Start Learning Free

Before You Begin

Question #44751

Difficulty: 8/10 | 🟢 Free

The Question

A 1.0 L1.0 \text{ L} aqueous solution of \ceH2O2\ce{H2O2} (initial concentration 1.0 M1.0 \text{ M}) is subjected to electrolysis at 298 K298 \text{ K} using inert electrodes and a constant current of 193 A193 \text{ A}. Simultaneously, in the presence of a homogeneous catalyst, \ceH2O2\ce{H2O2} undergoes a pseudo-first-order disproportionation into \ceH2O(l)\ce{H2O(l)} and \ceO2(g)\ce{O2(g)} with a rate constant k=10−3 s−1k = 10^{-3} \text{ s}^{-1}. The anodic electrolysis reaction strictly oxidizes \ceH2O2\ce{H2O2} to \ceO2(g)\ce{O2(g)} and \ceH+(aq)\ce{H+(aq)}. Assuming no volume change and 100%100\% current efficiency, what is the total moles of \ceO2(g)\ce{O2(g)} evolved from the solution by the time \ceH2O2\ce{H2O2} is completely depleted? (Given: F=96500 C mol−1F = 96500 \text{ C mol}^{-1})

Options

  • A.

    0.5(1+ln⁡2) mol0.5(1 + \ln 2) \text{ mol}

  • B.

    0.75 mol0.75 \text{ mol}

  • C.

    1.0 mol1.0 \text{ mol}

  • D.

    ln⁡2 mol\ln 2 \text{ mol}

Check Your Answer

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

Faraday's First and Second Laws|Pseudo-First Order Kinetics

Why This Question Is Tricky

Common Mistakes

This problem tests the coupling of Faraday's Laws of Electrolysis with First-Order Chemical Kinetics.

The 'Hidden Variable' is the time-dependent concentration of the reactant, which simultaneously undergoes a zero-order depletion (constant current electrolysis) and a first-order depletion (chemical decomposition).

Students often fall into the trap of treating the two processes independently or assuming the chemical decomposition rate is constant (zero-order), which leads to an incorrect total time and mass distribution (Option B).

Furthermore, failing to account for the different stoichiometric ratios of O2 production in the two parallel pathways leads to additional errors.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterElectrochemistry - Class 12
ConceptFaraday's First and Second Laws
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree