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20.2 Oxidation Numbers > Chapter 20 Oxidation-Reduction Reactions 20.1 The Meaning of Oxidation and Reduction 20.2 Oxidation Numbers 20.3 Describing Redox Equations 1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > CHEMISTRY & YOU Why does a sparkler have such a bright light? If you have ever seen or held a sparkler, then you know that sparklers give off very bright light. They are like handheld fireworks. 2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Assigning Oxidation Numbers Assigning Oxidation Numbers What is the general rule for assigning oxidation numbers? 3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Assigning Oxidation Numbers An oxidation number is a positive or negative number assigned to an atom to indicate its degree of oxidation or reduction. 4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Assigning Oxidation Numbers As a general rule, a bonded atom’s oxidation number is the charge that it would have if the electrons in the bond were assigned to the atom of the more electronegative element. 5 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Assigning Oxidation Numbers In binary ionic compounds, the oxidation numbers of the atoms equal their ionic charges. • The compound sodium chloride is composed of sodium ions (Na1+) and chloride ions (Cl1–). • The oxidation number of sodium is +1. • That of chlorine is –1. – Notice that the sign is put before the oxidation number. 6 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Assigning Oxidation Numbers Because water is a molecular compound, no ionic charges are associated with its atoms. • Oxygen is reduced in the formation of water. • Oxygen is more electronegative than hydrogen. • The two shared electrons in the H–O bond are shifted toward oxygen and away from hydrogen. – The oxidation number of oxygen is –2. – The oxidation number of each hydrogen is +1. 7 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Assigning Oxidation Numbers Oxidation numbers are often written above the chemical symbols in a formula. +1 –2 H 2O 8 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Assigning Oxidation Numbers Rules for Assigning Oxidation Numbers 9 1. The oxidation number of a monatomic ion is equal in magnitude and sign to its ionic charge. For example, the oxidation number of the bromide ion (Br1–) is –1; that of the Fe3+ ion is +3. 2. The oxidation number of hydrogen in a compound is +1, except in metal hydrides, such as NaH, where it is –1. 3. The oxidation number of oxygen in a compound is –2, except in peroxides, such as H2O2, where it is –1, and in compounds with the more electronegative fluorine, where it is positive. 4. The oxidation number of an atom in uncombined (elemental) form is 0. For example, the oxidation number of the potassium atoms in potassium metal (K) or of the nitrogen atoms in nitrogen gas (N2) is 0. 5. For any neutral compound, the sum of the oxidation numbers of the atoms in the compound must equal 0. 6. For a polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.2 Assigning Oxidation Numbers to Atoms What is the oxidation number of each kind of atom in the following ions and compounds? 10 a. SO2 c. Na2SO4 b. CO32– d. (NH4)2S Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.2 1 Analyze Identify the relevant concepts. Use the set of rules you just learned to assign and calculate oxidation numbers. 11 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.2 2 Solve Apply concepts to this situation. a. There are two oxygen atoms, and the oxidation number of each oxygen is –2 (Rule 3). The sum of the oxidation numbers for the neutral compound must be 0 (Rule 5). Therefore, the oxidation number of sulfur is +4, because +4 + (2 × (–2)) = 0. +4 –2 SO2 12 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.2 2 Solve Apply concepts to this situation. b. The oxidation number of each oxygen is –2 (Rule 3). ? –2 CO32– 13 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.2 2 Solve Apply concepts to this situation. b. The sum of the oxidation numbers of the carbon and oxygen atoms must equal the ionic charge, –2 (Rule 6). The oxidation number of carbon must be +4, because +4 + (3 × (–2)) = –2. +4 –2 CO32– 14 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.2 2 Solve Apply concepts to this situation. c. The oxidation number of each sodium ion, Na+, is the same as its ionic charge, +1 (Rule 1). The oxidation number of oxygen is –2 (Rule 3). +1 ? –2 Na2SO4 15 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.2 2 Solve Apply concepts to this situation. c. For the sum of the oxidation numbers in the compound to be 0 (Rule 5), the oxidation number of sulfur must be +6, because (2 × (+1)) + (+6) + (4 × (–2)) = 0. +1 +6 –2 Na2SO4 16 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.2 2 Solve Apply concepts to this situation. d. Ammonium ions, NH4+, have an ionic charge of +1, so the sum of the oxidation numbers of the atoms in the ammonium ion must be +1. The oxidation number of hydrogen is +1 in this ion. So, the oxidation number of nitrogen must be –3. ? +1 NH4+ ? + 4(+1) = +1 –3 + 4(+1) = +1 17 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.2 2 Solve Apply concepts to this situation. d. Two ammonium ions have a total charge of +2. Since the compound (NH4)2S is neutral, sulfur must have a balancing oxidation number of –2. –3 +1 –2 (NH4)2S 18 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.2 3 Evaluate Do the results make sense? • The results are consistent with the rules for determining oxidation numbers. • Also, addition of the oxidation numbers correctly gives the final overall charge for the ion and the three neutral compounds. 