Chemical Reactions PDF
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University of Idaho
Marisa Alviar-Agnew & Henry Agnew
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This document covers chemical reactions, including evidence of chemical reactions, writing chemical equations, and classifying types of reactions. The document also includes examples and exercises to test understanding.
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7: CHEMICAL REACTIONS CHAPTER OVERVIEW 7: Chemical Reactions 7.2: Evidence of a Chemical Reaction 7.3: The Chemical Equation 7.4: How to Write Balanced Chemical Equations 7.5: Aqueous Solutions and Solubility - Compounds Dissolved in Water 7.6: Precipitation Reactions 7.7: Writing Che...
7: CHEMICAL REACTIONS CHAPTER OVERVIEW 7: Chemical Reactions 7.2: Evidence of a Chemical Reaction 7.3: The Chemical Equation 7.4: How to Write Balanced Chemical Equations 7.5: Aqueous Solutions and Solubility - Compounds Dissolved in Water 7.6: Precipitation Reactions 7.7: Writing Chemical Equations for Reactions in Solution- Molecular, Complete Ionic, and Net Ionic Equations 7.8: Acid–Base and Gas Evolution Reactions 7.9: Oxidation–Reduction Reactions 7.10: Classifying Chemical Reactions 7.11: The Activity Series- Predicting Spontaneous Redox Reactions 7: Chemical Reactions is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 1 7.2: Evidence of a Chemical Reaction Learning Objectives Identify the evidence for chemical reactions. In a chemical change, new substances are formed. In order for this to occur, the chemical bonds of the substances break, and the atoms that compose them separate and rearrange themselves into new substances with new chemical bonds. When this process occurs, we call it a chemical reaction. A chemical reaction is the process in which one or more substances are changed into one or more new substances. Figure 7.2.1 : (a) Copper and nitric acid undergo a chemical change to form copper nitrate and brown, gaseous nitrogen dioxide. (b) During the combustion of a match, cellulose in the match and oxygen from the air undergo a chemical change to form carbon dioxide and water vapor. (c) Cooking red meat causes a number of chemical changes, including the oxidation of iron in myoglobin that results in the familiar red-to-brown color change. (d) A banana turning brown is a chemical change as new, darker (and less tasty) substances form. (Credit b: modification of work by Jeff Turner; credit c: modification of work by Gloria Cabada-Leman; credit d: modification of work by Roberto Verzo.) To identify a chemical reaction, we look for a chemical change. A chemical change always produces one or more types of matter that differ from the matter present before the change. The formation of rust is a chemical change because rust is a different kind of matter than the iron, oxygen, and water present before the rust formed. The explosion of nitroglycerin is a chemical change because the gases produced are very different kinds of matter from the original substance. Other examples of chemical changes include: reactions that are performed in a lab (such as copper reacting with nitric acid), all forms of combustion (burning), and food being cooked, digested, or rotting (Figure 7.2.1). 6:3- Evidence of Chemical Reactions Access for free at OpenStax 7.2.1 https://chem.libretexts.org/@go/page/47498 Video7.2.1 : Evidence of a Chemical Reaction Example 7.2.1: Evidence of a Chemical Reaction Which of the following is a chemical reaction? a. Freezing liquid mercury. b. Adding yellow to blue to make green. c. Cutting a piece of paper into two pieces. d. Dropping a sliced orange into a vat of sodium dydroxide. e. Filling a balloon with natural air. Solution A, B, C, & E involve only physical changes. A sliced orange has acid (citric acid) that can react with sodium hydroxide, so the answer is D. Exercise 7.2.1 Which of the following is a chemical reaction? a. Painting a wall blue. b. A bicycle rusting. c. Ice cream melting. d. Scratching a key across a desk. e. Making a sand castle. Answer B Example 7.2.2: Evidence of a Chemical Reaction Which of the following is not a chemical reaction? a. Shattering glass with a baseball. b. Corroding metal. c. Fireworks exploding. d. Lighting a match. e. Baking a cake. Solution Shattering glass with a baseball results in glass broken into many pieces but no chemical change happens, so the answer is A. Exercise 7.2.2 Which of the following is NOT a chemical reaction? a. Frying an egg. b. Slicing carrots. c. A Macbook falling out of a window. d. Creating ATP in the human body. e. Dropping a fizzy tablet into a glass of water. Answer B and C Access for free at OpenStax 7.2.2 https://chem.libretexts.org/@go/page/47498 Summary Chemical reactions can be identified via a wide range of different observable factors including change in color, energy change (temperature change or light produced), gas production, formation of precipitate and change in properties. This page titled 7.2: Evidence of a Chemical Reaction is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform. Access for free at OpenStax 7.2.3 https://chem.libretexts.org/@go/page/47498 7.3: The Chemical Equation Learning Objectives Identify the reactants and products in any chemical reaction. Convert word equations into chemical equations. Use the common symbols, (s) , (l), (g) , (aq) , and → appropriately when writing a chemical reaction. In a chemical change, new substances are formed. In order for this to occur, the chemical bonds of the substances break, and the atoms that compose them separate and rearrange themselves into new substances with new chemical bonds. When this process occurs, we call it a chemical reaction. A chemical reaction is the process in which one or more substances are changed into one or more new substances. Reactants and Products To describe a chemical reaction, we need to indicate what substances are present at the beginning and what substances are present at the end. The substances that are present at the beginning are called reactants and the substances present at the end are called products. Sometimes when reactants are put into a reaction vessel, a reaction will take place to produce products. Reactants are the starting materials, that is, whatever we have as our initial ingredients. The products are just that—what is produced—or the result of what happens to the reactants when we put them together in the reaction vessel. If we think about baking chocolate chip cookies, our reactants would be flour, butter, sugar, vanilla, baking soda, salt, egg, and chocolate chips. What would be the products? Cookies! The reaction vessel would be our mixing bowl. Flour + Butter + Sugar + Vanilla + Baking Soda + Eggs + Chocolate Chips → Cookies Ingredients = Reactants Product Writing Chemical Equations When sulfur dioxide is added to oxygen, sulfur trioxide is produced. Sulfur dioxide and oxygen, SO 2 +O 2 , are reactants and sulfur trioxide, SO , is the product. 3 2 SO (g) + O (g) → 2 SO (g) 2 2 3 Reactants Products In chemical reactions, the reactants are found before the symbol "→" and the products are found after the symbol "→". The general equation for a reaction is: Reactants → Products There are a few special symbols that we need to know in order to "talk" in chemical shorthand. In the table below is the summary of the major symbols used in chemical equations. Table 7.3.1 shows a listing of symbols used in chemical equations. Table 7.3.1 : Symbols Used in Chemical Equations Symbol Description Symbol Description used to separate multiple reactants reactant or product in the solid + (s) or products state yield sign; separates reactants (l) reactant or product in the liquid → from products state replaces the yield sign for ⇌ reversible reactions that reach (g) reactant or product in the gas state equilibrium Pt formula written above the arrow is reactant or product in an aqueous → (aq) used as a catalyst in the reaction solution (dissolved in water) 7.3.1 https://chem.libretexts.org/@go/page/47500 Symbol Description Symbol Description Δ triangle indicates that the reaction → is being heated Chemists have a choice of methods for describing a chemical reaction. 1. They could draw a picture of the chemical reaction. 2. They could write a word equation for the chemical reaction: "Two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of water vapor." 3. They could write the equation in chemical shorthand. 2H (g) + O (g) → 2 H O (g) 2 2 2 In the symbolic equation, chemical formulas are used instead of chemical names for reactants and products, while symbols are used to indicate the phase of each substance. It should be apparent that the chemical shorthand method is the quickest and clearest method for writing chemical equations. We could write that an aqueous solution of calcium nitrate is added to an aqueous solution of sodium hydroxide to produce solid calcium hydroxide and an aqueous solution of sodium nitrate. Or in shorthand we could write: Ca (NO ) (aq) + 2NaOH (aq) → Ca (OH) (s) + 2 NaNO (aq) 3 2 2 3 How much easier is that to read? Let's try it in reverse. Look at the following reaction in shorthand and write the word equation for the reaction: Cu (s) + AgNO (aq) → Cu(NO ) (aq) + Ag (s) 3 3 2 The word equation for this reaction might read something like "solid copper reacts with an aqueous solution of silver nitrate to produce a solution of copper (II) nitrate with solid silver." To turn word equations into symbolic equations, we need to follow the given steps: 1. Identify the reactants and products. This will help you know which symbols go on each side of the arrow and where the + signs go. 2. Write the correct formulas for all compounds. You will need to use the rules you learned in Chapter 5 (including making all ionic compounds charge balanced). 3. Write the correct formulas for all elements. Usually this is given straight off of the periodic table. However, there are seven elements that are considered diatomic, meaning that they are always found in pairs in nature. They include those elements listed in the table. Table 7.3.1 : Diatomic Elements Element Name Hydrogen Nitrogen Oxygen Fluorine Chlorine Bromine Iodine Formula H2 N2 O2 F2 Cl2 Br2 I2 Example 7.3.1 Transfer the following symbolic equations into word equations or word equations into symbolic equations. a. HCl (aq) + NaOH (aq) → NaCl (aq) + H O (l) 2 b. Gaseous propane, C H , burns in oxygen gas to produce gaseous carbon dioxide and liquid water. 3 8 7.3.2 https://chem.libretexts.org/@go/page/47500 c. Hydrogen fluoride gas reacts with an aqueous solution of potassium carbonate to produce an aqueous solution of potassium fluoride, liquid water, and gaseous carbon dioxide. Solution a. An aqueous solution of hydrochloric acid reacts with an aqueous solution of sodium hydroxide to produce an aqueous solution of sodium chloride and liquid water. b. Reactants: propane (C 3 H 8 ) and oxygen (O )2 Product: carbon dioxide (CO ) and water (H 2 2 O ) C H (g) + O (g) → CO (g) + H O (l) 3 8 2 2 2 c. Reactants: hydrogen fluoride and potassium carbonate Products: potassium fluoride, water, and carbon dioxide HF (g) + K CO (aq) → KF (aq) + H O (l) + CO (g) 2 3 2 2 Exercise 7.3.1 Transfer the following symbolic equations into word equations or word equations into symbolic equations. a. Hydrogen gas reacts with nitrogen gas to produce gaseous ammonia. b. HCl (aq) + LiOH (aq) → LiCl (aq) + H O (l) 2 c. Copper metal is heated with oxygen gas to produce solid copper(II) oxide. Answer a H2 (g) + N2 (g) → N H3 (g) Answer b An aqueous solution of hydrochloric acid reacts with an aqueous solution of lithium hydroxide to produce an aqueous solution of lithium chloride and liquid water. Answer c C u(s) + O2 (g) → C uO(s) Summary A chemical reaction is the process by which one or more substances are changed into one or more new substances. Chemical reactions are represented by chemical equations. Chemical equations have reactants on the left, an arrow that is read as "yields", and the products on the right. 7.3: The Chemical Equation is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 7.3.3 https://chem.libretexts.org/@go/page/47500 7.4: How to Write Balanced Chemical Equations Learning Objectives Explain the roles of subscripts and coefficients in chemical equations. Balance a chemical equation when given the unbalanced equation. Explain the role of the Law of Conservation of Mass in a chemical reaction. Even though chemical compounds are broken up and new compounds are formed during a chemical reaction, atoms in the reactants do not disappear, nor do new atoms appear to form the products. In chemical reactions, atoms are never created or destroyed. The same atoms that were present in the reactants are present in the products—they are merely reorganized into different arrangements. In a complete chemical equation, the two sides of the equation must be present on the reactant and the product sides of the equation. Coefficients and Subscripts There are two types of numbers that appear in chemical equations. There are subscripts, which are part of the chemical formulas of the reactants and products; and there are coefficients that are placed in front of the formulas to indicate how many molecules of that substance is used or produced. Figure 7.4.1 : Balancing Equations. You cannot change subscripts in a chemical formula to balance a chemical equation; you can change only the coefficients. Changing subscripts changes the ratios of atoms in the molecule and the resulting chemical properties. For