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1 Faraday = _______________

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in Engineering Chemistry II by (789k points)
1 Faraday = _______________

(a) 1 coulomb

(b) 1000 coulomb

(c) 5000 coulomb

(d) 96500 coulomb

I got this question in final exam.

Query is from Faraday’s Laws of Electrolysis topic in chapter Corrosion, Electrolysis, Fuel Cells and Batteries of Engineering Chemistry II

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The correct answer is (d) 96500 coulomb

The best I can explain: 1 Faraday = 96,500 Coulomb. It is calculated mathematically and is widely used everywhere.

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