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Question #44500

Difficulty: 8/10 | 🟢 Free

The Question

An aqueous solution of potassium butanoate (\ce{CH3CH2CH2COOK}) is subjected to Kolbe's electrolysis using inert platinum electrodes. The electrolysis is carried out until exactly $4.30 \text{ g}$ of the major hydrocarbon product is formed at the anode. Assuming $100\%$ current efficiency and standard temperature and pressure (STP: $273 \text{ K}, 1 \text{ atm}$ ), what is the total volume and the total mass of all gases evolved at both electrodes?\n\n(Given: Molar volume of an ideal gas at STP = $22.4 \text{ L mol}^{-1}$ , Molar masses in $\text{g mol}^{-1}$ : $\ce{C}=12, \ce{H}=1, \ce{O}=16$ )

Options

A.
$3.36 \text{ L}$ and $4.50 \text{ g}$
B.
$4.48 \text{ L}$ and $8.80 \text{ g}$
C.
$2.24 \text{ L}$ and $4.40 \text{ g}$
D.
$3.36 \text{ L}$ and $8.70 \text{ g}$

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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

Kolbe’s Electrolysis Mechanism|Faraday's First and Second Laws|Stoichiometric Coefficient Division

Why This Question Is Tricky

Common Mistakes

This problem tests Faraday's Laws applied to Kolbe Electrolysis, specifically targeting two major student blind spots: multi-compartment stoichiometry and standard physical states.

Students frequently forget that reduction at the cathode simultaneously produces hydrogen gas (\ce{H2}) in a 1:2 molar ratio to the electrons passed.

Furthermore, the problem uses potassium butanoate to produce hexane, which is a liquid at STP.

A common trap is blindly treating all anodic products as gases, leading to incorrect volume and mass calculations.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Hydrocarbons - Class 11 Part 2
Concept	Kolbe’s Electrolysis Mechanism
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free