example, water (H2O) and hydrogen peroxide (H2O2) are chemically distinct substances. H2O2 decomposes to H2O and O2 gas when it comes in contact with the metal platinum, whereas no such reaction occurs between water and platinum. The subscripts are part of the formulas and once the formulas for the reactants and products are determined, the subscripts may not be changed. The coefficients indicate the number of each substance involved in the reaction and may be changed in order to balance the equation. The equation above indicates that one mole of solid copper is reacting with two moles of aqueous silver nitrate to produce one mole of aqueous copper (II) nitrate and two atoms of solid silver. Balancing a Chemical Equation Because the identities of the reactants and products are fixed, the equation cannot be balanced by changing the subscripts of the reactants or the products. To do so would change the chemical identity of the species being described, as illustrated in Figure 7.4.1. Original molecule H 2 O : if the coefficient 2 is added in front, that makes 2 water molecules; but if the subscript 2 is added to make H O , that's hydrogen peroxide. 2 2 The simplest and most generally useful method for balancing chemical equations is “inspection,” better known as trial and error. The following is an efficient approach to balancing a chemical equation using this method. 7.4.1 https://chem.libretexts.org/@go/page/47502 Steps in Balancing a Chemical Equation 1. Identify the most complex substance. 2. Beginning with that substance, choose an element(s) that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element(s) on both sides. 3. Balance polyatomic ions (if present on both sides of the chemical equation) as a unit. 4. Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients. 5. Count the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced. Example 7.4.1: Combustion of Heptane Balance the chemical equation for the combustion of Heptane (C 7 H 16 ). C H (l) + O (g) → CO (g) + H O(g) 7 16 2 2 2 Solution Solutions to Example 7.4.1 Steps Example The most complex substance is the one with the largest number of different atoms, which is C H. We will assume initially that the 7 16 1. Identify the most complex substance. final balanced chemical equation contains 1 molecule or formula unit of this substance. a. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO molecules, each of which contains 1 carbon atom, on the 2 right side: C H (l) + O (g) → 7CO (g) + H O(g) 7 16 2 – 2 2 7 carbon atoms on both reactant and product sides 2. Adjust the coefficients. b. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H O molecules, each of which contains 2 hydrogen atoms, 2 on the right side: C H (l) + O (g) → 7 CO (g) + 8H O(g) 7 16 2 2 – 2 16 hydrogen atoms on both reactant and product sides 3. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction. The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front 4. Balance the remaining atoms. of the least complex substance, O2, on the reactant side: C H (l) + 11O (g) → 7 CO (g) + 8 H O(g) 7 16 –– – 2 2 2 22 oxygen atoms on both reactant and product sides The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 5. Check your work. oxygen atoms on each side. Always check to be sure that a chemical equation is balanced. 7.4.2 https://chem.libretexts.org/@go/page/47502 Example 7.4.2: Combustion of Isooctane Combustion of Isooctane (C 8 H 18 ) C H (l) + O (g) ⟶ CO (g) + H O(g) 8 18 2 2 2 Solution The assumption that the final balanced chemical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start. The combustion of any hydrocarbon with oxygen produces carbon dioxide and water. Solutions to Example 7.4.2 Steps Example The most complex substance is the one with the largest number of different atoms, which is C H. We will assume initially that the 1. Identify the most complex substance. 8 18 final balanced chemical equation contains 1 molecule or formula unit of this substance. a. The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO2 molecules in the products: C H (l) + O (g) ⟶ 8CO (g) + H O(g) 8 18 2 – 2 2 8 carbon atoms on both reactant and product sides 2. Adjust the coefficients. b. 18 hydrogen atoms in isooctane means that there must be 9 H2O molecules in the products: C H (l) + O (g) ⟶ 8 CO (g) + 9H O(g) 8 18 2 2 – 2 18 hydrogen atoms on both reactant and product sides 3. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction. The carbon and hydrogen atoms are now balanced, but we have 25 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the least complex substance, O2, but because there are 2 oxygen atoms per O2 molecule, we must use a fractional 25 coefficient ( ) to balance the oxygen atoms: 2 25 C H (l) + O (g) → 8 CO (g) + 9 H O(g) 8 18 2 2 2 2 ––– 4. Balance the remaining atoms. 25 oxygen atoms on both reactant and product sides The equation is now balanced, but we usually write equations with whole number coefficients. We can eliminate the fractional coefficient by multiplying all coefficients on both sides of the chemical equation by 2: 2C H (l) + 25O (g) ⟶ 16CO (g) + 18H O(g) – 8 18 –– – 2 –– – 2 –– – 2 The balanced chemical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side. Balancing equations requires some practice on your part as well as 5. Check your work. some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly. 7.4.3 https://chem.libretexts.org/@go/page/47502 Example 7.4.3: Precipitation of Lead (II) Chloride Aqueous solutions of lead (II) nitrate and sodium chloride are mixed. The products of the reaction are an aqueous solution of sodium nitrate and a solid precipitate of lead (II) chloride. Write the balanced chemical equation for this reaction. Solution Solutions to Example 7.4.3 Steps Example The most complex substance is lead (II) chloride. 