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A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The...

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 5.00×10−2 M and 1.70 M, respectively. What is the cell potential when the concentration of Cu2+ has fallen to 0.250 M ? What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.37 V?

Solutions

Expert Solution

1)

Pb (s) ----------> Pb2+ + 2e-      Eo = - 0.13 V

Cu2+ (aq) + 2e- ----------> Cu (s)    Eo = 0.34 V

Eocell = Eocathode - Eoanode

          = 0.34 - (- 0.13)

          = 0.47 V

concentration of Cu2+ = 0.250 M

Cu2+ (aq) + Pb (s)   ------------>   Pb2+ (aq) + Cu (s)

1.70                                              5.0x 10^-2

1.70-x                                            5.0x10^-2+x

Cu2+ concentration = 0.250

1.70 - x = 0.250

x = 1.45

Ecell = Eo - 0.05916 / 2 log [Pb2+ / Cu2+]

        = 0.47 - 0.05916 / 2 log [1.5 / 0.25]

Ecell = 0.447 V

2)

Ecell = 0.37 V

Cu2+ (aq) + Pb (s)   ------------>   Pb2+ (aq) + Cu (s)

0.25                                            1.45

0.25 - x                                          1.45 + x

Ecell = Eo - 0.05916 / 2 log [Pb2+ / Cu2+]

0.37 = 0.47 - 0.05916 / 2 log [1.45 + x / 0.25 - x]

2402.7 = [1.45 + x / 0.25 - x]

600.62 - 2402.5 x = 1.45 + x

599.17 = 2403.5 x

x = 0.249

concentration of Pb2+ = 1.70 M

concentration of Cu2+ = 0.001 M

queen_honey_blossom answered 2 years ago
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