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 1.50 M , respectively. What is the cell potential when the concentration of Cu2+ has fallen to 0.210 M ? I got 0.45 V which is correct but i want part B to solve it

B- What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V ? M

Enter your answers numerically separated by a comma. Express your answer using two significant figures.

Explanation / Answer

cell reaction : Pb + Cu+2 ---> Pb+2 + Cu

E0Cu2+/Cu = +0.34 v

E0Pb2+/Pb = -0.13

E0cell = E0cathode - E0anode

       = 0.34--0.13

       = 0.47 v

Nernst equation:

in part A , [Cu2+] change = 1.5 - x M

            [Pb2+] change = 0.053 + x M

two electrons are transferred. n = 2

E = E0 – (0.059/n) log [Pb+2]/[Cu+2]

0.37 = 0.47 - (0.0591/2)log(x)

[Pb+2]/[Cu+2] = x = 2421.55

(0.053+x)/(1.5-x) = 2421.55

x = 1.5 M

[Pb2+] = 0.053+1.5 = 1.553 M

[Cu2+] = 1.5-1.5 = 0 M

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