w SG2216 Assignmer X c Quantitative Chem x CI2.2s2032-21- x G consider the follo
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w SG2216 Assignmer X c Quantitative Chem x CI2.2s2032-21- x G consider the follo x C Chemist question X C chegg study IGui x CIOX SG2216 Ass x x C D www.webassign.net/web/Student/Assignment-Responses submit?dep 14204282 7. points My Notes Ask Your Teacher Suppose that the concentrations of NaFand KCI were each 0.15 M in the cell Pb(s PbF2(s) l F (aq) ll Cl aq) I AgCl(s) l Ag(s) (a) Using the half-reactions 2 AgCl(s) 2 e Ag(s) 2 Cl (aq) and PbF2(s) 2 e Pb(s) 2 F (aq), calculate the cell voltage. (Assume that the Nernst potential is 0.05916 V. 4.0 (b) In which direction do electrons flow? O Electrons will flow clockwise through the wire from the left half-cell to the right half-cell. O Electrons will flow counterclockwise through the wire from the right half-cell to the left half-ce (c) Now calculate the cell voltage by using the re 2 Ag+(aq) 2 e e 2 Ag(s) and Pb2+(aq) 2 e pb(s). (Assume Ksp of pbF2 3.6x10-9, Ksp of Agcl 1.8x 10 10 4.0 Supporting Materials Periodic Table Physical Constants Supplemental Materials 8. 3/12 points I Previous Answers My Notes Ask Your Teacher A solution contains 0.311 M Ce and 3.11 M HCIO4. (The Faraday's constant is 9.649 x104 Coulombs/mole 3.11 x 10 M Ce 11 x 10 M Mn 0.311 M Mno4 9:06 PM A 4 Ask me anything /21/2016 1Explanation / Answer
The reduction reaction is PbF2 + 2e ==> Pb(s) + 2F-
The oxidation reaction is : Ag(s) + 2Cl- ==> 2AgCl(s)
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overall reaction = PbF2 + Ag(s) + 2Cl- ==>Pb(s) + 2F- + 2AgCl(s)
E= E^0 - 0.059/n ln [(Pb)(AgCl)(F-)^2/PbF2) *(Ag)*(Cl-)^2]
= 0.05916 - 0.059/2 ln 1
= 0.029 V
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electrons will flow from right half cell to left half cell.
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Ag(s) ---> Ag^2+ + 2e
Pb^2+ + 2e ---> Pb (s)
overall : Ag (s) + Pb^2+ ==> Ag^+ + Pb (s)
E = E^0 - 0.059/2 ln [Ag^+/Pb^2+]
Calculate the concentration of Ag+ and Pb^2+ from Ksp values.
Ksp (AgCl) = [Ag+][Cl-]
or, [Ag+] = 1.8*10^-10/0.15 = 6.67*10^-10 M
Ksp (PbF2) = [Pb^2+][F-]2
[Pb^2+] = 3.6*10^-8/(0.15)^2 =1.6*10^-6 M
E^0 = E (Pb2+/Pb) + E(Ag+/Ag) = 0.80 - (-0.13) = 0.93
E = E^0 - 0.059/2 ln [Ag^+/Pb^2+]
=0.93 - 0.059/2 ln (6.67*10^-10 M/ 1.6*10^-6 M) = 1.029 V
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