The solubility of oxygen in water is 0.2 mM, and the diffusion coefficient (D) in solution is 10-5 cm² s-1. What is the limiting current due to the two electron-reduction of oxygen at a micro- electrode of radius 100 μm ? The value for > = 50um. iL = -nFAD (Co/s) Answer: The unit of diffusion is (cm² s-¹) 8 = 0.005 cm Concentration = 0.2 mM or 0.2 μmol cm-³ A = 3 x 104 cm² İL = -nFAD (C/ 8) i = 2 x 96500 x 3 x 104 x 105 x 0.2 x10-6/5 x 10-3 = 23 nA

The solubility of oxygen in water is 0.2 mM, and the diffusion coefficient (D) in solution is 10-5 cm² s-1. What is the limiting current due to the two electron-reduction of oxygen at a micro- electrode of radius 100 μm ? The value for > = 50um. iL = -nFAD (Co/s) Answer: The unit of diffusion is (cm² s-¹) 8 = 0.005 cm Concentration = 0.2 mM or 0.2 μmol cm-³ A = 3 x 104 cm² İL = -nFAD (C/ 8) i = 2 x 96500 x 3 x 104 x 105 x 0.2 x10-6/5 x 10-3 = 23 nA

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Explain this step by step

The solubility of oxygen in water is 0.2 mM, and the diffusion coefficient (D.) in
solution is 10-5 cm² s-1.
What is the limiting current due to the two electron-reduction of oxygen at a micro-
electrode of radius 100 μm ?
The value for 8 = 50μm.
i₁ = -nFAD (Co/8)
Answer:
The unit of diffusion is (cm² s-¹)
8 0.005 cm
Concentration = 0.2 mM or 0.2 μmol cm-³
A 3 x 104 cm²
IL-nFAD (C/ 8)
2 x 96500 x 3 x 104 x 105 x 0.2 x10-6 / 5 x 10³ = 23 nA

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