1 207 Oxidation- reduction (redox) reactions Chapter 12: Oxidation and Reduction. At different times, oxidation and reduction (redox) have had different, but complimentary, definitions. Compare the following definitions: Oxidation is: Gaining oxygen Losing hydrogen Losing electrons (+ charge increases) Reduction is: Losing oxygen Gaining hydrogen Gaining electrons (+ charge decreases) Oxidation and reduction are opposite reactions. They are also paired reactions: in order for one to occur, the other must also occur simultaneously. While the first two definitions of oxidation- reduction are correct, the most useful definition is the third - involving the gain or loss of electrons. I will use this particular definition for the rest of our discussion. How do we know whether or not an element has gained or lost electrons? To determine which elements have been oxidized or reduced, we look at changes in the oxidation number of the element. English scientist Michael Faraday ( ), one of the greatest pioneers in electrochemistry and electromagnetism of the 19 th century, developed the system of oxidation numbers to follow these kinds of reactions. The oxidation number describes the oxidation state of the element in a compound, and these numbers are assigned following a relatively simple set of rules. Rules for assigning oxidation numbers. 1. The oxidation number of an element in its elemental form is 0 (zero). 2. The oxidation number of a simple ion is equal to the charge on the ion. Both the size and the polarity of the charge are part of the oxidation number: an ion can have a +2 oxidation number or a - 2 oxidation number. The + and - signs are just as important as the number! 3. The oxidation numbers of group 1A and 2A (group 1 and group 2) elements are +1 and +2 respectively. 4. In compounds, the oxidation number of hydrogen is almost always +1. The most common exception occurs when hydrogen combines with metals; in this case the oxidation number of hydrogen is typically 1.
2 In compounds, the oxidation number of oxygen is almost always 2. The most common exception is in peroxides, when the oxidation number is 1. Peroxides are compounds having two oxygen atoms bonded together. For example, hydrogen peroxide is H- O- O- H. In hydrogen peroxide, each oxygen atom has a - 1 oxidation number. When oxygen is bonded to fluorine, as in hypofluorous acid (HOF), the oxidation number of oxygen is 0. Oxygen- fluorine compounds are relatively rare and not too terribly important for our studies. 6. In compounds, the oxidation number of fluorine is always 1. The oxidation number of other halogens (Cl, Br, I) is also 1, except when they are combined with oxygen. The oxidation number of halides (except fluorine) combined with oxygen is typically positive. For example, in ClO -, chlorine s oxidation number is For a complex ion, the sum of the positive and negative oxidation numbers of all elements in the ion equals the charge on the ion. 8. For an electrically neutral compound, the sum of the positive and negative oxidation numbers of all elements in the compound equals zero. Identifying redox reactions. Now that you can assign proper oxidation numbers to the elements in substances, we can use changes in the oxidation numbers to identify oxidation and reduction reactions. Consider the following chemical equation: Zn(s) + 2HCl(aq) Zn +2 (aq) + 2Cl - (aq) + H2(g) The oxidation number for elemental zinc is 0. The oxidation number for zinc ion is +2. The oxidation number for hydrogen in hydrogen chloride is +1. The oxidation number for elemental hydrogen, H2, is 0. The oxidation number for chlorine in hydrogen chloride is 1. The oxidation number for chloride ion is 1. Zinc has lost two electrons, and therefore developed a +2 charge. Zinc has been oxidized its oxidation number has become more positive (from 0 to +2). Hydrogen has gained an electron, and its positive charge has decreased. Hydrogen has been reduced its oxidation number has become less positive (from +1 to 0). Chloride has not changed its oxidation state. It is a spectator ion in this reaction.
