A voltaic cell consists of a Pb/Pb^2+ half-cell and a Cu/Cu^2+ half-cell at 25 d

Question

A voltaic cell consists of a Pb/Pb^2+ half-cell and a Cu/Cu^2+ half-cell at 25 degree C. The initial concentrators of Pb^2+ and Cu^2+ are 5.30 times 10^-2 M and 1.40 M, respectively. What is the initial cell potential? Express your answer using two significant figures. What is the cell potential when the concentration of Cu^2+ has fallen to 0.230 M? Express your answer using two significant figures. What are the concentrations of Pb^2+and Cu^2+ when the cell potential falls to 0.36C V? Enter your answers numerically separated by a comma. Express your answer using two significant figures.

Explanation / Answer

A)

initially:

Ecell = E°cell - 0.0592/n*log(Q)

Pb2+ + 2 e Pb(s) 0.126

Cu2+ + 2 e Cu(s) +0.337

E°cell = 0.337- -0.126 = 0.473 V

so..

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

Ecell = 0.473 -0.0592/(2) * log((5.3*10^-2)/(1.4)) = 0.5150 V

b)

when Cu+2 -- >0.23 M

then.. 1.4-0.23 = 1.17 M reacted

assume Pb+2 --> 5.3*10^-2 + 1.17 = 1.223

so

Q = (1.223/0.23 ) = 5.31

Ecell = 0.473 -0.0592/(2) * log((5.31)) = 0.451 V

c)

Find

Ecell = 0.473 -0.0592/(2) * log([Pb+2]/[Cu+2])

ifEcell = 0.36

0.36 = 0.473 -0.0592/(2) * log([Pb+2]/[Cu+2])

(0.36 -0.473 ) * 2 /-0.0592 = log([Pb+2]/[Cu+2])

3.81 = log([Pb+2]/[Cu+2])

[Pb+2]/[Cu+2] = 10^3.81 = 6456.54

Pb+2 = [Cu+2]*6456.54

[Pb+2] = 5.3*10^-2 + x

[Cu+2] = 1.4 - x

substitue

5.3*10^-2 + x = (1.4 - x)(6456.54)

5.3*10^-2 + x = 1.4*6456.54 - 6456.54x

(5.3*10^-2 ) - 1.4*6456.54 = -6457.54x

9039.103 / 6457.54 = x

x = 1.39977

[Pb+2] = 5.3*10^-2 + 1.39977 = 1.45277 V

[Cu+2] = 1.4 - 1.39977 = 0.00023

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