Start Learning Free

Before You Begin

Question #44662

Difficulty: 8/10 | 🟢 Free

The Question

A 100 mL100 \text{ mL} aqueous solution containing 0.1 M \ceK2Cr2O70.1 \text{ M } \ce{K2Cr2O7} and excess \ceH2SO4\ce{H2SO4} is treated with V mLV \text{ mL} of 0.4 M \ceH2O20.4 \text{ M } \ce{H2O2} at 273 K273 \text{ K} in the presence of diethyl ether. The organic layer, containing the blue-colored complex \ceCrO5\ce{CrO5}, is completely separated. The extracted \ceCrO5\ce{CrO5} is then titrated against a 0.2 M0.2 \text{ M} acidified \ceKMnO4\ce{KMnO4} solution. It requires exactly 50 mL50 \text{ mL} of the \ceKMnO4\ce{KMnO4} solution to completely oxidize the peroxy linkages of \ceCrO5\ce{CrO5} to \ceO2(g)\ce{O2(g)}, leaving chromium in the +6+6 oxidation state as \ceCr2O72−\ce{Cr2O7^{2-}}.

Assuming ideal stoichiometry and 100%100\% extraction efficiency, determine the actual value of VV. Furthermore, what would be the calculated value of VV if one incorrectly assumes that the average oxidation state of Cr in \ceCrO5\ce{CrO5} dictates its nn-factor during the titration?

Options

  • A.

    62.5 mL62.5 \text{ mL} and 41.7 mL41.7 \text{ mL}

  • B.

    62.5 mL62.5 \text{ mL} and 35.7 mL35.7 \text{ mL}

  • C.

    31.3 mL31.3 \text{ mL} and 41.7 mL41.7 \text{ mL}

  • D.

    125.0 mL125.0 \text{ mL} and 83.3 mL83.3 \text{ mL}

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

Limiting Reagent Stoichiometry|Average vs. Actual Oxidation State|Peroxy Linkage Identification

Why This Question Is Tricky

Common Mistakes

The core concept tests the limitation of the 'Average Oxidation State' method in redox stoichiometry.

When a molecule contains peroxy linkages (like \ceCrO5\ce{CrO5}), calculating the nn-factor using the average oxidation state of the central atom leads to an incorrect value if the peroxy bonds undergo a specific redox transformation while the central atom does not.

Here, the actual nn-factor is 4 (derived from the oxidation of 4 peroxy oxygens), but the average oxidation state method yields a mathematically skewed nn-factor of 6.

This directly impacts the stoichiometric calculations in titrations, serving as a lethal trap for students who rely on algebraic oxidation state formulas instead of structural redox analysis.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterSome Basic Concepts of Chemistry - Class 11 Part 1
ConceptLimiting Reagent Stoichiometry
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree