Question

A galvanic cell Pb│PbCl2│NaCl (0.0100M)││AgNO3 (0.500 M)│Ag is setup in the lab. Ksp PbCl2 = 87×...

  1. A galvanic cell Pb│PbCl2│NaCl (0.0100M)││AgNO3 (0.500 M)│Ag is setup in the lab. Ksp PbCl2 = 87× 10-5.
    1. What is the concentration of Pb2+ in the Pb│PbCl2 half-cell? (10 pts.)
    2. Calculate Ecell. (20 pts.)
    3. E0 red=-0.13 Pb2+ + 2e- --> Pb
    4. E0 red= 0.80 Ag+ + e= -->Ag
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Answer #1

Ans:) The given problem can be solved by using the concepts of solubility product and Nernst equation from which the Ecell is determined. Please like if it helps.

Ahus! Giyen cell equatwn is Ablebele land (0-01N) || nyalog Co-Fennel an - ksp (ebab) = 878 10-5 o we know that Pol Pbela hay

Now the Ecell is given as 1 Eceu - E-cathode - Eanode - Eron - Engdn Eren = 0.80 - (- 0.13) = 0.93 v. now the Eceu as given a

Thank you !

Ahus! Giyen cell equatwn is Ablebele land (0-01N) || nyalog Co-Fennel an - ksp (ebab) = 878 10-5 o we know that Pol Pbela hay cell equation can be given as - Pbal Pb2+ + 2t - + x + 2x then a Ksp = [e] [an]2 = 443 4003 - 87 X 105 ² = 21.75 x 10-5 = 217.5 x 10-6 x = 5.90 x 10 = 0.059 M Concentration of Pb2+ is 0.059M -- haly cell reaction for the above is given asi- The Cell ( onication : Pb - Pb2+ + 26- reduction in Ag+ + e - Ag The net reaction is as :- Pb + 2 Ag+ → 244 + Pb2+ 2 +

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