Explanation For the cell that contains the two Cu / Cu + 2 half-cells, the cell potential at 25°C is E cell = 0.22 V. It is also given that the concentration of C u 2 in one half-cell = 1.5 × 10 − 3 M . This low concentration indicates that this is the concentration of the solution present in the oxidation half-cell. For concentration of cells at standard conditions, E cell º = 0 V . Thus, from Nernst equation, E cell = E cell º - 0.0592 V n log [ Cu 2 + ] unknown [ Cu 2 + ] known Substituting values 0.22 V = 0 V − 0.0592 V 2 log [ C u 2 + ] u n k n o w n 1.5 × 10 − 3 M log [ C u 2 + ] u n k n o w n 1.5 × 10 − 3 M = − 0.22 V × 2 0.0592 × V = − 7

Chemistry: A Molecular Approach Plus Mastering Chemistry with Pearson eText -- Access Card Package (4th Edition) (New Chemistry Titles from Niva Tro)

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Chapter 19, Problem 82E

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Interpretation: The concentration of Cu2+ in the other half-cell is to be determined when concentration of Cu2+ in one of the half-cell is given.

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Chapter 19, Problem 81E

Chapter 19, Problem 83E

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Chemistry: A Molecular Approach Plus Mastering Chemistry with Pearson eText -- Access Card Package (4th Edition) (New Chemistry Titles from Niva Tro)

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The concentration of C u 2 + in the other half-cell is to be determined when concentration of C u 2 + in one of the half-cell is given.

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Interpretation: The concentration of Cu2+ in the other half-cell is to be determined when concentration of Cu2+ in one of the half-cell is given.

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