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1 st you need the BALANCED EQUATION

1 st you need the BALANCED EQUATION. Pb 2+ + 2 e- Pb (s) E 0 red = -0.13 V Cu 2+ + 2 e- Cu (s) E 0 red = 0.34 V You need an OXIDATION and a REDUCTION. E 0 cell >0 or nothing happens!. 1 st you need the BALANCED EQUATION. Pb (s)  Pb 2+ + 2 e- E 0 ox = + 0.13 V

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1 st you need the BALANCED EQUATION

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  1. 1st you need the BALANCED EQUATION Pb2+ + 2 e- Pb(s)E0red = -0.13 V Cu2+ + 2 e- Cu(s) E0red = 0.34 V You need an OXIDATION and a REDUCTION. E0cell >0 or nothing happens!

  2. 1st you need the BALANCED EQUATION Pb(s)Pb2+ + 2 e- E0ox = +0.13 V Cu2+ + 2 e- Cu(s) E0red = 0.34 V Pb(s) + Cu2+ Pb2+ + Cu(s) E0cell = 0.13+0.34V That assumes 1 M concentrations. Since they aren’t 1 M: NERNST EQUATION

  3. 1st you need the BALANCED EQUATION Pb(s) + Cu2+ Pb2+ + Cu(s) E0cell = 0.47 v Ecell = E0cell – 0.0592/n log Q Ecell = E0cell – 0.0592 log Pb2+ n Cu2+ Ecell = 0.47V – 0.0592 log 0.050 2 1.50 Ecell = 0.51 V

  4. What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M? Ecell = E0cell – 0.0592 log Pb2+ n Cu2+ Ecell = 0.47V – 0.0592 log 1.35 2 0.2 Ecell = 0.45 V Pb(s) + Cu2+ Pb2+ + Cu(s)

  5. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.35 V? Ecell = E0cell – 0.0592 log Pb2+ n Cu2+ 0.35 = 0.47V – 0.0592 log 0.05+x 2 1.5-x -0.12 = -0.0296 log (0.05+x/1.5-x) 4.054 = log (0.05+x/1.5-x) 104.054= 0.05+x/1.5-x 1.1324x104 = 0.05+x/1.5-x 1.6986x104 – 1.1324x104 x = 0.05 +x 1.6986x104 = 1.1325x104 x X = 1.4999 Pb(s) + Cu2+ Pb2+ + Cu(s)

  6. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.35 V? Ecell = E0cell – 0.0592 log Pb2+ n Cu2+ 0.35 = 0.47V – 0.0592 log 0.05+x 2 1.5-x -0.12 = -0.0296 log (0.05+x/1.5-x) 4.054 = log (0.05+x/1.5-x) 104.054= 0.05+x/1.5-x 1.1324x104 = 0.05+x/1.5-x 1.6986x104 – 1.1324x104 x = 0.05 +x 1.6986x104 = 1.1325x104 x X = 1.4999 Pb(s) + Cu2+ Pb2+ + Cu(s)

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