19 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Chromium in its uncombined state is a dull silvery color. Orange potassium dichromate (K2Cr2O7) and purple chromium(III) potassium sulfate (CrK(SO4)2·12H2O) are both compounds of chromium. What is the oxidation number of chromium in each compound? 20 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Chromium in its uncombined state is a dull silvery color. Orange potassium dichromate (K2Cr2O7) and purple chromium(III) potassium sulfate (CrK(SO4)2·12H2O) are both compounds of chromium. What is the oxidation number of chromium in each compound? +1 +6 –2 K2Cr2O7 21 +3 +1 +6 –2 CrK(SO4)2·12H2O Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Oxidation-Number Changes in 20.2 Oxidation Numbers > Chemical Reactions Oxidation-Number Changes in Chemical Reactions How are oxidation and reduction defined in terms of a change in oxidation number? 22 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Oxidation-Number Changes in 20.2 Oxidation Numbers > Chemical Reactions The figure below shows what happens when copper wire is placed in a solution of silver nitrate. • The oxidation number of silver decreases from +1 to 0 as each silver ion (Ag1+) gains an electron and is reduced to silver metal (Ag0). 23 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Oxidation-Number Changes in 20.2 Oxidation Numbers > Chemical Reactions The figure below shows what happens when copper wire is placed in a solution of silver nitrate. • Copper’s oxidation number increases from 0 to +2 as each atom of copper metal (Cu0) loses two electrons and is oxidized to copper(II) ion (Cu2+). 24 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Oxidation-Number Changes in 20.2 Oxidation Numbers > Chemical Reactions The figure below shows what happens when copper wire is placed in a solution of silver nitrate. +1 +5 –2 0 +2 +5 –2 0 2AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2Ag(s) 25 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Oxidation-Number Changes in 20.2 Oxidation Numbers > Chemical Reactions This figure illustrates a redox reaction that shows what occurs when a shiny iron nail is dipped into a solution of copper(II) sulfate. • The iron reduces Cu2+ ions in solution and is simultaneously oxidized to Fe2+. • The iron becomes coated with metallic copper. 26 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Oxidation-Number Changes in 20.2 Oxidation Numbers > Chemical Reactions You can define oxidation and reduction in terms of a change in oxidation number. An increase in the oxidation number of an atom or ion indicates oxidation. A decrease in the oxidation number of an atom or ion indicates reduction. 27 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > CHEMISTRY & YOU What happens to the oxidation numbers of metals as they burn in a sparkler? 28 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > CHEMISTRY & YOU What happens to the oxidation numbers of metals as they burn in a sparkler? As the metals burn, they gain oxygen or undergo oxidation. A substance that undergoes oxidation has an increase in oxidation number. Therefore, as metals burn, their oxidation numbers increase. 29 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.3 Identifying Oxidized and Reduced Atoms Use changes in oxidation number to identify which atoms are oxidized and which are reduced in the following reactions. Also identify the oxidizing agent and the reducing agent. a. Cl2(g) + 2HBr(aq) → 2HCl(aq) + Br2(l) b. C(s) + O2(g) → CO2(g) 30 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.3 1 Analyze Identify the relevant concepts. • An increase in oxidation number indicates oxidation. • A decrease in oxidation number indicates reduction. • The substance that is oxidized in a redox reaction is the reducing agent. • The substance that is reduced is the oxidizing agent. 31 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.3 2 Solve Apply concepts to this situation. a. Use the rules to assign oxidation numbers to each atom in the equation. 0 +1 –1 +1 –1 0 Cl2(g) + 2HBr(aq) → 2HCl(aq) + Br2(l) The oxidation number of each chlorine in Cl2 is 0 because of Rule 4. 32 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.3 2 Solve Apply concepts to this situation. a. Then use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced. 0 +1 –1 +1 –1 0 Cl2(g) + 2HBr(aq) → 2HCl(aq) + Br2(l) • The element chlorine is reduced because its oxidation number decreases (0 to –1). • The bromide ion from HBr(aq) is oxidized because its oxidation number increases (–1 to 0). 33 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.3 2 Solve Apply concepts to this situation. a. Finally, identify the oxidizing and reducing agents. 0 +1 –1 +1 –1 0 Cl2(g) + 2HBr(aq) → 2HCl(aq) + Br2(l) • Chlorine is reduced, so Cl2 is the oxidizing agent. • The bromide ion from HBr(aq) is oxidized, so Br– is the reducing agent. 34 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.3 2 Solve Apply concepts to this situation. b. Use the rules to assign oxidation numbers to each atom in the equation. 0 0 +4 –2 C(s) + O2(g) → CO2(g) 35 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.3 2 Solve Apply concepts to this situation. b. Then use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced. 0 0 +4 –2 C(s) + O2(g) → CO2(g) • The element carbon is oxidized because its oxidation number increases (0 to +4). • The element oxygen is reduced because its oxidation number decreases (0 to –2). 36 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.3 2 Solve Apply concepts to this situation. b. Finally, identify the oxidizing and reducing agents. 0 0 +4 –2 C(s) + O2(g) → CO2(g) • Carbon is oxidized, so C is the reducing agent. • Oxygen is reduced, so O2 is the oxidizing agent. 37 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.3 3 Evaluate Do the results make sense? • It makes sense that what is oxidized in a chemical reaction is the reducing agent because it loses electrons—it becomes the agent by which the atom that is reduced gains electrons. • Conversely, it makes sense that what is reduced in a chemical reaction is the oxidizing agent because it gains electrons—it is the agent by which the atom that is oxidized loses electrons. 38 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.4 Identifying Oxidized and Reduced Atoms Use changes in oxidation number to identify which atoms are oxidized and which are reduced in the following reaction. Also identify the oxidizing agent and the reducing agent. Zn(s) + 2MnO2(s) + 2NH4Cl(aq) → ZnCl2(aq) + Mn2O3(s) + 2NH3(g) + H2O(l) 39 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.4 1 Analyze Identify the relevant concepts. • An increase in oxidation number indicates oxidation. • A decrease in oxidation number indicates reduction. • The substance that is oxidized in a redox reaction is the reducing agent. • The substance that is reduced is the oxidizing agent. 40 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.4 2 Solve Apply concepts to this situation. Use the rules to assign oxidation numbers to each atom in the equation. 0 +4 –2 –3 +1 –1 +2 –1 +3 –2 –3 +1 +1 –2 Zn(s) + 2MnO2(s) + 2NH4Cl(aq) → ZnCl2(aq) + Mn2O3(s) + 2NH3(g) + H2O(l) 41 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.4 2 Solve Apply concepts to this situation. Then use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced. 0 +4 –2 –3 +1 –1 +2 –1 +3 –2 –3 +1 +1 –2 Zn(s) + 2MnO2(s) + 2NH4Cl(aq) → ZnCl2(aq) + Mn2O3(s) + 2NH3(g) + H2O(l) • The element zinc is oxidized because its oxidation number increases (0 to +2). • The manganese ion is reduced because its oxidation number decreases (+4 to +3). 42 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Sample Problem 20.4 2 Solve Apply concepts to this situation. Finally, identify the oxidizing and reducing agents. 0 +4 –2 –3 +1 –1 +2 –1 +3 –2 –3 +1 +1 –2 Zn(s) + 2MnO2(s) + 2NH4Cl(aq) → ZnCl2(aq) + Mn2O3(s) + 2NH3(g) + H2O(l) • Zinc is oxidized, so Zn is the reducing agent. • Manganese (in MnO2) is reduced, so Mn4+ is the oxidizing agent. 43 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Use changes in oxidation number to identify which atoms are oxidized and which are reduced in the following reaction. 2HNO3(aq) + 3H2S(g) → 2NO(g) + 4H2O(l) + 3S(s) 44 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Use changes in oxidation number to identify which atoms are oxidized and which are reduced in the following reaction. 2HNO3(aq) + 3H2S(g) → 2NO(g) + 4H2O(l) + 3S(s) +1 +5 –2 +1 –2 +2 –2 +1 –2 0 2HNO3(aq) + 3H2S(g) → 2NO(g) + 4H2O(l) + 3S(s) Sulfur is oxidized because its oxidation number increases (–2 to 0). Nitrogen is reduced because its oxidation number decreases (+5 to +2). 45 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Key Concepts As a general rule, a bonded atom’s oxidation number is the charge that it would have if the electrons in the bond were assigned to the atom of the more electronegative element. An increase in the oxidation number of an atom or ion indicates oxidation. A decrease in the oxidation number of an atom or ion indicates reduction. 46 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > Glossary Term oxidation number: a positive or negative number assigned to an atom to indicate its degree of oxidation or reduction; the oxidation number of an uncombined element is zero 47 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > BIG IDEA Reactions Redox reactions are identified by changes in oxidation number. 48 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20.2 Oxidation Numbers > END OF 20.2 49 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.