Outline 6.1 Chemical Equations 6.2 Balancing Chemical Equations 6.3 Avogadro s Number and the Mole 6.4 Gram Mole Conversions 6.5 Mole Relationships and Chemical Equations 6.6 Mass Relationships and Chemical Equations 6.7 Percent Yield 6.8 Classes of Chemical Reactions 6.9 Precipitation Reactions and Solubility Guidelines 6.10 Acids, Bases, and Neutralization Reactions 6.11 Redox Reactions 6.12 Recognizing Redox Reactions 6.13 Net Ionic Equations Prentice Hall 2007 Chapter Six 1 6.1 Chemical Equations Chemical equation: An expression in which symbols are used to represent a chemical reaction. Reactant: A substance that undergoes change in a chemical reaction and is written on the left side of the reaction arrow in a chemical equation. Product: A substance that is formed in a chemical reaction and is written on the right side of the reaction arrow in a chemical equation. Prentice Hall 2007 Chapter Six 2 The numbers and kinds of atoms must be the same on both sides of the reaction arrow. Numbers in front of formulas are called coefficients; they multiply all the atoms in a formula. The symbol 2 NaHCO 3 indicates two units of sodium bicarbonate, which contains 2 Na, 2 H, 2 C, and 6 O. Substances involved in chemical reactions may be solids, liquids, gases, or they may be in solution. This information is added to an equation by placing the appropriate symbols after the formulas: Solid=(s) Liquid=(l) Gas=(g) Aqueous solution=(aq) 6.2 Balancing Chemical Equations Balancing chemical equations can be done using four basic steps: STEP 1: Write an unbalanced equation, using the correct formulas for all reactants and products. STEP 2: Add appropriate coefficients to balance the numbers of atoms of each element. Prentice Hall 2007 Chapter Six 3 Prentice Hall 2007 Chapter Six 4 A polyatomic ion appearing on both sides of an equation can be treated as a single unit. STEP 3: Check the equation to make sure the numbers and kinds of atoms on both sides of the equation are the same. STEP 4: Make sure the coefficients are reduced to their lowest whole-number values. The equation: 2 H 2 SO 4 + 4 NaOH 2 Na 2 SO 4 + 4 H 2 O is balanced, but can be simplified by dividing all coefficients by 2: H 2 SO 4 + 2 NaOH Na 2 SO 4 + 2 H 2 O Hint: If an equation contains a pure element as a product or reactant it helps to assign that element s coefficient last. Prentice Hall 2007 Chapter Six 5 Prentice Hall 2007 Chapter Six 6 1
Symbols Used in Equations Chemical Equations are Balanced Symbols used in chemical equations show the states of the reactants. the states of the products. the reaction conditions. -TABLE 5.2 In a balanced chemical reaction atoms are not gained or lost. the number of reactant atoms is equal to the number of product atoms. 7 8 A Balanced Chemical Equation In a balanced chemical equation, there must be the same number of each type of atom on the reactant side and on the product side of a balanced equation. numbers called coefficients are used in front of one or more formulas. Determine if each equation is balanced or not. A. Na(s) + N 2 (g) Na 3 N(s) B. C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) Al + S Al 2 S 3 Not Balanced 2Al + 3S Al 2 S 3 Balanced 2 Al = 2 Al 3 S = 3 S 9 10 -Guide to Balancing a Chemical Equation Determine if each equation is balanced or not. A. Na(s) + N 2 (g) Na 3 N(s) No. 2 N on reactant side, 1 N on product side. 1 Na on reactant side, 3 Na on product side. B. C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) Yes. 2 C = 2 C 6 H = 6 H 1 O = 1 O -Copyright 2005 by Pearson Education, Inc. -Publishing as Benjamin Cummings 11 12 2
