Explanation Nitric acid ( H N O 3 ) can oxidize metals through the reduction half-reaction shown below: NO 3 - ( aq ) + 4H + ( aq ) + 3e - → NO ( g ) + 2H 2 O ( l ) E red º = 0.96 V The substances with large positive reduction half-cell potentials show a strong tendency to gain electrons (reduce) and act as strong oxidizing agents. So, nitric acid can oxidize (dissolve) metals that have smaller positive reduction half-cell potentials than 0.96 V. a. The standard reduction potential and reduction half-reaction for Cu are Cu + 2 ( aq ) + 2e - → Cu ( s ) E red º = 0.34 V As the reduction half-cell potential of copper is smaller than 0.96 V, copper undergoes oxidation and can dissolve in 1 M nitric acid. Multiply oxidation half-reaction for Cu by three and reduction half-reaction for H N O 3 by two, then add to obtain balanced redox equation as follows: 3 C u ( s ) → 3 C u + 2 ( a q )

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: A Molecular Approach (4th Edition)

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ISBN: 9780134162430

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Chapter 19, Problem 59E

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Interpretation: It is to be determined whether each of the given metal will dissolve in 1 M HNO3 solution or not. Further, for the metals that dissolve in 1 M HNO3 solution, the redox reaction depicting the reaction of metal and HNO3 solution is to be written.

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Chapter 19, Problem 58E

Chapter 19, Problem 60E

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Chapter 19 Solutions

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: A Molecular Approach (4th Edition)

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It is to be determined whether each of the given metal will dissolve in 1 M H N O 3 solution or not. Further, for the metals that dissolve in 1 M H N O 3 solution, the redox reaction depicting the reaction of metal and H N O 3 solution is to be written.

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Interpretation: It is to be determined whether each of the given metal will dissolve in 1 M HNO3 solution or not. Further, for the metals that dissolve in 1 M HNO3 solution, the redox reaction depicting the reaction of metal and HNO3 solution is to be written.

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Chapter 19 Solutions