3 209 Figure 12.2 may be useful in deciding if an element has been oxidized or reduced. If an elements oxidation number increases (moves towards the right), then the element is oxidized. If an elements oxidation number decreases (moves towards the left), then the element is reduced. NOTE: an element doesn t have to become positive or negative for oxidation or reduction to occur. Instead, the element has to become more positive or more negative. A change from - 3 to - 1 is still oxidation, while a change from +3 to +1 is still reduction. Figure Graphic description of oxidation and reduction. Substances that cause changes in the oxidation state are called oxidizing agents or reducing agents. An oxidizing agent causes oxidation number to occur. How does the oxidizing agent cause oxidation? To increase the positive oxidation number of an element, the oxidizing agent must take one or more electrons from the element. As the element being oxidized loses electron(s), its oxidation number becomes more positive. However, the electrons don t disappear! The oxidizing agent has taken these electrons, and therefore the oxidizing agent becomes more negative it is reduced! In our reaction above, hydrogen in hydrogen chloride takes an electron from the zinc metal. The zinc metal becomes more positive; it is oxidized. By taking the electron from the zinc metal, the hydrogen in hydrogen chloride becomes less positive; it is reduced. Hydrogen chloride is the oxidizing agent, because it contains the element that causes oxidation to occur. Similarly, the zinc metal donates electrons to the hydrogen in hydrogen chloride, causing the oxidation state of hydrogen to decrease from +1 to 0. By
4 210 providing the electrons necessary to reduce the oxidation number of hydrogen, the oxidation number of zinc increases from 0 to +2; zinc is oxidized. While being oxidized, zinc reduced the oxidation number of hydrogen. Therefore, zinc is a reducing agent. Oxidizing agents are substances containing the element(s) that accept electrons, allowing another element(s) to be oxidized. By accepting electrons, the element(s) in the oxidizing agent are reduced. Reducing agents are substances containing the element(s) that donate electrons, allowing another element(s) to be reduced. By donating electrons, the element(s) in the reducing agent are oxidized. As you can see, there is a great deal of symmetry between oxidizing and reducing agents, and between oxidation and reduction. Whenever one process happens, the other process MUST also happen, because we are transferring electrons from one material to another material. Electrons must come from someplace, and must go someplace; electrons cannot simply appear and disappear. There are two general considerations to keep in mind when discussing oxidizing/reducing agents. First, in most cases there will only be one element in each agent that is oxidized or reduced. In our reaction, only hydrogen in hydrogen chloride was reduced the chloride wasn t changed. Second, the oxidizing or reducing agent is the compound containing the element that is oxidized or reduced. The reducing agent is hydrogen chloride, not just hydrogen ion. Similarly, the oxidizing agent is the compound containing the element that is reduced. Oxidizing and reducing agents are ALWAYS reactants, NEVER products! Balancing redox reactions. We can use our knowledge of redox reactions to help us balance chemical reactions. The simplest process to use is the half- reaction method. Consider the following chemical reaction (unbalanced): MnO4 - + C2O4-2 Mn +2 + CO2 First, we assign oxidation numbers to all elements shown in the reaction: MnO4 - + C2O4-2 Mn +2 + CO2 Comparing oxidation numbers, we see that manganese goes from +7 to +2, and has been reduced, while carbon goes from +3 to +4 and has been oxidized. Oxygen doesn t change its oxidation state. The oxidizing agent is permanganate, and the reducing agent is oxalate.
5 211 The half- reaction method involves balancing the oxidation reaction as if it were an isolated reaction. Then the reduction half- reaction is balanced as if it were an isolated reaction. Finally, the two half- reactions are combined. The oxidation half- reaction is: C2O4-2 CO2 The first step is to balance elements, other than oxygen and hydrogen, using the normal methods of balancing chemical equations. We have two carbons in oxalate, so we need a 2 in front of carbon dioxide to balance the carbons: C2O4-2 2CO2 By comparing reactants and products, we see that we have the same number of carbon and oxygen atoms on each side of the reaction. There aren t any other elements present, so we don t need to change any more coefficients. Charge balance requires that the net charge of reactants and products must be equal. We equalize the charges by adding 2 electrons to the products side: C2O4-2 2CO2 + 2e - All elements are identical on both sides, the number of atoms of each element is the same on both sides, and the total charge is the same on both sides: - 2 for oxalate and - 2 for the two electrons. This is a balanced oxidation half- reaction. The reduction half- reaction is: MnO4 - Mn +2 The manganese atoms are balanced, but what are we to do about oxygen? To balance oxygen, we add water molecules. Four oxygen atoms in permanganate require four water molecules as products: MnO4 - Mn H2O Adding water molecules balanced oxygen, but now we have hydrogen atoms. We balance 8 hydrogen atoms from 4 water molecules by adding 8 hydrogen ions to the reaction: 8H + + MnO4 - Mn H2O
6 212 Now we have the same number and kinds of atoms on both sides. However, charge balance requires that the net charge of reactants and products must be equal. We have a net electrical charge of +7 for the reactants, and a net electrical charge of +2 for the products. We add 5 electrons to the reactant side of the equation: 8H + + MnO e - Mn H2O This is a balanced reduction half- reaction. (NOTE: in an oxidation half- reaction, electrons are produced, while in a reduction half- reaction, electrons are consumed.) We are ready to combine the two half- reactions. To combine these reactions, the same number of electrons must be produced as are consumed. When different numbers of electrons are produced and consumed, you need to find the least common multiple. Since 5 electrons are consumed in the reduction reaction, and 2 electrons are produced in the oxidation reaction, the least common multiple is 10. We need to multiply the reduction reaction by 2, and the oxidation reaction by 5: 16H + + 2MnO e - 2Mn H2O 5C2O4-2 10CO2 + 10e - Now that we have the same number of electrons produced and consumed, and we can add the reactions: 16H + + 2MnO e - + 5C2O4-2 2Mn H2O + 10CO2 + 10e - We eliminate substances that are identical on both sides of the equation: 16H + + 2MnO C2O4-2 2Mn H2O + 10CO2 Finally, we check the equation to insure that mass and charge balance have been achieved. Reactants Products 16 H 16 H 2 Mn 2 Mn 28 O 28 O 10 C 10 C +4 +4
7 213 This is an example of balancing a redox reaction in acid solution, by the half- reaction method. We can also balance the reaction in basic solution. The initial work is the same as shown above. Once the reaction is balanced in acid solution, we neutralize hydrogen ion by adding an equal number of hydroxide ions to both sides of the reaction: 16OH H + + 2MnO C2O4-2 2Mn H2O + 10CO2 + 16OH - Hydroxide ion combines with hydrogen ion, forming water molecules, and we eliminate equal numbers of water molecules from both sides of the equation: 16H2O + 2MnO C2O4-2 2Mn H2O + 10CO OH - 8H2O + 2MnO C2O4-2 2Mn CO2 + 16OH - Finally, we check to insure mass and charge balance. Reactants Products 16 H 16 H 2 Mn 2 Mn 36 O 36 O 10 C 10 C
8 214 Vocabulary. The following terms are defined and explained in the text. Make sure that you are familiar with the meanings of the terms as used in chemistry. Understand that you may have been given incomplete or mistaken meanings for these terms in earlier courses. The meanings given in the text are correct and proper. Oxidation Reduction Oxidation number Oxidizing agent Reducing agent Assign correct oxidation numbers to all elements in the following substances. 1. H2S 2. BaO 3. P2O5 4. Cr2O CH2O 6. HBr 7. Fe(NO3)3 8. CO 9. HOCl 10. H2SO4 In the following reactions, identify the oxidizing agent and reducing agent. 1. Fe + Ni(NO3)2 Fe(NO3)2 + Ni 2. Cu + 4HNO3 Cu(NO3)2 + 2H2O + 2NO2 3. CH4 + 2O2 CO2 + 2H2O 4. C2H4 + H2 C2H6 5. H2SO4 + 2NaOH 2Na + + SO H2O
