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Question #45011

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The Question

A $500 \text{ mL}$ aqueous solution containing $40 \text{ mmol}$ of $\ce{KMnO4}$ and $50 \text{ mmol}$ of $\ce{MnSO4}$ is allowed to react at $298 \text{ K}$ in the presence of an excess of $\ce{ZnO}$ (to maintain a neutral pH). The mixture undergoes a complete comproportionation reaction to yield a dark brown precipitate of $\ce{MnO2(s)}$ .\n\nOnce the reaction is complete, the suspension is filtered. The precipitate is washed thoroughly and then treated with an excess of acidified $\ce{KI}$ solution. The liberated iodine requires $V_1 \text{ mL}$ of $0.1 \text{ M } \ce{Na2S2O3}$ for complete titration.\n\nThe filtrate is similarly treated with an excess of acidified $\ce{KI}$ solution, and the liberated iodine requires $V_2 \text{ mL}$ of $0.1 \text{ M } \ce{Na2S2O3}$ .\n\nAssuming no other side reactions occur, what is the integer value of the ratio $V_1 / V_2$ ?

Options

A.
2
B.
3
C.
4
D.
5

Check Your Answer

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

Comproportionation Reversibility|Equivalent Concept in Redox|Stoichiometric Coefficient Division

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Equivalent Concept' in a comproportionation reaction, specifically targeting the trap of n-factor misattribution and stoichiometric coefficient division.

Students often assume that the equivalents of the product ( $\ce{MnO2}$ ) equal the equivalents of the limiting reagent ( $100 \text{ mEq}$ ), and then incorrectly use the standard redox n-factor of $\ce{MnO2}$ ( $n=2$ ) to find its moles ( $100/2 = 50 \text{ mmol}$ ).

This leads to a trap ratio of 3.

The hidden variable is that $\ce{MnO2}$ is formed from both the oxidizing and reducing agents.

Its effective n-factor for this specific comproportionation is $1.2$ ( $6 \text{ e}^- / 5 \text{ moles}$ ), not $2$ .

Mass balance is the safest route, but requires recognizing that the limiting reagent's equivalents dictate the consumption of the excess reagent.

Quick Facts About This Question

Question Information

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