Steps in Balancing an Equation Balancing Chemical Equations To balance the following equation, Fe 3 O 4 (s) + H 2 (g) Fe(s) + H 2 O(l) 1. Write the equation with the correct formulas. NH 3 (g) + O 2 (g) NO(g) + H 2 O(g) work on one element at a time. use only coefficients in front of formulas. do not change any subscripts. Fe: Fe 3 O 4 (s) + H 2 (g) 3Fe(s) + H 2 O(l) 2. Determine if the equation is balanced. No, not all atoms are balanced. 3. Balance with coefficients in front of formulas. 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) O: Fe 3 O 4 (s) + H 2 (g) 3Fe(s) + 4H 2 O(l) 4. Check that atoms of each element are equal in reactants and products. 4 N (4 x 1 N) = 4 N (4 x 1 N) 12 H (4 x 3 H) = 12 H (6 x 2 H) 10 O (5 x 2 O) = 10 O (4 O + 6 O) H: Fe 3 O 4 (s) + 4H 2 (g) 3Fe(s) + 4H 2 O(l) 13 14 Check the balance of atoms in the following. Fe 3 O 4 (s) + 4H 2 (g) 3Fe(s) + 4H 2 O(l) A. Number of H atoms in products. 1) 2 2) 4 3) 8 B. Number of O atoms in reactants. 1) 2 2) 4 3) 8 C. Number of Fe atoms in reactants. 1) 1 2) 3 3) 4 Balance each equation and list the coefficients in the balanced equation going from reactants to products: A. Mg(s) + N 2 (g) Mg 3 N 2 (s) 1) 1, 3, 2 2) 3, 1, 2 3) 3, 1, 1 B. Al(s) + Cl 2 (g) AlCl 3 (s) 1) 3, 3, 22) 1, 3, 1 3) 2, 3, 2 15 16 A. 3) 3, 1, 1 3Mg(s) + 1N 2 (g) 1Mg 3 N 2 (s) B. 3) 2, 3, 2 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 6.3 Avogadro s Number and the Mole Molecular weight: The sum of atomic weights of all atoms in a molecule. Formula weight: The sum of atomic weights of all atoms in one formula unit of any compound. Mole: One mole of any substance is the amount whose mass in grams (molar mass) is numerically equal to its molecular or formula weight. Avogadro s number: The number of molecules or formula units in a mole. N A = 6.022 x 10 23 17 Prentice Hall 2007 Chapter Six 18 3
6.4 Gram Mole Conversions Molar mass = Mass of 1 mole of a substance. = Mass of 6.022 x 10 23 molecules of a substance. = Molecular (formula) weight of substance in grams. Molar mass serves as a conversion factor between numbers of moles and mass. If you know how many moles you have, you can calculate their mass; if you know the mass of a sample, you can calculate the number of moles. The molar mass of water is 18.0 g. The conversion factor between moles of water and mass of water is 18.0 g/mol and the conversion factor between mass of water and moles of water is 1 mol/18.0 g: Prentice Hall 2007 Chapter Six 19 Prentice Hall 2007 Chapter Six 20 Collection Terms A Mole of Atoms A collection term states a specific number of items. 1 dozen donuts 1 ream of paper = 12 donuts = 500 sheets 1 case = 24 cans -Copyright 2005 by Pearson Education, Inc. -Publishing as Benjamin Cummings A mole is a collection that contains the same number of particles as there are carbon atoms in 12.0 g of carbon. 6.02 x 10 23 atoms of an element (Avogadro s number). 1 mole element Number of Atoms 1 mole C = 6.02 x 10 23 C atoms 1 mole Na = 6.02 x 10 23 Na atoms 1 mole Au = 6.02 x 10 23 Au atoms 21 22 A mole A Mole of a Compound of a covalent compound has Avogadro s number of molecules. 1 mole CO 2 = 6.02 x 10 23 CO 2 molecules 1 mole H 2 O = 6.02 x 10 23 H 2 O molecules of an ionic compound contains Avogadro s number of formula units. 1 mole NaCl = 6.02 x 10 23 NaCl formula units 1 mole K 2 SO 4 = 6.02 x 10 23 K 2 SO 4 formula units Avogadro s Number Avogadro s number 6.02 x 10 23 can be written as an equality and two conversion factors. Equality: 1 mole = 6.02 x 10 23 particles Conversion Factors: 6.02 x 10 23 particles and 1 mole 1 mole 6.02 x 10 23 particles 23 24 4
