quantitativeelectrolysis

quantitativeelectrolysis - .0389 mole Cu 63.5 g l mole =...

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Cut to the chase: Quantitative Electrolysis Problems One Faraday is a mole of electrons. If you think of it this way, then electrolysis problems will be very simple. Example: A current of 2.5 amps is run through a solution containing Cu 2+ ions for 50 minutes. What mass of copper is plated out? Conversions: 1 F = 96500 Coulombs = 1 mole of electrons Equations: I = q/t or It = q [ I = amps, t= time in seconds, q = charge(coulombs) ] Step one: It = coulombs It = 3000s X 2.5 amps = 7500C (7500C / 1) (1F/96500 C) (1 mole Cu / 2F) =
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Unformatted text preview: .0389 mole Cu ( 63.5 g / l mole) = 2.47g [ The 2F = l mole Cu comes from the charge of Cu which is 2+. It takes two moles of electrons ( or two F) to change one mole of Cu 2+ to one mole of Cu. A formula you might like that will work this type of problem is : it = nFe i = amps, t= seconds, n = moles, F = 96500C, and e = electrons transferred For the previous problem: it = nFe 2.5 amps X 3000s = n X 96500C X 2e n = .0389 moles e = 2 because for copper to reduce to copper metal: Cu 2+ + 2e Cu...
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This note was uploaded on 10/13/2009 for the course CHEM 1442 taught by Professor Rogers during the Fall '08 term at UT Arlington.

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