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Question #45023

Difficulty: 8/10 | 🟢 Free

The Question

Consider the Latimer diagram for a transition metal MM in acidic medium (pH=0pH = 0) at 298 K298\text{ K}: \ceMO4−(aq)−>[+0.90 V]MO42−(aq)−>[+2.20 V]MO2(s)−>[+0.95 V]M3+(aq)−>[+1.50 V]M2+(aq)−>[−0.20 V]M(s)\ce{MO_4^{-}(aq) ->[+0.90\text{ V}] MO_4^{2-}(aq) ->[+2.20\text{ V}] MO_2(s) ->[+0.95\text{ V}] M^{3+}(aq) ->[+1.50\text{ V}] M^{2+}(aq) ->[-0.20\text{ V}] M(s)} An aqueous solution containing 1 M1\text{ M} of \ceM2+\ce{M^{2+}} and 1 M1\text{ M} of \ceH+\ce{H+} is subjected to electrolysis at 298 K298\text{ K} using inert platinum electrodes. The kinetic overpotential for \ceO2(g)\ce{O_2(g)} evolution is 0.40 V0.40\text{ V}. Assume standard conditions for all other species. Given: \ce2H2O(l)<=>O2(g)+4H+(aq)+4e−Eox∘=−1.23 V\ce{2H_2O(l) <=> O_2(g) + 4H+(aq) + 4e^-} \quad E^\circ_{ox} = -1.23\text{ V}

Which of the following correctly identifies the standard reduction potential E\ceMO4−/M2+∘E^\circ_{\ce{MO_4^-/M^{2+}}} and the primary product formed at the anode during this electrolysis?

Options

  • A.

    +1.55 V+1.55\text{ V} and \ceMO2(s)\ce{MO_2(s)}

  • B.

    +5.55 V+5.55\text{ V} and \ceO2(g)\ce{O_2(g)}

  • C.

    +1.39 V+1.39\text{ V} and \ceM3+(aq)\ce{M^{3+}(aq)}

  • D.

    +1.11 V+1.11\text{ V} and \ceO2(g)\ce{O_2(g)}

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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

Electrode Potential Sign Conventions|Hess's Law Application|Electrolysis Product Prediction

Why This Question Is Tricky

Common Mistakes

The question tests the transition from intensive properties (Electrode Potentials) to extensive properties (Gibbs Free Energy) using Hess's Law in Latimer diagrams.

The primary trap is the direct addition or simple averaging of potentials, ignoring the varying electron transfer numbers (nn).

Furthermore, it challenges the rote assumption that water is always oxidized to \ceO2\ce{O_2} or that a metal cation simply oxidizes to its next adjacent higher state (\ceM3+\ce{M^{3+}}).

By requiring the calculation of a non-adjacent oxidation potential (\ceM2+−>MO2\ce{M^{2+} -> MO_2}), it reveals a hidden thermodynamic preference for \ceMO2\ce{MO_2} deposition, a principle utilized in the industrial electrosynthesis of Manganese Dioxide (EMD).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterRedox Reactions - Class 11 Part 2
ConceptElectrode Potential Sign Conventions
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree