Ch21_7 - Consider electrolysis of aqueous silver ion Ag(aq...

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Unformatted text preview: Consider electrolysis of aqueous silver ion. Ag+ (aq) + e- ---> Ag(s) aq) 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 Quantitative Aspects of Electrochemistry 1 Quantitative Aspects of Electrochemistry charge passing Current = time I (amps) = coulombs seconds 2 Quantitative Aspects of Electrochemistry Quantitative Aspects of Electrochemistry I (amps) = coulombs seconds 3 But how is charge related to moles of electrons? Charge on 1 mol of e- = (1.60 x 10 -19 C/e-)(6.02 x 10 23 e-/mol) C/e-)(6.02 e-/mol) = 96,500 96,500 1.50 amps flow thru a Ag+(aq) solution for aq) 15.0 min. What mass of Ag metal is deposited? Solution (a) Calc. charge Calc. Coulombs = amps x time = (1.5 amps)(15.0 min)(60 s/min) = 1350 C I (amps) = coulombs seconds C/mol e- = 1 Faraday C/mol Faraday Quantitative Aspects of Electrochemistry Quantitative Aspects of Electrochemistry I (amps) = coulombs seconds 4 Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2eaq) aq) If a battery delivers 1.50 amp, and you have 454 g of Pb, how Pb, long will the battery last? 5 Quantitative Aspects of Electrochemistry Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2eaq) aq) If a battery delivers 1.50 amp, and you have 454 g of Pb, how Pb, long will the battery last? 6 1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What aq) mass of Ag metal is deposited? Solution (a) Charge = 1350 C (b) Calculate moles of e- used 1350 C • 1 mol e96,500 C 0.0140 mol e- Solution a) 454 g Pb = 2.19 mol Pb b) Calculate moles of e2.19 mol Pb • 2 mol e= 4 .38 mol e1 mol Pb Solution a) 454 g Pb = 2.19 mol Pb b) Mol of e- = 4.38 mol c) Charge = 423,000 C d) Calculate time Time (s) = Charge (C) I (amps) (c) Calc. quantity of Ag Calc. 0.0140 mol Ag or 1.51 g Ag c) 1 mol Ag 0.0140 mol e- • 1 mol e- Calculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C C/mol Time (s) = 423,000 C = 2 82,000 s About 78 hours 1.50 amp Page 1 ...
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This note was uploaded on 12/14/2010 for the course CHEM 1e03 taught by Professor Jb during the Winter '10 term at Macalester.

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