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.

Part A

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

Ecell= ? V

Part B

What is the cell potential when the concentration of Cu2+ has fallen to 0.230 M ?

Express your answer using two significant figures.

Part B

What is the cell potential when the concentration of Cu2+ has fallen to 0.230 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 0.360 V ? Enter your answers numerically separated by a comma. Express your answer using two significant figures.

[Pb2+],[Cu2+]= ? M

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

        = 0.337- - 0.126

        = 0.463 v

Ecell = E0cell - (0.0591/n)log([Cu^2+]/[Pb2+])

       = 0.463- (0.0591/2)log(0.053/1.6)

       = 0.507 v

part B

cell reaction : Pb(s) + Cu^2+(aq) ----> Pb^2+(aq) + Cu(s)

E0cell = E0cathode- E0anode

        = 0.337- - 0.126

        = 0.463 v

Ecell = E0cell - (0.0591/n)log([Cu^2+]/[Pb2+])

       = 0.463- (0.0591/2)log(0.053/0.23)

       = 0.482 v

part C

Ecell = E0cell - (0.0591/n)log([Cu^2+]/[Pb2+])

0.36 = 0.463- (0.0591/2)log((0.053+x)/(1.6-x))

x = 1.59

[Pb^2+] = 0.053+1.59 = 1.64M

[Cu2+]= 1.6-1.59 = 0.01 M

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