Using Avogadro s Number Using Avogadro s Number Avogadro s number is used to convert moles of a substance to particles. Avogadro s number is used to convert particles of a substance to moles. How many Cu atoms are in 0.50 mole Cu? 0.50 mole Cu x 6.02 x 10 23 Cu atoms 1 mole Cu = 3.0 x 10 23 Cu atoms How many moles of CO 2 are in 2.50 x 10 24 molecules CO 2? 2.50 x 10 24 molecules CO 2 x 1 mole CO 2 6.02 x 10 23 molecules CO 2 -Copyright 2005 by Pearson Education, Inc. -Publishing as Benjamin Cummings = 4.15 mole CO 2 25 26 1. The number of atoms in 2.0 mole Al is A. 2.0 Al atoms. B. 3.0 x 10 23 Al atoms. C. 1.2 x 10 24 Al atoms. C. 1.2 x 10 24 Al atoms 2.0 mole Al x 6.02 x 10 23 Al atoms 1 mole Al 2. The number of moles of S in 1.8 x 10 24 atoms S is A. 1.0 mole S atoms. B. 3.0 mole S atoms. C. 1.1 x 10 48 mole S atoms. B. 3.0 mole S atoms 1.8 x 10 24 S atoms x 1 mole S 6.02 x 10 23 S atoms 27 28 6.5 Mole Relationships and Chemical Equations The coefficients in a balanced chemical equation tell how many molecules, and thus how many moles, of each reactant are needed and how many molecules, and thus moles, of each product are formed. See the example below: The coefficients can be put in the form of mole ratios, which act as conversion factors when setting up factor-label calculations. In the ammonia synthesis, the mole ratio of H 2 to N 2 is 3:1, the mole ratio of H 2 to NH 3 is 3:2, and the mole ratio of N 2 to NH 3 is 1:2 leading to the following conversion factors: (3 mol H 2 )/(1 mol N 2 ) (3 mol H 2 )/(2 mol NH 3 ) (1 mol N 2 )/(2 mol NH 3 ) Prentice Hall 2007 Chapter Six 29 Prentice Hall 2007 Chapter Six 30 5
Subscripts and Moles Factors from Subscripts The subscripts in a formula show the relationship of atoms in the formula. the moles of each element in 1 mole of compound. The subscripts are used to write conversion factors for moles of each element in 1 mole compound. For aspirin C 9 H 8 O 4, the following factors can be written: Glucose C 6 H 12 O 6 In 1 molecule: 6 atoms C 12 atoms H 6 atoms O In 1 mole: 6 mole C 12 mole H 6 mole O 31 9 mole C 8 mole H 4 mole O 1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4 and 1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4 9 mole C 8 mole H 4 mole O 32 A. How many mole O are in 0.150 mole aspirin C 9 H 8 O 4? A. How many mole O are in 0.150 mole aspirin C 9 H 8 O 4? 0.150 mole C 9 H 8 O 4 x 4 mole O = 0.600 mole O B. How many O atoms are in 0.150 mole aspirin C 9 H 8 O 4? 33 1 mole C 9 H 8 O 4 subscript factor B. How many O atoms are in 0.150 mole aspirin C 9 H 8 O 4? 0.150 mole C 9 H 8 O 4 x 4 mole O x 6.02 x 10 23 O atoms = 3.61 x 10 23 O atoms 1 mole C 9 H 8 O 4 1 mole O subscript Avogadro s factor Number 34 6.6 Mass Relationships and Chemical Equations Mole to mole conversions are carried out using mole ratios as conversion factors. The molar mass Molar Mass is the mass of one mole of an element or compound. is the atomic mass expressed in grams. -Copyright 2005 by Pearson Education, Inc. -Publishing as Benjamin Cummings Prentice Hall 2007 Chapter Six 35 36 6
Molar Mass from Periodic Table Molar mass is the atomic mass expressed in grams. Give the molar mass for each (to the tenths decimal place). A. 1 mole K atoms = B. 1 mol Sn atoms = - 1 mole Ag 1 mole C 1 mole S - = 107.9 g = 12.01 g = 32.07 g 37 38 Molar Mass of K 3 PO 4 Give the molar mass for each (to the tenths decimal place). A. 1 mole K atoms = 39.1 g B. 1 mole Sn atoms = 118.7 g Calculate the molar mass of K 3 PO 4. Element Number of Moles Atomic Mass Total Mass in K 3 PO 4 K 3 39.1 g/mole 117.3 g P 1 31.0 g/mole 31.0 g O 4 16.0 g/mole 64.0 g K 3 PO 4 212.3 g 39 40 Some One-mole Quantities -One-Mole Quantities What is the molar mass of each of the following? A. K 2 O B. Al(OH) 3 32.1 g 55.9 g 58.5 g 294.2 g 342.2 g 41 42 7