9 215 Use the half- reaction method and balance the following equations under the listed conditions. 1. CN - + MnO4 - CNO - + MnO2 (basic solution) 2. Cr2O7-2 + I - Cr +3 + IO3 - (acidic solution) 3. MnO4 - + CH3OH Mn +2 + HCO2H (acidic solution) 4. As + ClO3 - H3AsO3 + HClO (acidic solution) 5. H2O2 + ClO2 ClO2 - + O2 (basic solution)
10 216 Answers: Assigning oxidation numbers. 1. The oxidation number of hydrogen is +1. Since there are 2 hydrogen atoms in the compound, and no net electrical charge is shown, the oxidation number of sulfur must be 2: +1-2 H2S 2. The oxidation number of oxygen is 2. With no net electrical charge shown, the oxidation number of barium is +2. (Note: barium is a group IIA (2) element, and by rule 3 would have an oxidation number of +2.) 3. The oxidation number of oxygen is 2. There is no net electrical charge shown. Five oxygen atoms produce a total negative oxidation state of 10. There must be a total of +10 oxidation numbers on phosphorous. With two phosphorous atoms, we assign an oxidation number of +5 to each phosphorous. 4. The oxidation number of oxygen is 2. For seven oxygen atoms, this produces a total of 14. There is a - 2 charge shown for the ion; therefore two of the 14 oxidation numbers are assigned to the charge. This leaves 12, which must be balanced by the positive oxidation state of chromium. With two chromium atoms in the ion, we assign each an oxidation number of The oxidation number of oxygen is 2. The oxidation number of hydrogen is +1. Since the total positive oxidation state (2 x +1) exactly balances the total negative oxidation state, the oxidation state of carbon must be zero (0). NOTE: while the oxidation state of elements in their elemental form must be zero, it is also possible for an element to have a zero oxidation state when combined in a compound. 6. The oxidation number of hydrogen is +1, while that of bromine is It is important (and easiest) if you recognize this as an ionic compound composed of the simple ion Fe +3, and the polyatomic anion NO3 -. The oxidation number of the iron is equal to its charge, +3. In the polyatomic anion, the oxidation number of each oxygen atom is 2, for a total of 6. One of these negative oxidation numbers is the charge on the ion, - 1. Therefore,
11 217 there are five negative oxidation numbers to be balanced by the positive oxidation number of nitrogen, which must have a value of The oxidation state of oxygen is 2, therefore carbon must be The oxidation state of hydrogen is +1. The oxidation state of oxygen is 2, therefore the oxidation state of chlorine must be The oxidation state of hydrogen is +1. The oxidation state of oxygen is 2. Four oxygen atoms result in a total of 8, while two hydrogen atoms produce a total of +2. With no net charge shown, the oxidation state of sulfur must be +6. Identifying oxidizing and reducing agents. 1. Elemental iron has an oxidation number of 0. The oxidation number of iron in iron(ii) nitrate is +2. The oxidation number of nickel in nickel(ii) nitrate is +2. The oxidation number of elemental nickel is 0. The oxidation numbers of nitrogen and oxygen don t change; nitrogen is +5 and oxygen is - 2. Therefore, iron is oxidized (it is the reducing agent), while nickel ion is reduced (nickel(ii) nitrate is the oxidizing agent). 2. Elemental copper has an oxidation number of 0. The copper in copper(ii) nitrate has an oxidation number of +2. The oxidation number of hydrogen in all compounds shown is +1 (it did not change). The oxidation number of oxygen in all compounds shown is 2 (it also has not changed). The oxidation number of nitrogen in the nitric acid is +5. The oxidation number of nitrogen in the copper(ii) nitrate is also +5. However, the oxidation number of nitrogen in nitrogen dioxide is +4. Therefore, copper has lost electrons and been oxidized (it is the reducing agent). The nitrogen in nitric acid has gained electrons and has been reduced (nitric acid is oxidizing agent). 3. The oxidation number of carbon in methane (CH4) is 4, while its oxidation number in carbon dioxide is +4. The oxidation number of hydrogen in all compounds is +1. The oxidation number of oxygen in its elemental form is 0, while its value in carbon dioxide and water is 2. Therefore, carbon has been oxidized, and methane is the reducing agent. Oxygen has been reduced, and it is the oxidation agent. 4. The oxidation number of carbon in ethene (C2H4) is 2, while in ethane (C2H6) carbon has an oxidation number of 3. The oxidation number of hydrogen in ethene and ethane is +1. The oxidation number of elemental hydrogen is 0. Therefore, carbon in ethene has been reduced, and ethene is the oxidizing agent. Elemental hydrogen has been oxidized, and it is the reducing agent.
