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 5.1×102 M and 1.600 M , respectively.

What is the initial cell potential?

What is the cell potential when the concentration of Cu2+ has fallen to 0.240 M ?

Express your answer using two significant figures.

What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.350 V ?

Enter your answers numerically separated by a comma. Express your answer using three decimal places.

Explanation / Answer

First write the cell reaction as follows:

              Pb ---> Pb+2 + 2e- +0.13

Cu+2 + 2e- ---> Cu +0.34

Pb + Cu+2 == Pb+2 + Cu +0.47

From the cell reaction we go the cell potential +0.47.

Now use the Nernst equation to calculate initial cell potential as follows:

Here; n = 2

E = 0.47 – 0.059/2 log [Pb+2]/[Cu+2]

and [Pb+2]/= 5.1×102 M and [Cu+2] =1.600 M

E = 0.47 -0.03log(5.0x 10^-2) / 1.600

E = 0.47 - 0.03(-1.5)

E = 0.47 + 0.045

E = 0.515

E = 0.52

Now the cell potential when the concentration of Cu2+ has fallen to 0.240 M

[Cu+2] that reacts: 1.600 – 0.240 = 1.36

Additional [Pb+2] produced = 1.36

[Pb+2] that remain = 5.0x 10^-2+ 1.36 = 1.41

now put these new concentration of both ions as follows:

E = 0.47 – 0.03log1.41/0.240

E = 0.47 -0.03(0.77)

E = 0.47 – 0.023

E = 0.447 or 0.45

Now calculate the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.350 V

0.350 = 0.470 – 0.03log[Pb+2]/ [Cu+2]

-0.12 / -0.03 = log[Pb+2] / [Cu+2]

4 = log[Pb+2] /[Cu+2]

[Pb+2] / [Cu+2] = 10000

0.05 + x / 1.600 –x = 10000

0.05 + x = 16000 – 10000 x

10001 x = 16000

x = 1.599 which is approximately 1.600 means , all of the Cu+2 has been consumed.

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