# Ch21_7 - Quantitative Aspects of Electrochemistry 1...

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Unformatted text preview: Quantitative Aspects of Electrochemistry 1 Consider electrolysis of aqueous silver ion. Ag+ (aq) + e- ---> Ag(s) 1 mol e- ---> 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-? charge passing Current = time I (amps) = I (amps) = coulombs seconds 1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a) Charge = 1350 C (b) Calculate moles of e- used 1350 C • (c) 1 mol e96,500 C 1 mol Ag 0.0140 mol e- • 1 mol e- I (amps) = = 96,500 96,500 4 Quantitative Aspects of Electrochemistry Quantitative Aspects of Electrochemistry I (amps) = 3 coulombs seconds 1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a) Calc. charge Coulombs = amps x time = (1.5 amps)(15.0 min)(60 s/min) = 1350 C coulombs seconds But how is charge related to moles of electrons? Charge on 1 mol of e- = -19 C/e-)(6.02 (1.60 x 10 -19 C/e-)(6.02 x 10 23 e-/mol) C/ mol e- = 1 Faraday Faraday Quantitative Aspects of Electrochemistry 5 Quantitative Aspects of Electrochemistry Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2eIf a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a) 454 g Pb = 2.19 mol Pb b) Calculate moles of e- 6 The anode reaction in a lead storage battery is Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2eIf a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a) 454 g Pb = 2.19 mol Pb b) Mol of e- = 4.38 mol c) Charge = 423,000 C 2.19 mol Pb • 0.0140 mol e- Calc. quantity of Ag 2 charge passing Current = time coulombs seconds Quantitative Aspects of Electrochemistry Quantitative Aspects of Electrochemistry Quantitative Aspects of Electrochemistry c) 2 mol e= 4 .38 mol e1 mol Pb Calculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C 4.38 0.0140 mol Ag or 1.51 g Ag Page 1 d) Calculate time Time (s) = Time (s) = Charge (C) I (amps) 423,000 C = 2 82,000 s About 78 hours 1.50 amp ...
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## This note was uploaded on 10/19/2011 for the course CHM 2210 taught by Professor Reynolds during the Fall '01 term at University of Florida.

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