platinum electrode into a solution prepared by mixing 50 mL50 \text{ mL} of 0.20 M0.20 \text{ M} sodium thiosulfate (\ceNa2S2O3\ce{Na2S2O3}) and 50 mL50 \text{ mL} of 2.0×10−4 M2.0 \times 10^{-4} \text{ M} potassium tetrathionate (\ceK2S4O6\ce{K2S4O6}). The solution is maintained at 298 K298 \text{ K} and buffered at pH=5.0\text{pH} = 5.0.

The standard reduction potential for the tetrathionate/thiosulfate couple is E∘=+0.080 VE^\circ = +0.080 \text{ V}. Assuming ideal behavior, what is the equilibrium reduction potential of this half-cell?

(Given: 2.303RTF=0.060 V\frac{2.303RT}{F} = 0.060 \text{ V} at 298 K298 \text{ K}, log⁡2=0.30\log 2 = 0.30, log⁡5=0.70\log 5 = 0.70)

Options

  • A.

    +0.020 V+0.020 \text{ V}

  • B.

    −0.010 V-0.010 \text{ V}

  • C.

    −0.244 V-0.244 \text{ V}

  • D.

    −0.280 V-0.280 \text{ V}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Fractional Oxidation State Misconception|Nernstian Redox Equilibrium|E° (M³⁺/M²⁺) Redox Potential

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Fractional Oxidation State Trap' in Nernstian Equilibria.

Students are frequently taught to balance redox reactions using average oxidation states.

However, applying this average state directly to the Nernst equation as a pseudo-species (e.g., \ceS+2.5\ce{S^{+2.5}}) fundamentally alters the reaction quotient QQ and the electron transfer number nn.

The Nernst equation is derived from macroscopic Gibbs Free Energy (ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT \ln Q), which strictly requires the actual molecular species (\ceS4O62−\ce{S4O6^{2-}} and \ceS2O32−\ce{S2O3^{2-}}) and their correct stoichiometric coefficients.

The pH=5.0\text{pH} = 5.0 condition serves as an additional distractor to test if students recognize that oxygen is balanced without \ceH+\ce{H+} in this specific half-reaction.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterRedox Reactions - Class 11 Part 2
ConceptFractional Oxidation State Misconception
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree