Chapter19ISM

Chapter19ISM - CHAPTER 19 ELECTROCHEMISTRY Problem...

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CHAPTER 19 ELECTROCHEMISTRY Problem Categories Biological : 19.66, 19.68, 19.126. Conceptual : 19.67, 19.79, 19.85, 19.94, 19.103, 19.106, 19.108, 19.128. Descriptive : 19.13, 19.14, 19.17, 19.18, 19.46a, 19.52a, 19.61, 19.77, 19.87, 19.93, 19.99, 19.100, 19.101, 19.111, 19.113, 19.123. Environmental : 19.63. Industrial : 19.48, 19.56, 19.89, 19.109, 19.112, 19.120. Organic : 19.38. Difficulty Level Easy : 19.11, 19.12, 19.13, 19.14, 19.16, 19.17, 19.18, 19.22, 19.45, 19.63, 19.67, 19.79, 19.91, 19.97. Medium : 19.1, 19.2, 19.15, 19.21, 19.23, 19.24, 19.25, 19.26, 19.29, 19.30, 19.31, 19.32, 19.33, 19.34, 19.38, 19.46, 19.47, 19.48, 19.49, 19.50, 19.51, 19.52, 19.53, 19.55, 19.57, 19.58, 19.59, 19.60, 19.61, 19.62, 19.64, 19.65, 19.66, 19.69, 19.70, 19.72, 19.73, 19.74, 19.75, 19.77, 19.81, 19.82, 19.83, 19.84, 19.85, 19.86, 19.87, 19.89, 19.93, 19.94, 19.96, 19.98, 19.99, 19.100, 19.103, 19.104, 19.106, 19.107, 19.108, 19.109, 19.110, 19.113, 19.114, 19.115, 19.116, 19.117, 19.122, 19.124, 19.126. Difficult : 19.37, 19.54, 19.56, 19.68, 19.71, 19.76, 19.78, 19.80, 19.88, 19.90, 19.92, 19.95, 19.101, 19.102, 19.105, 19.111, 19.112, 19.118, 19.119, 19.120, 19.121, 19.123, 19.125, 19.127, 19.128. 19.1 We follow the steps are described in detail in Section 19.1 of the text. (a) The problem is given in ionic form, so combining Steps 1 and 2, the half-reactions are: oxidation: Fe 2 + Fe 3 + reduction: H 2 O 2 H 2 O Step 3: We balance each half-reaction for number and type of atoms and charges. The oxidation half-reaction is already balanced for Fe atoms. There are three net positive charges on the right and two net positive charges on the left, we add one electrons to the right side to balance the charge. Fe 2 + Fe 3 + + e Reduction half-reaction : we add one H 2 O to the right-hand side of the equation to balance the O atoms. H 2 O 2 2H 2 O To balance the H atoms, we add 2H + to the left-hand side. H 2 O 2 + 2H + 2H 2 O There are two net positive charges on the left, so we add two electrons to the same side to balance the charge. H 2 O 2 + 2H + + 2 e 2H 2 O Step 4: We now add the oxidation and reduction half-reactions to give the overall reaction. In order to equalize the number of electrons, we need to multiply the oxidation half-reaction by 2. 2(Fe 2 + Fe 3 + + e ) H 2 O 2 + 2H + + 2 e 2H 2 O 2Fe 2 + + H 2 O 2 + 2H + + 2 e 2Fe 3 + + 2H 2 O + 2 e
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CHAPTER 19: ELECTROCHEMISTRY 564 The electrons on both sides cancel, and we are left with the balanced net ionic equation in acidic medium. 2Fe 2 + + H 2 O 2 + 2H + 2Fe 3 + + 2H 2 O (b) The problem is given in ionic form, so combining Steps 1 and 2, the half-reactions are: oxidation: Cu Cu 2 + reduction: HNO 3 NO Step 3: We balance each half-reaction for number and type of atoms and charges. The oxidation half-reaction is already balanced for Cu atoms. There are two net positive charges on the right, so we add two electrons to the right side to balance the charge. Cu Cu 2 + + 2 e Reduction half-reaction : we add two H 2 O to the right-hand side of the equation to balance the O atoms. HNO 3 NO + 2H 2 O To balance the H atoms, we add 3H + to the left-hand side. 3H + + HNO 3 NO + 2H 2 O There are three net positive charges on the left, so we add three electrons to the same side to balance the charge. 3H + + HNO 3 + 3 e NO + 2H 2 O Step 4: We now add the oxidation and reduction half-reactions to give the overall reaction. In order to equalize the number of electrons, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2.
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Chapter19ISM - CHAPTER 19 ELECTROCHEMISTRY Problem...

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