Molar Mass Factors A. K 2 O 94.2 g/mole 2 mole K (39.1 g/mole) + 1 mole O (16.0 g/mole) 78.2 g + 16.0 g = 94.2 g B. Al(OH) 3 78.0 g/mole 1 mole Al (27.0 g/mole) + 3 mol O (16.0 g/mole) + 3 mol H (1.01 g/mole) 27.0 g + 48.0 g + 3.03 g = 78.0 g 43 Molar mass conversion factors are written from molar mass. relate grams and moles of an element or compound. Example: Write molar mass factors for methane, CH 4, used in gas stoves and gas heaters. Molar mass: 1 mol CH 4 = 16.0 g Conversion factors: 16.0 g CH 4 and 1 mole CH 4 1 mole CH 4 16.0 g CH 4 44 Calculations Using Molar Mass Molar mass factors are used to convert between the grams of a substance and the number of moles. Allyl sulfide C 6 H 10 S is a compound that has the odor of garlic. How many moles of C 6 H 10 S are in 225 g? - Grams -Molar mass factor -Moles 45 46 Grams, Moles, and Particles Calculate the molar mass of C 6 H 10 S. (6 x 12.0) + (10 x 1.01) + (1 x 32.1) = 114.2 g/mole Set up the calculation using a mole factor. 225 g C 6 H 10 S x 1 mole C 6 H 10 S = 1.97 mole C 6 H 10 S 114.2 g C 6 H 10 S molar mass factor (inverted) A molar mass factor and Avogadro s number convert grams to particles. molar mass Avogadro s number g mole particles particles to grams. Avogadro s molar mass number particles mole g 47 48 8
How many H 2 O molecules are in 24.0 g H 2 O? How many H 2 O molecules are in 24.0 g H 2 O? 1) 4.52 x 10 23 2) 1.44 x 10 25 3) 8.03 x 10 23 3) 8.02 x 10 23 24.0 g H 2 O x 1 mole H 2 O x 6.02 x 10 23 H 2 O molecules 18.0 g H 2 O 1 mole H 2 O = 8.03 x 10 23 H 2 O molecules 49 50 Mole to mass and mass to mole conversions are carried out using molar mass as a conversion factor. Mass to mass conversions are frequently needed, but cannot be carried out directly. Overall, there are four steps for determining mass relationships among reactants and products. Prentice Hall 2007 Chapter Six 51 Mass to mass conversions: STEP 1: Write the balanced chemical equation. STEP 2: Choose molar masses and mole ratios to convert known information into needed information. STEP 3: Set up the factorlabel expression, and calculate the answer. STEP 4: Estimate or check the answer using a ballpark solution. Prentice Hall 2007 Chapter Six 52 Reading Equations In Moles Consider the following equation: 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) This equation can be read in moles by placing the word moles between each coefficient and formula. 4 moles Fe + 3 moles O 2 2 moles Fe 2 O 3 Writing Mole-Mole Factors A mole-mole factor is a ratio of the moles for any two substances in an equation. 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Fe and O 2 4 moles Fe and 3 moles O 2 3 moles O 2 4 moles Fe Fe and Fe 2 O 3 4 moles Fe and 2 moles Fe 2 O 3 2 moles Fe 2 O 3 4 moles Fe O 2 and Fe 2 O 3 3 moles O 2 and 2 moles Fe 2 O 3 2 moles Fe 2 O 3 3 moles O 2 53 54 9
Consider the following equation: 3 H 2 (g) + N 2 (g) 2 NH 3 (g) A. A mole-mole factor for H 2 and N 2 is 1) 3 moles N 2 2) 1 mole N 2 3) 1 mole N 2 1 mole H 2 3 moles H 2 2 moles H 2 B. A mole-mole factor for NH 3 and H 2 is 1) 1 mole H 2 2) 2 moles NH 3 3) 3 moles N 2 2 moles NH 3 3 moles H 2 2 moles NH 3 3H 2 (g) + N 2 (g) 2NH 3 (g) A. A mole-mole factor for H 2 and N 2 is 2) 1 mole N 2 3 moles H 2 B. A mole-mole factor for NH 3 and H 2 is 2) 2 moles NH 3 3 moles H 2 55 56 Calculations with Mole Factors -Guide to Using Mole Factors How many moles of Fe 2 O 3 can form from 6.0 mole O 2? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Relationship: 3 mole O 2 = 2 mole Fe 2 O 3 Write a mole-mole factor to determine the moles of Fe 2 O 3. 6.0 mole O 2 x 2 mole Fe 2 O 3 = 4.0 mole Fe 2 O 3 3 mole O 2 57 58 How many moles of Fe are needed for the reaction of 12.0 moles O 2? 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) 1) 3.00 moles Fe 2) 9.00 moles Fe 3) 16.0 moles Fe 3) 16.0 moles Fe 12.0 moles O 2 x 4 moles Fe = 16.0 moles Fe 3 moles O 2 59 60 10
