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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 5.20×10?2 M and 1.40 M , respectively.

Part A I have part A answered but need help with Part B and C...below.

What is the initial cell potential?

Express your answer using two significant figures.

Ecell =
0.51

V
Part B

What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M?

Express your answer using two significant figures.

Ecell =
V

Part 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.

Respuesta :

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

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

The reaction at the anode is equal to:

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

The number of moles of the electrons that are involved is equal to n = 2

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

 The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

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

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

lhmarianateixeira

In this exercise it is necessary to calculate the cellular potential energy, thus we have:

a) [tex]Ecell = 0.5123 V[/tex]

b) [tex]Ecell = 0.4695 V[/tex]

c) [[tex]Pb2^+] = 4.75 M[/tex]

a)The reaction at the cathode is represented as follows:

[tex]Cu2^+ + 2e^- \rightarrow Cu (s) \\Eocathode = 0.34 V[/tex]

The reaction at the anode is equal to:

[tex]Pb (s) \rightarrow Pb2 ^+ + 2e^-\\Eoanode = -0.13 V[/tex]

The number of moles of the electrons that are involved is equal to n = 2. Standard cell potential equals:

[tex]E_o = Eocathode - Eoanode \\= 0.34 V- (-0.13 V) \\= 0.47 V[/tex]

The initial cell potential can be calculated with the following formula:

[tex]Ecell = Eocell \rightarrow (0.0592 / n) log ([(Pb2^+)] / [(Cu2^+)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V[/tex]

b)The reaction in the cell is equal to:

[tex]Cu2^+ + Pb (s) \rightarrow Cu (s) + Pb2^+[/tex]

The concentration of Cu2 that gives the exercise is equal 0.2 M. Therefore, the change in concentration for [tex]Cu2^+[/tex] is equal to:

[tex]Cu2^+ = 1.4 M - 0.2 M = 1.2 M[/tex]

We use the formula from part a):

[tex]Ecell = Eocell \rightarrow (0.0592 / n) log ([(Pb2^+)] / [(Cu2^+)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V[/tex]

c)To find the concentration of [tex]Pb2^+[/tex] when there is a potential change in the cell of 0.37 V, we must clear the concentration of [tex]Pb2^+[/tex] from the following formula:

[tex]Eccell = Echocell - (0.0592 / n) log (([Pb2^+]) / ([Cu2^+]))[/tex]

[tex]0.0296 log ([Pb2^+] / [Cu2^+]) = (Eocélula - Ecélula / 0.0296)[/tex]

Clearing [tex]Pb2^+[/tex]:

[tex][Pb2^+] = 4.75 M[/tex]

Learn more: brainly.com/question/22139821

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