1. Identify the most complex substance. Pb(NO ) (aq) + NaCl(aq) → NaNO (aq) + PbCl (s) 3 2 3 2 There are twice as many chloride ions in the reactants as there are in the products. Place a 2 in front of the NaCl in order to balance the chloride ions. 2. Adjust the coefficients. Pb(NO ) (aq) + 2NaCl(aq) → NaNO (aq) + PbCl (s) 3 2 – 3 2 1 Pb atom on both reactant and product sides 2 Na atoms on reactant side, 1 Na atom on product side 2 Cl atoms on both reactant and product sides The nitrate ions are still unbalanced. Place a 2 in front of the NaNO3. The result is: Pb(NO ) (aq) + 2 NaCl(aq) → 2NaNO (aq) + PbCl (s) 3 2 – 3 2 3. Balance polyatomic ions as a unit. 1 Pb atom on both reactant and product sides 2 Na atoms on both reactant and product sides 2 Cl atoms on both reactant and product sides 2 NO3- atoms on both reactant and product sides There is no need to balance the remaining atoms because they are 4. Balance the remaining atoms. already balanced. Pb(NO ) (aq) + 2 NaCl(aq) → 2 NaNO (aq) + PbCl (s) 3 2 3 2 1 Pb atom on both reactant and product sides 5. Check your work. 2 Na atoms on both reactant and product sides 2 Cl atoms on both reactant and product sides 2 NO3- atoms on both reactant and product sides Exercise 7.4.1 Is each chemical equation balanced? + a. 2 Hg(ℓ) O (g) → Hg O (s) 2 2 2 b. C H (g) + 2 O (g) → 2 CO (g) + 2 H O(g) 2 4 2 2 2 c. Mg(NO ) (s) + 2 Li(s) → Mg(s) 2 LiNO (s) 3 2 + 3 Answer a yes Answer b no Answer c yes 7.4.4 https://chem.libretexts.org/@go/page/47502 Exercise 7.4.2 Balance the following chemical equations. a. N (g) + O (g) → NO (g) 2 2 2 b. Pb(NO ) (aq) + FeCl (aq) → Fe(NO ) 3 2 3 3 3 (aq) + PbCl (s) 2 c. C H (l) + O (g) → CO (g) + H O(g) 6 14 2 2 2 Answer a N2 (g) + 2O2 (g) → 2NO2 (g) Answer b 3Pb(NO3)2(aq) + 2FeCl3(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s) Answer c 2C6H14(l) + 19O2(g)→ 12CO2(g) + 14H2O(g) Summary To be useful, chemical equations must always be balanced. Balanced chemical equations have the same number and type of each atom on both sides of the equation. The coefficients in a balanced equation must be the simplest whole number ratio. Mass is always conserved in chemical reactions. Vocabulary Chemical reaction - The process in which one or more substances are changed into one or more new substances. Reactants - The starting materials in a reaction. Products - Materials present at the end of a reaction. Balanced chemical equation - A chemical equation in which the number of each type of atom is equal on the two sides of the equation. Subscripts - Part of the chemical formulas of the reactants and products that indicate the number of atoms of the preceding element. Coefficient - A small whole number that appears in front of a formula in a balanced chemical equation. 7.4: How to Write Balanced Chemical Equations is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 7.4.5 https://chem.libretexts.org/@go/page/47502 7.5: Aqueous Solutions and Solubility - Compounds Dissolved in Water Learning Objectives Define and give examples of electrolytes. When some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called electrolytes. Substances that do not yield ions when dissolved are called nonelectrolytes. If the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte. If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, it is called a weak electrolyte. Substances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit (Figure 7.5.1). Figure 7.5.1 : Solutions of nonelectrolytes, such as ethanol, do not contain dissolved ions and cannot conduct electricity. Solutions of electrolytes contain ions that permit the passage of electricity. The conductivity of an electrolyte solution is related to the strength of the electrolyte. This diagram shows three separate beakers. Each has a wire plugged into a wall outlet. In each case, the wire leads from the wall to the beaker and is split resulting in two ends. One end leads to a light bulb and continues on to a rectangle labeled with a plus sign. The other end leads to a rectangle labeled with a minus sign. The rectangles are in a solution. In the first beaker, labeled “Ethanol No Conductivity,” four pairs of linked small green spheres suspended in the solution between the rectangles. In the second beaker, labeled “K C l Strong Conductivity,” six individual green spheres, three labeled plus and three labeled minus are suspended in the solution. Each of the six spheres has an arrow extending from it pointing to the rectangle labeled with the opposite sign. In the third beaker, labeled “Acetic acid solution Weak conductivity,” two pairs of joined green spheres and two individual spheres, one labeled plus and one labeled minus are shown suspended between the two rectangles. The plus labeled sphere has an arrow pointing to the rectangle labeled minus and the minus labeled sphere has an arrow pointing to the rectangle labeled plus. Ionic Electrolytes Water and other polar molecules are attracted to ions, as shown in Figure 7.5.2. The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction. These attractions play an important role in the dissolution of ionic compounds in water. 7.5.1 https://chem.libretexts.org/@go/page/47504 Figure 7.5.2 : As potassium chloride (KCl) dissolves in water, the ions are hydrated. The polar water molecules are attracted by the charges on the K+ and Cl− ions. Water molecules in front of and behind the ions are not shown. The diagram shows eight purple spheres labeled K superscript plus and eight green spheres labeled C l superscript minus mixed and touching near the center of the diagram. Outside of this cluster of spheres are seventeen clusters of three spheres, which include one red and two white spheres. A red sphere in one of these clusters is labeled O. A white sphere is labeled H. Two of the green C l superscript minus spheres are surrounded by three of the red and white clusters, with the red spheres closer to the green spheres than the white spheres. One of the K superscript plus purple spheres is surrounded by four of the red and white clusters. The white spheres of these clusters are closest to the purple spheres. When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation. Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes. Let us consider what happens at the microscopic level when we add solid KCl to water. Ion-dipole forces attract the positive (hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative (oxygen) ends to the positive potassium ions. The water molecules penetrate between individual K+ and Cl− ions and surround them, reducing the strong interionic forces that bind the ions together and letting them move off into solution as solvated ions, as Figure 7.5.2 shows. The reduction of the electrostatic attraction permits the independent motion of each hydrated ion in a dilute solution, resulting in an increase in the disorder of the system, as the ions change from their fixed and ordered positions in the crystal to mobile and much more disordered states in solution. This increased disorder is responsible for the dissolution of many ionic compounds, including KCl, which dissolve with absorption of heat. In other cases, the electrostatic attractions between the ions in a crystal are so large, or the ion-dipole attractive forces between the ions and water molecules are so weak, that the increase in disorder cannot compensate for the energy required to separate the ions, and the crystal is insoluble. Such is the case for compounds such as calcium carbonate (limestone), calcium phosphate (the inorganic component of bone), and iron oxide (rust). Solubility Rules Some combinations of aqueous reactants result in the formation of a solid precipitate as a product. However, some combinations will not produce such a product. If solutions of sodium nitrate and ammonium chloride are mixed, no reaction occurs. One could write a molecular equation showing a double-replacement reaction, but both products, sodium chloride and ammonium nitrate, are soluble and would remain in the solution as ions. Every ion is a spectator ion and there is no net ionic equation at all. It is useful to be able to predict when a precipitate will occur in a reaction. To do so, you can use a set of guidelines called the solubility rules (Tables 7.5.1 and 7.5.2). Table 7.5.1 : Solubility Rules for Soluble Substances Soluble in Water Important Exceptions (Insoluble) All Group IA and NH4+ salts none All nitrates, chlorates, perchlorates and acetates none All sulfates CaSO4, BaSO4, SrSO4, PbSO4 All chlorides, bromides, and iodides AgX, Hg2X2, PbX2 (X= Cl, Br, or I) Table 7.5.2 : Solubility Rules for Sparingly Soluble Substances 7.5.2 https://chem.libretexts.org/@go/page/47504 Sparingly Soluble in Water Important Exceptions (Soluble) All carbonates and phosphates Group IA and NH4+ salts All hydroxides Group IA and NH4+ salts; Ba2+, Sr2+, Ca2+ sparingly soluble All sulfides Group IA, IIA and NH4+ salts; MgS, CaS, BaS sparingly soluble All oxalates Group IA and NH4+ salts Special note: The following electrolytes are of only moderate solubility in water: CH3COOAg, Ag2SO4, KClO4. They will precipitate only if rather concentrated solutions are used. As an example on how to use the solubility rules, predict if a precipitate will form when solutions of cesium bromide and lead (II) nitrate are mixed. + − 2 + − Cs (aq) + Br (aq) + Pb (aq) + 2 NO (aq) →? 3 The potential precipitates from a double-replacement reaction are cesium nitrate and lead (II) bromide. According to the solubility rules table, cesium nitrate is soluble because all compounds containing the nitrate ion, as well as all compounds containing the alkali metal ions, are soluble. Most compounds containing the bromide ion are soluble, but lead (II) is an exception. Therefore, the cesium and nitrate ions are spectator ions and the lead (II) bromide is a precipitate. The balanced net ionic reaction is: 2 + − Pb (aq) + 2 Br (aq) → PbBr (s) 2 Example 7.5.1: Solubility Classify each compound as soluble or insoluble a. Zn(NO3)2 b. PbBr2 c. Sr3(PO4)2 Solution a. All nitrates are soluble in water, so Zn(NO3)2 is soluble. b. All bromides are soluble in water, except those combined with Pb2+, so PbBr2 is insoluble. c. All phosphates are insoluble, so Sr3(PO4)2 is insoluble. Exercise 7.5.1: Solubility Classify each compound as soluble or insoluble. a. Mg(OH)2 b. KBr c. Pb(NO3)2 Answer a insoluble Answer b soluble Answer c soluble Summary Substances that dissolve in water to yield ions are called electrolytes. Nonelectrolytes are substances that do not produce ions when dissolved in water. Solubility rules allow prediction of what products will be insoluble in water. 7.5.3 https://chem.libretexts.org/@go/page/47504 Contributions & Attributions Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193- [email protected]). 7.5: Aqueous Solutions and Solubility - Compounds Dissolved in Water is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 7.5.4 https://chem.libretexts.org/@go/page/47504 7.6: Precipitation Reactions Learning Objectives To identify a precipitation reaction and predict solubility. A precipitation reaction is a reaction that yields an insoluble product—a precipitate—when two solutions are mixed. When a colorless solution of silver nitrate is mixed with a yellow-orange solution of potassium dichromate, a reddish precipitate of silver dichromate is produced. AgNO (aq) + K Cr O (aq) → Ag Cr O (s) + KNO (aq) (7.6.1) 3 2 2 7 2 2 7 3 This unbalanced equation has the general form of an exchange reaction: AC + BD → AD + BC (7.6.2) insoluble Thus precipitation reactions are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble. Because both components of each compound change partners, such reactions are sometimes called double-displacement reactions. Precipitation reactions are used to isolate metals that have been extracted from their ores, and to recover precious metals for recycling. Precipitation of silver(I) dichromate (Pe… (Pe… Video: Mixing potassium dichromate and silver nitrate together to initiate a precipitation reaction (Equation 7.6.1 ). Just as important as predicting the product of a reaction is knowing when a chemical reaction will not occur. Simply mixing solutions of two different chemical substances does not guarantee that a reaction will take place. For example, if 500 mL of aqueous N aC l solution is mixed with 500 mL of aqueous KBr solution, the final solution has a volume of 1.00 L and contains Na (aq), Cl (aq) , K (aq) , and Br (aq). As you will see in (Figure 7.6.1), none of these species reacts with any of the others. + − + − When these solutions are mixed, the only effect is to dilute each solution with the other. 7.6.1 https://chem.libretexts.org/@go/page/83655 Figure 7.6.1 : The Effect of Mixing Aqueous KBr and NaCl Solutions. Because no net reaction occurs, the only effect is to dilute each solution with the other. (Water molecules are omitted from molecular views of the solutions for clarity.) Predicting Precipitation Reactions A precipitation reaction occurs when a solid precipitate forms after mixing two strong electrolyte solutions. As stated previously, if none of the species in the solution reacts then no net reaction occurred. Predict what will happen when aqueous solutions of barium chloride and lithium sulfate are mixed. Change the partners of the anions and cations on the reactant side to form new compounds (products): Chemical equation of the reactants barium chloride and lithium sulfate forming the products barium sulfate and lithium chloride. Because barium chloride and lithium sulfate are strong electrolytes, each dissociates completely in water to give a solution that contains the constituent anions and cations. Mixing the two solutions initially gives an aqueous solution that contains Ba , Cl , 2 + − Li , and SO ions. The only possible exchange reaction is to form LiCl and BaSO. + 2 − 4 4 Correct the formulas of the products based on the charges of the ions. No need to correct the formula as both compounds already have their charges balanced. BaCl (aq) + Li SO (aq) → BaSO + LiCl 2 2 4 4 Refer to the solubility rules table to determine insoluble products which will therefore form a precipitate. BaCl (aq) + Li SO (aq) → BaSO (s) + LiCl(aq) 2 2 4 4 Table 7.5.1 from the previous section shows that LiCl is soluble in water, but BaSO is not soluble in water. 