Steps in Finding the Moles and Masses in a Chemical Reaction Moles to Grams Suppose we want to determine the mass (g) of NH 3 that can form from 2.50 moles N 2. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) The plan needed would be moles N 2 moles NH 3 grams NH 3 61 The factors needed would be: mole factor NH 3 /N 2 and the molar mass NH 3 62 Moles to Grams The setup for the solution would be: 2.50 mole N 2 x 2 moles NH 3 x 17.0 g NH 3 1 mole N 2 1 mole NH 3 given mole-mole factor molar mass = 85.0 g NH 3 How many grams of O 2 are needed to produce 0.400 mole Fe 2 O 3 in the following reaction? 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) 1) 38.4 g O 2 2) 19.2 g O 2 3) 1.90 g O 2 63 64 2) 19.2 g O 2 0.400 mole Fe 2 O 3 x 3 mole O 2 x 32.0 g O 2 = 19.2 g O 2 2 mole Fe 2 O 3 1 mole O 2 mole factor molar mass 6.7 Percent Yield The amount of product actually formed in a chemical reaction is somewhat less than the amount predicted by theory. Unwanted side reactions and loss of product during handling prevent one from obtaining a perfect conversion of all the reactants to desired products. The amount of product actually obtained in a chemical reaction is usually expressed as a percent yield. 65 Prentice Hall 2007 Chapter Six 66 11
Percent yield is defined as: (Actual yield Theoretical yield) x 100% The actual yield is found by weighing the product obtained. The theoretical yield is found by a mass-to-mass calculation. 6.8 Classes of Chemical Reactions When learning about chemical reactions it is helpful to group the reactions of ionic compounds into three general classes: precipitation reactions, acid base neutralization reactions, and oxidation reduction reactions. Precipitation reactions are processes in which an insoluble solid called a precipitate forms when reactants are combined in aqueous solution. Prentice Hall 2007 Chapter Six 67 Prentice Hall 2007 Chapter Six 68 Acid base neutralization reactions are processes in which H + ions from an acid react with OH - ions from a base to yield water. An ionic compound called a salt is also produced. The salt produced need not be common table salt. Any ionic compound produced in an acid base reaction is called a salt. Oxidation reduction reactions, or redox reactions, are processes in which one or more electrons are transferred between reaction partners (atoms, molecules, or ions). As a result of this transfer, the charges on atoms in the various reactants change. 6.9 Precipitation Reactions and Solubility Guidelines Reaction of aqueous Pb(NO 3 ) 2 with aqueous KI gives a yellow precipitate of PbI 2. To predict whether a precipitation reaction will occur on mixing aqueous solutions of two ionic compounds, you must know the solubility of the potential products. Prentice Hall 2007 Chapter Six 69 Prentice Hall 2007 Chapter Six 70 If a potential product does not contain at least one of the ions listed below, it is probably not soluble and will precipitate from solution when formed. 6.10 Acids, Bases, and Neutralization Reactions When acids and bases are mixed together in correct proportion, acidic and basic properties disappear. A neutralization reaction produces water and a salt. HA (aq) + MOH (aq) H 2 O (l) + MA (aq) acid + base water + salt The reaction of hydrochloric acid with potassium hydroxide to produce potassium chloride is an example: HCl (aq) + KOH (aq) H 2 O (l) + KCl (aq) Prentice Hall 2007 Chapter Six 71 Prentice Hall 2007 Chapter Six 72 12
