Exercise 18.78 What is the initial cell potential? Express your answer using two

Question

Exercise 18.78 What is the initial cell potential? Express your answer using two significant figures A voltaic cell consists of a Pb/Pb hail-cell and a Cu/Cu2 halif-cell at 25 C. The initial concentrations of Pb2+ and Cu2+ are 5.30-10-2 M and 160 M respectively Submit My Answers Give Up Part B What is the cell potential when the concentration of Cu3 has tailen to 0.230 M Express your answer using two significant figures. Submit My Answers Glive Up Part C What are the concentrations of Pb and Cu2 when the cel potential falls to 0.370 V? Enter your answers numericaly separated by a comma.Express your answer using two significant figures

Explanation / Answer

Cu2+ + 2e Cu(s)

E0red = 0.337 V

Pb(s) Pb2+ + 2e

E0ox = 0.126 V

Cell reaction:

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

E0cell = E0red + E0ox

E0cell = 0.337 V + 0.126 V

= 0.463 V

At 25 °C,

Ecell = E0cell – 0.05912 / n * log([Pb2+]/[Cu2+])

n = 2

A)

[Pb2+] = 5.3 x 10-2 M

[Cu2+] = 1.6 M

Ecell = E0cell – 0.05912 / 2 * log([Pb2+]/[Cu2+])

= 0.463 – 0.02956 * log(5.3 x 10-2 / 1.6)

= 0.51 V

B)

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

(1.6 – x) + Pb(s) Cu(s) + (5.3 x 10-2 + x)

[Cu2+] = 0.23 M = 1.6 – x

x = 1.37

[Pb2+] = 5.3 x 10-2 M + x

[Pb2+] = 1.423 M

Ecell = E0cell – 0.05912 / 2 * log([Pb2+]/[Cu2+])

= 0.463 – 0.02956 * log(1.423 / 0.23)

= 0.44 V

C)

[Cu2+] = (1.6 – x) M

[Pb2+] = (5.3 x 10-2 + x) M

Ecell = E0cell – 0.05912 / 2 * log([Pb2+]/[Cu2+])

0.37 = 0.463 – 0.02956 * log((5.3 x 10-2 + x) / (1.6 – x))

log((5.3 x 10-2 + x) / (1.6 – x)) = 3.146 V

x = 1.59882 M

[Pb2+] = (5.3 x 10-2 + x) M = 1.7 M

[Cu2+] = (1.6 – x) M = 1.2 x 10-3 M

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