4 Balance the equation: BaCl (aq) + Li SO (aq) → BaSO (s) + 2 LiCl(aq) 2 2 4 4 Although soluble barium salts are toxic, BaSO is so insoluble that it can be used to diagnose stomach and intestinal problems 4 without being absorbed into tissues. An outline of the digestive organs appears on x-rays of patients who have been given a “barium milkshake” or a “barium enema”—a suspension of very fine BaSO particles in water. 4 7.6.2 https://chem.libretexts.org/@go/page/83655 Figure 7.6.2 : An x-ray of the digestive organs of a patient who has swallowed a “barium milkshake.” A barium milkshake is a suspension of very fine BaSO4 particles in water; the high atomic mass of barium makes it opaque to x-rays. (Public Domain; Glitzy queen00 via Wikipedia). Example 7.6.1 Predict what will happen if aqueous solutions of rubidium hydroxide and cobalt(II) chloride are mixed. Solution Solutions to Example 7.6.1 Steps Example Change the partners of the anions and cations on the reactant side to form new compounds (products). Chemical equation of the reactants rubidium hydroxide and cobalt(II) chloride forming the products rubidium chloride and cobalt hydroxide. Correct the formulas of the products based on the charges of the ions. RbOH(aq) + CoCl (aq) → RbCl + Co (OH) 2 2 Refer to the solubility rules table to determine insoluble products RbOH(aq) + CoCl (aq) → RbCl(aq) + Co (OH) (s) which will therefore form a precipitate. 2 2 Coefficients already balanced. Balance the equation. RbOH(aq) + CoCl (aq) → RbCl(aq) + Co (OH) (s) 2 2 Example 7.6.2 Predict what will happen if aqueous solutions of strontium bromide and aluminum nitrate are mixed. Solution Solutions for Example 7.6.2 Steps Example Change the partners of the anions and cations on the reactant side to form new compounds (products). Chemical equation of the reactants strontium bromide and aluminum nitrate forming the products strontium nitrate and aluminum bromide. 7.6.3 https://chem.libretexts.org/@go/page/83655 Steps Example Correct the formulas of the products based on the charges of the ions. SrBr 2 (aq) + Al(NO ) (aq) → Sr(NO ) 3 3 3 2 + AlBr 3 SrBr (aq) + Al(NO ) (aq) → Sr(NO ) (aq) + AlBr (aq) Refer to the solubility rules table to determine insoluble products 2 3 3 3 2 3 According to Table 7.5.1 from the previous section, both AlBr (rule which will therefore form a precipitate. 3 4) and Sr(NO ) (rule 2) are soluble. 3 2 SrBr (aq) + Al(NO ) (aq) → If all possible products are soluble, then no net reaction will occur. 2 3 3 NO REACTION Exercise 7.6.2 Using the information in Table 7.5.1 from the previous section, predict what will happen in each case involving strong electrolytes. a. An aqueous solution of strontium hydroxide is added to an aqueous solution of iron(II) chloride. b. Solid potassium phosphate is added to an aqueous solution of mercury(II) perchlorate. c. Solid sodium fluoride is added to an aqueous solution of ammonium formate. d. Aqueous solutions of calcium bromide and cesium carbonate are mixed. Answer a Fe(OH)2 precipitate is formed. Answer b Hg3(PO4)2 precipitate is formed. Answer c No Reaction. Answer d CaCO3 is precipitate formed. Summary In a precipitation reaction, a subclass of exchange reactions, an insoluble material (a precipitate) forms when two electrolyte solutions are mixed. To predict the product of a precipitation reaction, all species initially present in the solutions are identified, as are any combinations likely to produce an insoluble salt. Contributions & Attributions Modified by Joshua Halpern (Howard University) 7.6: Precipitation Reactions is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 7.6.4 https://chem.libretexts.org/@go/page/83655 7.7: Writing Chemical Equations for Reactions in Solution- Molecular, Complete Ionic, and Net Ionic Equations A typical precipitation reaction occurs when an aqueous solution of barium chloride is mixed with one containing sodium sulfate. The complete chemical equation can be written to describe what happens, and such an equation is useful in making chemical calculations. BaCl (aq) + Na SO (aq) ⟶ BaSO (s) + 2 NaCl(aq) (7.7.1) 2 2 4 4 Complete Chemical Equation However, Equation 7.7.1 does not really represent the microscopic particles (that is, the ions) present in the solution. Below is the complete ionic equation: spectator spectator spectator spectator 2 + − + 2 − + − Ba (aq) + 2 Cl (aq) + 2 Na (aq) + SO (aq) ⟶ BaSO (s) + 2 Na (aq) + Cl (aq) (7.7.2) 4 4 Complete Ionic Equation Equation 7.7.2 is rather cumbersome and includes so many different ions that it may be confusing. In any case, we are often interested in the independent behavior of ions, not the specific compound from which they came. A precipitate of BaSO (s) will 4 form when any solution containing Ba (aq) is mixed with any solution containing SO (aq) (provided concentrations are not 2 + 2 – 4 extremely small). This happens independently of the Cl (aq) and Na (aq) ions in Equation 7.7.2. These ions are called spectator – + ions because they do not participate in the reaction. When we want to emphasize the independent behavior of ions, a net ionic equation is written, omitting the spectator ions. For precipitation of BaSO the net ionic equation is 4 2 + 2 − Ba (aq) + SO (aq) ⟶ BaSO (s) (7.7.3) 4 4 Net Ionic Equation Example 7.7.1 a. When a solution of AgNO is added to a solution of CaCl , insoluble AgCl precipitates. Write three equations (complete 3 2 chemical equation, complete ionic equation, and net ionic equation) that describe this process. b. Write the balanced net ionic equation to describe any reaction that occurs when the solutions of Na SO and NH I are 2 4 4 mixed. Solution Solutions to Example 7.7.1 Equation Type Example 7.7.1a Example 7.7.1b 2 AgNO (aq) + CaCl (aq) ⟶ 3 2 Na SO (aq) + 2 NH I(aq) ⟶ 2 4 4 2 AgCl(s) + Ca (NO ) (aq) 3 2 2 NaI(aq) + (NH ) SO (aq) Complete Chemical Equation The proper states and formulas of all 4 2 4 Both products are aqueous so there is no net products are written and the chemical ionic equation that can be written. equation is balanced. + − 2 + − 2 Ag (aq) + 2 NO (aq) + Ca (aq) + 2 Cl (aq) ⟶ 3 2 + − 2 AgCl(s) + Ca (aq) + 2 NO (aq) Complete Ionic Equation 3 AgCl is a solid so it does not break up into ions in solution. + − Ag (aq) + Cl (aq) ⟶ AgCl(s) NaI and (NH ) SO are both soluble. All spectator ions are removed. 