# 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 0.0520 M and 1.60 M , respectively. a) What is the initial cell potential? Express your answer using two significant figures.b) What is the cell potential when the concentration of Cu2+ has fallen to 0.230 M ? Express your answer using two significant figures.c) What is the concentration of Pb2+ when the cell potential falls to 0.370 V ? Express your answer using three significant figures.

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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 0.0520 M and 1.60 M , respectively.

a) What is the initial cell potential? Express your answer using two significant figures.

b) What is the cell potential when the concentration of Cu2+ has fallen to 0.230 M ? Express your answer using two significant figures.

c) What is the concentration of Pb2+ when the cell potential falls to 0.370 V ? Express your answer using three significant figures.

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Step 1

a)

Reaction at cathode:

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

Reaction at anode:

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

Number of moles of electrons involved, n = 2

Standard cell potential, Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V.

Initial cell potential can be calculated as follows,

Step 2

b)

The cell reaction is given below.

Cu2+ + Pb (s)  --> Cu(s) + Pb2+

The new concentration of Cu2 = 0.23 M

Change in concentration of Cu2 = 1.6 M – 0.23 M = 1.37 M

Theref...

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