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# Describing The Process Of Electorlysis And How The Number Of Coloumbs Can Verify The Amount Of Current Given Off

806 words - 3 pages

Electrolysis*First LawThe quantity of a substance produced by electrolysis is proprotional to the quantity of electricty used.*Second LawFor a given quantity of electricity the quantity of substance produced is proportional to its weight.*The quantity of electricity or charge contained in a current running for a specified time can be calculated: Q = I x tQ = quantity of electricity or charge in coulombs (C)I = current in amps (A)t = time (seconds)*The Faraday constant, F, is the quantity of electricity carried by one mole of electrons.F = Avogadro's Number x charge on electron in coulombsF = 6.022 x 1023 mol-1 x 1.602192 x 10-19 CF = 96,484 C mol-1This is usually rounded off to 96,500 C mol-1 for calculations in chemistry.*The quantity of electricity required to deposit an amount of metal can be calculated: Q = n(e) x FQ = quantity of electricity in coulombs (C)n(e) = moles of electronsF = Faraday constant = 96, 500 C mol-1*Electrical Energy, E, can be calculated: E = Q x VE = electrical energy in joules (J)Q = quantity of electricity in coulombs (C)V = voltage (or EMF) in volts (V)*1 kilowatt-hour, kWH, is a unit of electrical energy.1 kWH = 3.6 x 106 JExamples*Q = I x tCalculate the quantity of electricity, Q, obtained when a current of 25 amps runs for 1 minute.Q = ? CI = 25 At = 1 minute = 60 secondsQ = 25 x 60 = 1,500 CCalculate the current needed to 30,000 coulombs of electricity in 5 minutes.Q = 30,000 CI = ? At = 5 minutes = 5 x 60 = 300 secondsI = Q ÷ t = 30,000 ÷ 300 = 100 amps*t = Q ÷ ICalculate the time required to produce 12,000 C of electricity using a current of 10 amps.Q = 12,000 CI = 10 At = ?t = Q ÷ t = 12,000 ÷ 10 = 1,200 seconds = 1,200 ÷ 60 = 20 minutes*Q = n(e) x FCalculate the quantity of electricity obtained from 2 moles of electronsQ = n x FQ = ?n = 2 molF = 96,500 C mol-1Q = 2 x 96,500 = 193,000 C*n(e) = Q ÷ FCalculate the moles of electrons obtained from 250 C of electricityn(e) = ? molQ = 250 CF = 96,500 C mol-1n(e) = 250 ÷ 96,500 = 2.59 x 10-3 mol*Calculate the time required to deposit 56g of...

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