4 2 4 Net Ionic Equation No net ionic equation. The chemical equation is written using the There is no reaction. lowest common coefficients. The occurrence or nonoccurrence of precipitates can be used to detect the presence or absence of various species in solution. A BaCl solution, for instance, is often used as a test for the presence of SO (aq) ions. There are several insoluble salts of Ba , but 2 – 2 4 7.7.1 https://chem.libretexts.org/@go/page/47506 they all dissolve in dilute acid except for BaSO. Thus, if BaCl solution is added to an unknown solution which has previously 4 2 been acidified, the occurrence of a white precipitate is proof of the presence of the SO ion. 2 – 4 Figure 7.7.1 : The three common silver halide precipitates: AgI, AgBr and AgCl (left to right). The silver halides precipitate out of solution, but often form suspensions before settling. (CC BY-SA 3.0; Cychr). AgNO 3 solutions are often used in a similar way to test for halide ions. If AgNO solution is added to an acidified unknown 3 solution, a white precipitate indicates the presence of Cl ions, a cream-colored precipitate indicates the presence of Br ions, and – – a yellow precipitate indicates the presence of I ions (Figure 7.7.1). Further tests can then be made to see whether perhaps a – mixture of these ions is present. When AgNO is added to tap water, a white precipitate is almost always formed. The Cl ions in 3 – tap water usually come from the Cl which is added to municipal water supplies to kill microorganisms. 2 Exercise 7.7.1 Write balanced net ionic equations to describe any reaction that occurs when the following solutions are mixed. a. K CO + SrCl 2 3 2 b. FeSO + Ba(NO 4 3 ) 2 Answer a 2 + 2 − Sr (aq) + CO (aq) ⟶ SrCO (s) 3 3 Answer b 2 + 2 − Ba (aq) + SO (aq) ⟶ Ba(SO )(s) 4 4 Precipitates are also used for quantitative analysis of solutions, that is, to determine the amount of solute or the mass of solute in a given solution. For this purpose it is often convenient to use the first of the three types of equations described above. Then the rules of stoichiometry may be applied. Contributions & Attributions Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn. 7.7: Writing Chemical Equations for Reactions in Solution- Molecular, Complete Ionic, and Net Ionic Equations is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 7.7.2 https://chem.libretexts.org/@go/page/47506 7.8: Acid–Base and Gas Evolution Reactions Learning Objectives Identify when a reaction will evolve a gas. Neutralization Reactions Acids and bases react chemically with each other to form salts. A salt is a general chemical term for any ionic compound formed from an acid and a base. In reactions where the acid is a hydrogen-ion-containing compound and the base is a hydroxide-ion-containing compound, water is also a product. The general reaction is as follows: acid + base → water + salt The reaction of acid and base to make water and a salt is called neutralization. Like any chemical equation, a neutralization chemical equation must be properly balanced. For example, the neutralization reaction between sodium hydroxide and hydrochloric acid is as follows: NaOH(aq) + HCl(aq) → NaCl(aq) + H O(ℓ) (7.8.1) 2 with coefficients all understood to be one. The neutralization reaction between sodium hydroxide and sulfuric acid is as follows: 2 NaOH(aq) + H SO (aq) → Na SO (aq) + 2 H O(ℓ) (7.8.2) 2 4 2 4 2 Example 7.8.1: Neutralizing Nitric Acid Nitric acid (HNO3(aq)) can be neutralized by calcium hydroxide (Ca(OH)2(aq)). Write a balanced chemical equation for the reaction between these two compounds and identify the salt that it produces. Solution Solutions to Example 7.8.1 Steps Explanation Equation This is a double displacement reaction, so the Ca(OH)2(aq) + HNO3(aq) → Ca(NO3)2(aq) + Write the unbalanced equation. cations and anions swap to create new products. H2O(ℓ) Because there are two OH− ions in the formula for Ca(OH)2(aq) + 2HNO3(aq) → Ca(NO3)2(aq) + Balance the equation. Ca(OH)2, we need two moles of HNO3 to provide 2H2O(ℓ) H+ ions Additional step: identify the salt. The salt formed is calcium nitrate. Exercise 7.8.1 Hydrocyanic acid (HCN(aq)) can be neutralized by potassium hydroxide (KOH(aq) ). Write a balanced chemical equation for the reaction between these two compounds and identify the salt that it produces. Answer \[\ce{KOH (aq) + HCN(aq) → KCN (aq) + H2O(ℓ)} \nonumber \] Gas Evolving Reactions A gas evolution reaction is a chemical process that produces a gas, such as oxygen or carbon dioxide. In the following examples, an acid reacts with a carbonate, producing salt, carbon dioxide, and water, respectively. For example, nitric acid reacts with sodium carbonate to form sodium nitrate, carbon dioxide, and water (Table 7.8.1): + + + 2 HNO (aq) Na CO (aq) → 2 NaNO (aq) CO (g) H O(l) 3 2 3 3 2 2 Sulfuric acid reacts with calcium carbonate to form calcium sulfate, carbon dioxide, and water: + H SO (aq) + CaCO (aq) → CaSO (aq) + CO (g) H O(l) 2 4 3 4 2 2 Hydrochloric acid reacts with calcium carbonate to form calcium chloride, carbon dioxide, and water: 2 HCl(aq) + CaCO (aq) → CaCl (aq) + CO (g) + H O(l) 3 2 2 2 7.8.1 https://chem.libretexts.org/@go/page/47510 Figure 7.8.1 demonstrates this type of reaction: Figure 7.8.1 : Reaction of acids with carbonates. In this reaction setup, lime water (water + calcium hydroxide) is poured into one of the test tubes and sealed with a stopper. A small amount of hydrochloric acid is carefully poured into the remaining test tube. A small amount of sodium carbonate is added to the acid, and the tube is sealed with a rubber stopper. The two tubes are connected. As a result of the acid-carbonate reaction, carbon dioxide is produced and the lime water turns milky. In this reaction setup, lime water, a dilute calcium hydroxide (C a(OH ) ) solution, is poured into one of the test tubes and sealed with a stopper. A 2 small amount of hydrochloric acid is carefully poured into the remaining test tube. A small amount of sodium carbonate is added to the acid, and the tube is sealed with a rubber stopper. The two tubes are connected. As a result of the acid-carbonate reaction, carbon dioxide is produced and the lime water turns milky. Table 7.8.1 : Types of Compounds That Undergo Gas-Evolution Reactions Reactant Type Intermediate Product Gas Evolved Example sulfide none H S 2 2 HCl(aq) + K S → H S(g) + 2 KCl(aq) 2 2 carbonates and bicarbonates H CO 2 3 CO 2 2 HCl(aq) + K CO 2 2 → H O(l) + CO (g) + 2 2 sulfites and bisulfites H SO 2 3 SO 2 2 HCl(aq) + K SO 2 2 → H O(l) + SO (g) + 2 2 \(\ce{NH4Cl(aq) + KOH \rightarrow ammonia NH OH NH 4 3 H2O (l) + NH3(g) + 2KCl (aq)}\) The gas-evolving experiment lime water is illustrated in the following video: Science Experiment : Carbon Dioxide (Co₂) & L… L… Video 7.8.1 : Carbon Dioxide (C O ) & Limewater (Chemical Reaction). As the reaction proceeds, the limewater on the turns from clear to milky; 2 this is due to the C O (g) reacting with the aqueous calcium hydroxide to form calcium carbonate, which is only slightly soluble in water. 