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Question #44131
The Question
A aqueous solution at contains a mixture of \ce{Na2S2O3} and \ce{Na2S4O6}. A aliquot of this solution requires of \ce{I2} solution for complete reaction in a mildly acidic medium. Another aliquot of the original solution is made strongly alkaline (pH = 14) and subjected to exhaustive anodic electrolysis at a constant current of . The electrolysis requires for the complete oxidation of all sulfur species to sulfate (\ce{SO4^{2-}}). Assuming 100% current efficiency, what is the molarity of \ce{Na2S4O6} in the original solution?
Options
- A.
0.010 \text{ M}
- B.
0.020 \text{ M}
- C.
0.025 \text{ M}
- D.
0.014 \text{ M}
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
Average vs. Actual Oxidation State|Electrolysis Product Prediction|Half-Reaction Media Sensitivity
Why This Question Is Tricky
Common Mistakes
This question tests 'Half-Reaction Media Sensitivity' and the 'Average vs Actual O.S.' trap.
Students often memorize the n-factor of thiosulfate as 1 (from iodometry) and blindly apply it to all redox scenarios, failing to recognize that in strongly alkaline electrolysis, it undergoes exhaustive oxidation to sulfate (n=8).
Furthermore, when calculating the n-factor for tetrathionate, students who rely on the average oxidation state (+2.5) often misapply the stoichiometric multiplier, calculating a change of 3.5 and multiplying by 2 (habit from thiosulfate) instead of 4, yielding an incorrect n-factor of 7.
This creates a mismatch in Faraday's law equivalents, leading directly to the distractor 0.020 M.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Redox Reactions - Class 11 Part 2 |
| Concept | Average vs. Actual Oxidation State |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |