Unit 9: Topic 10 - Electrolysis and Faraday’s Law Flashcards

1
Q

State Faraday’s law of electrolysis and explain it using dimensional analysis.

A

By Faraday’s first law, the mass of the elements deposited at the electrode is proportional to the charge, i.e. m ~ Q. With constant current electrolysis, Q = It, where Q is charge, I is current, and t is the time. Then we have
m = ItM/Fv.
m is the mass liberated from the electrode (in grams), I is the current applied (in Amperes), t is the amount of time the current is applied (in seconds), M is the molar mass of the substance (in g/mol), F is the Faraday’s constant (approximately 96485 s A/mol), and v is the number of free electrons.

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2
Q

A current of 0.15A is passed through an aqueous solution of K2PtCl4. How long will it take to deposit 1.00g of Pt(s)? The molar mass of Pt is 195.1 g/mol.

A

We use the formula m = ItM/Fv. Since we want to solve for time t, we rearrange to get t = mFv/IM. m = 1.00, M = 195.1, I = 0.15, F = 96485, and v = 2 since the charge of platinum here is +2. Solving gives t = 6594s.

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