12 No element changed its oxidation number in this reaction. Nothing has been oxidized, and nothing has been reduced. This is not a redox reaction.
13 219 Balancing reactions by half reaction method. 1. CN - + MnO4 - CNO - + MnO2 (basic solution) a. CN - CNO - oxidation reaction b. H2O + CN - CNO - balance O with water c. H2O + CN - CNO - + 2H + balance H with H + d. H2O + CN - CNO - + 2H + + 2e - balance charge with e - e. MnO4 - MnO2 reduction reaction f. MnO4 - MnO2 + 2H2O balance O with water g. 4H + + MnO4 - MnO2 + 2H2O balance H with H + h. 4H + + MnO e - MnO2 + 2H2O balance charge with e - i. multiply d x 3: multiply h x 2 3H2O + 3CN - 3CNO - + 6H + + 6e - 8H + + 2MnO e - 2MnO2 + 4H2O j. add two equations together 3H2O + 3CN - + 8H + + 2MnO e - 3CNO - + 6H + + 6e - +2MnO2 + 4H2O k. cancel common substances on both sides 3CN - + 2H + + 2MnO4-3CNO - + 2MnO2 + H2O l. add OH - to both sides to neutralize H + 3CN - + 2H + + 2OH - + 2MnO4-3CNO - + 2MnO2 + H2O + 2OH - m. combine H+ with OH-, cancel out waters. 3CN - + H2O + 2MnO4-3CNO - + 2MnO2 + 2OH - n. check reactants/products for mass/charge balance
14 220 Reactants Products C 3 3 N 3 3 H 2 2 O 9 9 Mn 2 2 Charge Cr2O7-2 + I - Cr +3 + IO3 - (acidic solution) a. Cr2O7-2 Cr +3 reduction reaction b. Cr2O7-2 2Cr +3 balance Cr c. Cr2O7-2 2Cr H2O balance O with water d. Cr2O H + 2Cr H2O balance H with H + e. Cr2O H + + 6e - 2Cr H2O balance charge with e - f. I - IO3 - oxidation reaction g. I - + 3H2O IO3 - balance O with water h. I - + 3H2O IO H + balance H with H + i. I - + 3H2O IO H + + 6e - balance charge with e - j. Add e and i together Cr2O H + + 6e - + I - + 3H2O 2Cr H2O + IO H + + 6e - k. cancel common substances on both sides of equation Cr2O H + + I - 2Cr H2O + IO3 -
15 221 l. check reactants/products for mass/charge balance Reactants Products Cr 2 2 O 7 7 H 8 8 I 1 1 Charge MnO4 - + CH3OH Mn +2 + HCO2H (acidic solution) a, MnO4 - Mn +2 reduction reaction b. MnO4 - Mn H2O balance O with water c. 8H + + MnO4 - Mn H2O balance H with H + d. 8H + + 5e - + MnO4 - Mn H2O balance charge with e - e. CH3OH HCO2H oxidation reaction f. H2O + CH3OH HCO2H balance O with water g. H2O + CH3OH HCO2H + 4H + balance H with H + h. H2O + CH3OH HCO2H + 4H + + 4e - balance charge with e - i. multiply d by 4, multiply h by 5 32H e - + 4MnO4-4Mn H2O 5H2O + 5CH3OH 5HCO2H + 20H e - j. add two equations together 32H e - + 4MnO H2O + 5CH3OH 4Mn H2O + 5HCO2H + 20H e - k. cancel out common substances on both sides of equation 12H + + 4MnO CH3OH 4Mn H2O + 5HCO2H
16 222 l. check reactants/products for mass/charge balance Reactants Products H Mn 4 4 O C 5 5 Charge As + ClO3 - H3AsO3 + HClO (acidic solution) a. As H3AsO3 oxidation reaction b. 3H2O + As H3AsO3 balance O with water c. 3H2O + As H3AsO3 + 3H + balance H with H + d. 3H2O + As H3AsO3 + 3H + + 3e - balance charge with e - e. ClO3 - HClO reduction reaction f. ClO3 - HClO + 2H2O balance O with water g. 5H + + ClO3 - HClO + 2H2O balance H with H + h. 5H + + 4e - + ClO3 - HClO + 2H2O balance charge with e - i. multiply d by 4, multiply h by 3 12H2O + 4As 4H3AsO3 + 12H e - 15H e - + 3ClO3-3HClO + 6H2O j. add two equations together 12H2O + 4As + 15H e - + 3ClO3-4H3AsO3 + 12H e - + 3HClO + 6H2O k. cancel out common substances on both sides of equation 6H2O + 4As + 3H + + 3ClO3-4H3AsO3 + 3HClO
17 223 l. check reactants/products for mass/charge balance Reactants Products H O As 4 4 Cl 3 3 Charge H2O2 + ClO2 ClO2 - + O2 (basic solution) a, ClO2 ClO2 - reduction equation b. e - + ClO2 ClO2 - balance charge with e - c. H2O2 O2 oxidation reaction d. H2O2 O2 + 2H + balance H with H + e. H2O2 O2 + 2H + + 2e - balance charge with e - f. multiply b by 2 and add to e 2e - + 2ClO2 + H2O2 2ClO2 - + O2 + 2H + + 2e - g. cancel out common substances on both sides of equation 2ClO2 + H2O2 2ClO2 - + O2 + 2H + h. add OH - to both sides to neutralize H + 2ClO2 + H2O2 + 2OH - 2ClO2 - + O2 + 2H + + 2OH - 2ClO2 + H2O2 + 2OH - 2ClO2 - + O2 + 2H2O i. check reactants/products for mass/charge balance Reactants Products Cl 2 2 O 8 8 H 4 4 Charge - 2-2
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Oxidation / Reduction Handout Chem 2 WS11 The original concept of oxidation applied to reactions where there was a union with oxygen. The oxygen was either furnished by elemental oxygen or by compounds
Reactions in solution A subset of chemical reactions Learning objectives Define solution and its components Distinguish among strong, weak and non-electrolyte Identify strong acids and strong bases Apply
Chapter 4: Reactions in Aqueous Solution (Sections 4.1-4.12) Chapter Goals Be able to: Classify substances as electrolytes or nonelectrolytes. Write molecular, ionic, and net ionic equations for precipitation,
Chemical Formulas Chapter 7 I. Reviewing Ions 1. Ion = atom or a group of bonded atoms with a positive or negative charge. I. Reviewing Ions 2. Cation = ion with a POSITIVE (+) charge. a) Forms when an
AP Chemistry A. Allan Chapter 4 Notes - Types of Chemical Reactions and Solution Chemistry 4.1 Water, the Common Solvent A. Structure of water 1. Oxygen's electronegativity is high (3.5) and hydrogen's
SNC2D Chemistry Review Name: Draw the Bohr-Rutherford diagram for each of the following atoms and determine the valence charge of their ions: Sodium: Aluminum: Sulfur: Fluorine: From your periodic table/chemistry
Chem 100 - Week 6 Notes Page 1 of 7 Writing and Naming Inorganic Compounds Naming IUPAC is an agency that makes rules for naming chemicals Many compounds also have common names Inorganic compounds Compounds
Chapter 3 - Molecules, compounds, and chemical equations Elements and compounds Chemical bonding is the true difference between compounds and mixtures Atomic elements: Ionic bond: attraction of oppositely
CHEM 101 HOUR EXAM II 13-OCT-98 Directions: show all work for each question only on its corresponding numbered blue book page. 1. Write a balanced net ionic equation for each reaction or process shown:
3 Formulae, stoichiometry and the mole concept Content 3.1 Symbols, Formulae and Chemical equations 3.2 Concept of Relative Mass 3.3 Mole Concept and Stoichiometry Learning Outcomes Candidates should be
