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Chapter 20 Assigned Homework Answers

Chapter 20 Assigned Homework Answers - Chapter 20 Assigned...

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Chapter 20 Assigned Homework Answers Exercises: 4, 8, 10, 14, 22, 24, 26, 30, 34, 38, 44, 48, 50, 52, 58, 72, 74, 76, 80 4. (a) PBr 3 (l) + 3 H 2 O (l) H 3 PO 3 (aq) + 3 HBr (aq) There are no changes in oxidation numbers for any elements in this reaction. This is not an oxidation-reduction reaction. (b) NaI (aq) + 3 HOCl (aq) NaIO 3 (aq) + 3 HCl (aq) In this reaction, the oxidation number of I in NaI is equal to –1. The oxidation number of I in NaIO 3 is equal to +5. Therefore, NaI is oxidized to NaIO 3 . The oxidation number of Cl in HOCl is equal to +1. The oxidation number of Cl in HCl is equal to –1. Therefore, HOCl is reduced to HCl. (c) 3 SO 2 (g) + 2 HNO 3 (aq) + 2 H 2 O (l) 3 H 2 SO 4 (aq) + 2 NO (g) In this reaction, the oxidation number of S in SO 2 is +4. The oxidation number of S in H 2 SO 4 is +6. Therefore, SO 2 is oxidized to H 2 SO 4 . The oxidation number of N in HNO 3 is +5. The oxidation number of N in NO is +2. Therefore, HNO 3 is reduced to NO. (d) 2 H 2 SO 4 (aq) + 2 HBr (s) Br 2 (l) + SO 2 (g) + Na 2 SO 4 (g) + 2 H 2 O (l) In this reaction, the oxidation number of S in H 2 SO 4 is +6. The oxidation number of S in SO 2 is +4. Therefore, H 2 SO 4 is reduced to SO 2 . The oxidation number of Br in HBr is –1. The oxidation number of Br in Br 2 is 0. Therefore, HBr is oxidized to Br 2 . 8. Complete and balance the following half-reactions: (a) Mo 3+ (aq) Mo (s) (acidic solution) The non-H and non-O atoms are already balanced on both sides. There are no H nor O atoms to balance in this half-reaction. All that is left is to balance charge by adding e¯: Mo 3+ (aq) + 3 e¯ Mo (s) Since electrons appear as reactants, this is a reduction. (b) H 2 SO 3 (aq) SO 4 2- (acidic solution) The non-H and non-O atoms are already balanced on both sides. Balance the number of O atoms by adding 1 H 2 O on the left-hand side: H 2 SO 3 + H 2 O SO 4 2- Balance the number of H atoms by placing 4 H + ions on the right-hand side: H 2 SO 3 + H 2 O SO 4 2- + 4 H + Balance the charge on both sides by placing 2 e¯ on the right-hand side: H 2 SO 3 + H 2 O SO 4 2- + 4 H + + 2 e¯ Since electrons appear as products, this is an oxidation. (c) NO 3 ¯ NO (basic solution)
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The non-H and non-O atoms are already balanced on both sides. Balance the number of O atoms by placing 2 H 2 O molecules on the right-hand side: NO 3 ¯ NO + 2 H 2 O Balance the number of H atoms by placing 4 H + on the left-hand side: NO 3 ¯ + 4 H + NO + 2 H 2 O Balance the charge by placing 3 e¯ on the right-hand side: NO 3 ¯ + 4 H + + 3 e¯ NO + 2 H 2 O Convert to basic solution by adding 4 OH¯ to both sides: NO 3 ¯ + 4 H 2 O + 3 e¯ NO + 2 H 2 O + 4 OH¯ Cancel out 2 H 2 O molecules from both sides: NO 3 ¯ + 2 H 2 O + 3 e¯ NO + 4 OH¯ Since electrons appear as reactants, this is a reduction.
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