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Question #44777
The Question
A cylindrical corrosion pit of depth m and cross-sectional area m is formed on an iron surface. The pit is filled with an aqueous solution of a weak monoprotic acid \ce{HA} at a uniform concentration of M. Due to a localized gradient, the degree of dissociation of \ce{HA} varies with depth (where at the pit opening and at the iron base) according to the relation , with . The limiting molar conductivity of \ce{HA} is . A constant potential difference of V drives the anodic dissolution of iron (\ce{Fe -> Fe^{2+} + 2e^-}) at the base of the pit. Assuming the resistance of the electrolyte in the pit is the only resistance in the circuit, calculate the mass of iron lost due to corrosion after s. \n(Given: Atomic mass of Fe = 56 g/mol, C/mol, assume for the weak acid).
Options
- A.
28 mg
- B.
42 mg
- C.
56 mg
- D.
29.1 mg
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
Corrosion as Electrochemical Process|Faraday's First and Second Laws|Resistance and Resistivity Interconversion
Why This Question Is Tricky
Common Mistakes
This problem tests the student's ability to model a non-uniform electrolyte where resistivity varies spatially.
The primary trap is treating the electrolyte with a bulk resistance formula using surface resistivity, or incorrectly averaging the conductivity.
Students must recognize that the medium acts as a series of infinitesimal resistors, requiring the integration of the resistivity function over the depth of the pit.
Additionally, it tests the crucial unit conversion of concentration from M (mol/L) to SI units (mol/m) when calculating electrolytic conductivity, seamlessly integrating electrochemistry with calculus.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Electrochemistry - Class 12 |
| Concept | Corrosion as Electrochemical Process |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |