A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 C.
ID: 103399 • Letter: A
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 5.00×102M and 1.70 M , respectively.
Part A
What is the initial cell potential?
Express your answer using two significant figures.
0.52
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The cell potential can be calculated using the equation
Ecell=Ecell0.0592VnlogQ
where Ecell is the cell potential in volts, Ecell is the standard cell potential in volts, n is the number of moles of electrons transferred in the redox reaction, and Q is the reaction quotient. Inserting the values into the equation gives
Ecell=0.47V0.0592V2log5.00×102M1.70M
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.
0.44
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The balanced equation for this redox reaction is
Pb(s)+Cu2+(aq)Pb2+(aq)+Cu(s)
and since the concentration of Cu2+ drops from 1.70 M to 0.200 M , or a change of 1.50 M , then the concentration of Pb2+ will increase proportionally by 1.50 M as the stoichiometry is one to one. The cell potential can be calculated using the equation
Ecell=Ecell0.0592VnlogQ
where Ecell is the cell potential in volts, Ecell is the standard cell potential in volts, n is the number of moles of electrons transferred in the redox reaction, and Q is the reaction quotient. Inserting the values into the equation then gives
Ecell=0.47V0.0592V2log(5.00×102M+1.50M)0.200M
Part C
What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.36 V ?
Enter your answers numerically separated by a comma. Express your answers using two significant figures.
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 5.00×102M and 1.70 M , respectively.
You may want to reference ( pages 910 - 914)Section 19.6 while completing this problem.
Part A
What is the initial cell potential?
Express your answer using two significant figures.
Ecell =0.52
VSubmitMy AnswersGive Up
Correct
The cell potential can be calculated using the equation
Ecell=Ecell0.0592VnlogQ
where Ecell is the cell potential in volts, Ecell is the standard cell potential in volts, n is the number of moles of electrons transferred in the redox reaction, and Q is the reaction quotient. Inserting the values into the equation gives
Ecell=0.47V0.0592V2log5.00×102M1.70M
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 =0.44
VSubmitMy AnswersGive Up
Correct
The balanced equation for this redox reaction is
Pb(s)+Cu2+(aq)Pb2+(aq)+Cu(s)
and since the concentration of Cu2+ drops from 1.70 M to 0.200 M , or a change of 1.50 M , then the concentration of Pb2+ will increase proportionally by 1.50 M as the stoichiometry is one to one. The cell potential can be calculated using the equation
Ecell=Ecell0.0592VnlogQ
where Ecell is the cell potential in volts, Ecell is the standard cell potential in volts, n is the number of moles of electrons transferred in the redox reaction, and Q is the reaction quotient. Inserting the values into the equation then gives
Ecell=0.47V0.0592V2log(5.00×102M+1.50M)0.200M
Part C
What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.36 V ?
Enter your answers numerically separated by a comma. Express your answers using two significant figures.
Explanation / Answer
part A
Cu2+ + 2e- ---> Cu(s) E0 = +0.337 v (involves in reduction)
Pb2+ + 2e- ---> Pb(s) E0 = -0.126 v (involves in oxidation)
cell reaction : Pb(s) + Cu^2+(aq) ----> Pb^2+(aq) + Cu(s)
E0cell = E0cathode- E0anode
= 0.34- - 0.13
= 0.47 v
Ecell = E0cell - (0.0592/n)log([Cu^2+]/[Pb2+])
= 0.47- (0.0592/2)log(0.05/1.7)
= 0.52 v
part B
cell reaction : Pb(s) + Cu^2+(aq) ----> Pb^2+(aq) + Cu(s)
E0cell = E0cathode- E0anode
= 0.34- - 0.13
= 0.47 v
Ecell = E0cell - (0.0592/n)log([Cu^2+]/[Pb2+])
= 0.47- (0.0592/2)log(1.55/0.2)
= 0.44 v
part C
Ecell = E0cell - (0.0591/n)log([Cu^2+]/[Pb2+])
0.36 = 0.47- (0.0592/2)log((0.05+x)/(1.7-x))
x = 1.69
[Pb^2+] = 0.05+1.69 = 1.74 M
[Cu2+]= 1.7-1.69 = 0.01 M
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