18.78C A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at

Question

18.78C

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 5.10×102 M and 1.60 M , respectively.

PART A: What is the initial cell potential? (2 significant figures)

Ecell= V

PART B: What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M? (2 significant figures)   Ecell= V

PART C: What are the concentration of Pb2+ And Cu2+ when the cell potential falls to 0.350 V? (Separate anwers numerically by a comma)

Explanation / Answer

First we need to know the Eo of both solution:

EoPb = -0.13 V; EoCu = 0.34 V

Eo = 0.34 - (-0.13) = 0.47 V

The Nernst Equation: E = Eo - RT/nF lnQ Q = [Pb+2] / [Cu2+]

At 25 °C and with the value of R and F as constant, we have: E = Eo - 0.0592/n logQ n = 2 electrons

Now Finally:

Part A:

E = 0.47 - (0.0592/2) log(5.1x10-2/1.6) E = 0.51 V

Part B:

E = 0.47 - (0.0592/2)log(5.1x10-2/0.2)   E = 0.49 V

Part C:

0.350 = 0.47 - (0.059/2)log(0.051 + x / 1.6 - x)

-0.12 = -0.0296log(0.051 + x / 1.6 - x)

-4.054 = -log(0.051 + x / 1.6 - x)

8.83x10-5 = (5.1x10-2 + x / 1.6 - x)

1.41x10-4 - 8.83x10-5x = 0.051 + x

-0.051 = -1.0000883x

x = 0.05

[Cu] = 1.6 - 0.05 = 1.55 M

[Pb] = 0.051 + 0.05 = 0.056 M

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