6.11 Redox Reactions Oxidation reduction (redox) reaction: A reaction in which electrons transfer from one atom to another. Oxidation: Loss of one or more electrons by an atom. Reduction: Gain of one or more electrons by an atom. Oxidation and reduction always occur together. A substance that is oxidized gives up an electron, causes reduction, and is called a reducing agent. A substance that is reduced gains an electron, causes oxidation, and is called an oxidizing agent. The charge on the reducing agent increases during the reaction, and the charge on the oxidizing agent decreases. Prentice Hall 2007 Chapter Six 73 Prentice Hall 2007 Chapter Six 74 Reducing agent: Loses one or more electrons Causes reduction Undergoes oxidation Becomes more positive (or less negative) Oxidizing agent: Gains one or more electrons Causes oxidation Undergoes reduction Becomes more negative (or less positive) 6.12 Recognizing Redox Reactions One can determine whether atoms are oxidized or reduced in a reaction by keeping track of changes in electron sharing by the atoms. Each atom in a substance is assigned a value called an oxidation number or oxidation state. The oxidation number indicates whether the atom is neutral, electron-rich, or electron-poor. By comparing the oxidation state of an atom before and after reaction, we can tell whether the atom has gained or lost electrons. Prentice Hall 2007 Chapter Six 75 Prentice Hall 2007 Chapter Six 76 Rules for assigning oxidation numbers: An atom in its elemental state has an oxidation number of zero. In a molecular compound, an atom usually has the same oxidation number it would have if it were a monatomic ion. Examples: H often has an oxidation number of +1, oxygen often has an oxidation number of -2, halogens often have an oxidation number of -1. A monatomic ion has an oxidation number equal to its charge. Prentice Hall 2007 Chapter Six 77 Prentice Hall 2007 Chapter Six 78 13
For compounds with more than one nonmetal element, such as SO 2, NO, and CO 2, the more electronegative element oxygen in these examples has its preferred negative oxidation number. The less electronegative element is assigned a positive oxidation number so that the sum of the oxidation numbers in a neutral compound is 0. 6.13 Net Ionic Equations Ionic equation: An equation in which ions are explicitly shown. Spectator ion: An ion that appears unchanged on both sides of a reaction arrow. Net ionic equation: An equation that does not include spectator ions. Prentice Hall 2007 Chapter Six 79 Prentice Hall 2007 Chapter Six 80 Chapter Summary Chemical equations must be balanced; the numbers and kinds of atoms must be the same in both the reactants and the products. To balance an equation, coefficients are placed before formulas but the formulas themselves cannot be changed. A mole refers to Avogadro s number of formula units of a substance. One mole of any substance has a mass equal to its formula weight in grams. Molar masses act as conversion factors between numbers of molecules and masses in grams. Prentice Hall 2007 Chapter Six 81 Chapter Summary Cont. The coefficients in a balanced chemical equation represent the numbers of moles of reactants and products in a reaction. Mole ratios relate amounts of reactants and/or products. Using molar masses and mole ratios in factor-label calculations relates unknown masses to known masses or molar amounts. The yield is the amount of product obtained. The percent yield is the amount of product obtained divided by the amount theoretically possible and multiplied by 100%. Prentice Hall 2007 Chapter Six 82 Chapter Summary Cont. Precipitation reactions are processes in which an insoluble solid called a precipitate is formed. In acid base neutralization reactions an acid reacts with a base to yield water plus a salt. Oxidation reduction (redox) reactions are processes in which one or more electrons are transferred between reaction partners. Oxidation is the loss of electrons by an atom, and reduction is the gain of electrons by an atom. Oxidation numbers are assigned to provide a measure of whether an atom is neutral, electron-rich, or electron-poor. Prentice Hall 2007 Chapter Six 83 14