Chapter 5 - Molecules and Compounds How do we represent molecules? In pictures In formula In name Ionic compounds Molecular compounds On the course website, you will find a list of ions that I would like
H 2 O Chapter 9 Chemical Names and Formulas Section 9.1: Naming Ions OBJECTIVES: Identify the charges on monatomic ions by using the periodic table, and name the ions. Define a polyatomic ion and write
30 WRITING REDOX EQUATIONS : HALF-EQUATION METHOD Comparing different methods for balancing redox equations: If the objective is simply to balance an equation, and the equation can be balanced easily BY
PURPOSE EXPERIMENT 5: CHEMICAL REACTIONS AND EQUATIONS To perform and observe simple chemical reactions. To identify the products of chemical reactions and write balanced equations for those reactions.
SEATTLE CENTRAL COMMUNITY COLLEGE DIVISION OF SCIENCE AND MATHEMATICS OxidationReduction Oxidation is loss of electrons. (Oxygen is EN enough to grab e away from most elements, so the term originally meant
78 Chapter 6: Writing and Balancing Chemical Equations. It is convenient to classify chemical reactions into one of several general types. Some of the more common, important, reactions are shown below.
Chem 170 Stoichiometric Calculations Module Four Balancing Chemical Reactions DePauw University Department of Chemistry and Biochemistry Page 1 Introduction to Module Four When making a cheeseburger you
Name: Class: Date: Chapter 8 Review and Study Guide Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Knowledge about what products are produced in a chemical
Decomposition 1. Solid ammonium carbonate is heated. 2. Solid calcium carbonate is heated. 3. Solid calcium sulfite is heated in a vacuum. Composition 1. Barium oxide is added to distilled water. 2. Phosphorus
Solutions for Practice Problems p. 114 Ch.4 Solutions 1. Consider the following reaction. 2H 2(g) + O 2(g) 2H 2 O (l) (a) Write the ratio of H 2 molecules: O 2 molecules: H 2 Omolecules. The ratio is given
Using the Periodic table: Main Group Main Group Elements 1 Using the Periodic Table:Transition Group Transition Metals Form various cations by loss of electrons e.g. Fe 2+ / Fe 3+, Cu + / Cu 2+ Cr 2+ /
Topics 1. Reaction Types a. Combustion b. Synthesis c. Decomposition d. Single replacement i. Metal activity series ii. Nonmetal activity series e. Double replacement i. Precipitates and solubility rules
Chapter 5 Chemical Reactions and Equations What is a chemical reaction? How do we know a chemical reaction occurs? Writing chemical equations Predicting chemical reactions Representing reactions in aqueous
Name: Exam 2 Chemistry 65 Summer 2015 Score: Instructions: Clearly circle the one best answer 1. Valence electrons are electrons located A) in the outermost energy level of an atom. B) in the nucleus of
Chapter 9 Answers to Questions 1. Word equation: Silicon Tetrachloride + Water Silicon Dioxide + Hydrogen Chloride Formulas: Next, the chemical formulas are needed. As these are all covalent compounds,