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

Question

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 5.30×102 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 0.200 M ?

Express your answer using two significant figures.

c. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V ?

Enter your answers numerically separated by a comma. Express your answer using two significant figures.

Explanation / Answer

First we need the standard potential reduction which are:
E°Pb = -0.13 V;
E°Cu = +0.34 V

This means that Pb is being oxidized by Cu, and Cu is being reduced by Pb. Both of them are losing/gaining 2 electrons, so the E° would be:

E° = 0.34+0.13 = 0.47 V

Using the nerst equation, the initial cell potential is:
E = E° - 0.059/n logQ

Q = [Pb2+]/[Cu2+] = (0.053)/1.6 = 0.0331

E = 0.47 - 0.059/2 log(0.0331)
E = 0.5137 V

If concentration of Cu falls to 0.2 then:
E = 0.47 - 0.059/2 log(0.053/0.2)
E = 0.487 V

The last part, all you have to do is solve for Q in the nerst equation. That will give you the ratio of Pb/Cu. As we don't know which value to take as reference, we cannot calculte the concentrations of these ions, only the ratio for now.

Hope this helps

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