Exercise 19.78 Enhanced with Feedback A voltaic cellconsists of a Pb/Pb halfcell

Question

Exercise 19.78 Enhanced with Feedback A voltaic cellconsists of a Pb/Pb halfcell and a Cu/Cu half cell at 25 °C. The initial concentrations of Pb2+ and Cu are 5.30x10-2 Mand 1.70 M respectively. You may want to reference (D pages910-914 Section 19.6 while completing this problem. Part A What is the initial cell potential? Express your answer using two significant figures. IVO AE Submit My Answers Give Up Part B What is the cell potential when the concentration of Cu has fallen to 0.200 M? Express your answer using two significant figures. My Answers Give UR Submit Part C What are the concentrations ofPb and Cu when the cell potential falls to 0.37 V? Enter your answersnumerically separated by a comma. Express your answersusing two significant figures.

Explanation / Answer

Part A )

Pb(s) ----------> Pb2+(aq) + 2e E°(red)= -0.126V

Cu2+ (aq) + 2e -------> Cu(s) E°(red) = 0.337V

  Nerst equation is

E(cell) = E° - (0.0592/n)log([OX]/[Red])

[Ox]= [Pb2+] = 5.30×10^-2M

[Red]=[ Cu2+] = 1.70M

E°(cell) = E°(red,cathode) - E°(red,anode)

= 0.337 - (-0.126)

= 0.463V

No of electron transfer , n = 2

E(cell) = 0.463V - (0.0592/2)log(0.0530/1.70)

= 0.463 + 0.045

= 0.51V

Part B)

When [Cu2+] is fallen to 0.2M, decrease in [ Cu2+] is 1.5M

Decrease in [ Cu2+ ] = Increase in [Pb2+]

Therefore, [ Pb2+] = 1.553V

E(cell) = 0.463 - (0.0592/2)log(1.553/0.20)

= 0.463 - 0.026

= 0.44V

Part C)

When cell volt reach 0.37V

0.37V= 0.463V - (0.0592V/2)log([Ox]/[Red]))

log([Ox]/[Red])= 3.14

[Ox] / [ Red] = 1×10^3.14

= 1380.38

[Pb2+] = [Cu2+] × 1380.38

Total concentration [Cu2+] + [Pb2+]= 1.753M

   [ Cu2+] + 1380.38[Cu2+] = 1.753M

1381.38[Cu2+] = 1.753

   [Cu2+] =1.753/1381.38

   = 0.00126M

   [ Pb2+] = 1.753 - 0.00126 =1.7517M

  

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