2 When this experiment is repeated with nitric or sulfuric acid instead of H C l, it yields the same results: the clear limewater turns milky, indicating the production of carbon dioxide. Another method to chemically generate gas is the oxidation of metals in acidic solutions. This reaction will yield a metal salt and hydrogen gas. 2 HCl(aq) + Zn(s) → ZnCl (aq) + H (g) 2 2 Here, hydrochloric acid oxidizes zinc to produce an aqueous metal salt and hydrogen gas bubbles. Recall that oxidation refers to a loss of electrons, and reduction refers to the gain of electrons. In the above redox reaction, neutral zinc is oxidized to Zn , and the acid, H , is reduced 2+ + to H (g). The oxidation of metals by strong acids is another common example of a gas evolution reaction. 2 Contributors & Affiliations Boundless (www.boundless.com) Wikipedia (CC-BY-SA-3.0) 7.8.2 https://chem.libretexts.org/@go/page/47510 Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected]). 7.8: Acid–Base and Gas Evolution Reactions is shared under a CK-12 license and was authored, remixed, and/or curated by Marisa Alviar-Agnew & Henry Agnew. 7.8.3 https://chem.libretexts.org/@go/page/47510 7.9: Oxidation–Reduction Reactions Learning Objectives Define oxidation and reduction. Assign oxidation numbers to atoms in simple compounds. Recognize a reaction as an oxidation-reduction reaction. In the course of a chemical reaction between a metal and a nonmetal, electrons are transferred from the metal atoms to the nonmetal atoms. For example, when zinc metal is mixed with sulfur and heated, the compound zinc sulfide is produced. Two valence electrons from each zinc atom are transferred to each sulfur atom. Figure 7.9.1 : Reaction between zinc and sulfur. Since the zinc is losing electrons in the reaction, it is being oxidized. The sulfur is gaining electrons and is thus being reduced. An oxidation-reduction reaction is a reaction that involves the full or partial transfer of electrons from one reactant to another. Oxidation is the full or partial loss of electrons or the gain of oxygen. Reduction is the full or partial gain of electrons or the loss of oxygen. A redox reaction is another term for an oxidation-reduction reaction. Each of these processes can be shown in a separate equation called a half-reaction. A half-reaction is an equation that shows either the oxidation or the reduction reaction that occurs during a redox reaction. 2 ++ − Zn → Zn e (7.9.1) 2 Oxidation + − 2 − S 2e → S (7.9.2) Reduction It is important to remember that the two half-reactions occur simultaneously. The resulting ions that are formed are then attracted to one another in an ionic bond. Another example of an oxidation-reduction reaction involving electron transfer is the well-known combination of metallic sodium and chlorine gas to form sodium chloride: + 2 Na Cl → 2 NaCl (7.9.3) 2 The half reactions are as follows: + − 2 Na → 2 Na +2 e (7.9.4) Oxidation − − Cl +2 e → 2 Cl (7.9.5) 2 Reduction We will concern ourselves with the balancing of these equations at another time. Oxidation Numbers Redox reactions require that we keep track of the electrons assigned to each atom in a chemical reaction. How do we do that? We use oxidation numbers to keep track of electrons in atoms. Oxidation numbers are assigned to atoms based on four rules. Oxidation numbers are not necessarily equal to the charge on the atom (although sometimes they can be); we must keep the concepts of charge and oxidation numbers separate. Assigning Oxidation Numbers The rules for assigning oxidation numbers to atoms are as follows: 1. Atoms in their elemental state are assigned an oxidation number of 0. 7.9.1 https://chem.libretexts.org/@go/page/47511 In H2, both H atoms have an oxidation number of 0. 2. Atoms in monatomic (i.e., one-atom) ions are assigned an oxidation number equal to their charge. Oxidation numbers are usually written with the sign first, then the magnitude, to differentiate them from charges. In MgCl2, magnesium has an oxidation number of +2, while chlorine has an oxidation number of −1. 3. In compounds, fluorine is assigned a −1 oxidation number; oxygen is usually assigned a −2 oxidation number (except in peroxide compounds [where it is −1] and in binary compounds with fluorine [where it is positive]); and hydrogen is usually assigned a +1 oxidation number [except when it exists as the hydride ion (H−), in which case rule 2 prevails]. In H2O, the H atoms each have an oxidation number of +1, while the O atom has an oxidation number of −2, even though hydrogen and oxygen do not exist as ions in this compound (rule 3). By contrast, by rule 3, each H atom in hydrogen peroxide (H2O2) has an oxidation number of +1, while each O atom has an oxidation number of −1. 4. In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species (which is zero if the species is neutral). In SO2, each O atom has an oxidation number of −2; for the sum of the oxidation numbers to equal the charge on the species (which is zero), the S atom is assigned an oxidation number of +4. Does this mean that the sulfur atom has a 4+ charge on it? No, it means only that the S atom is assigned a +4 oxidation number by our rules of apportioning electrons among the atoms in a compound. Example 7.9.1: Assigning Oxidation States Assign oxidation numbers to the atoms in each substance. a. Cl2 b. GeO2 c. Ca(NO3)2 Solution a. Cl2 is the elemental form of chlorine. Rule 1 states that each atom has an oxidation number of 0. b. By rule 3, oxygen is normally assigned an oxidation number of −2. For the sum of the oxidation numbers to equal the charge on the species (zero), the Ge atom is assigned an oxidation number of +4. c. Ca(NO3)2 can be separated into two parts: the Ca2+ ion and the NO3− ion. Considering these separately, the Ca2+ ion has an oxidation number of +2 by rule 2. Now consider the NO3− ion. Oxygen is assigned an oxidation number of −2, and there are three of them. According to rule 4, the sum of the oxidation numbers on all atoms must equal the charge on the species, so we have the simple algebraic equation x + 3(−2) = −1 where x is the oxidation number of the N atom and the −1 represents the charge on the species. Evaluating for x, x + (−6) = −1x = +5 Thus the oxidation number on the N atom in the NO3− ion is +5. Exercise 7.9.1: Assigning Oxidation States Assign oxidation numbers to the atoms in the following: a. H3PO4 b. MgO Answer a H: +1; O: −2; P: +5 7.9.2 https://chem.libretexts.org/@go/page/47511 Answer b Mg: +2, O: −2 All redox reactions occur with a simultaneous change in the oxidation numbers of some atoms. At least two elements must change their oxidation numbers. When an oxidation number of an atom is increased in the course of a redox reaction, that atom is being oxidized. When an oxidation number of an atom is decreased in the course of a redox reaction, that atom is being reduced. Thus oxidation and reduction can also be defined in terms of increasing or decreasing oxidation numbers, respectively. Example 7.9.2: Formation of Sodium Bromide Identify what is being oxidized and reduced in the following redox reaction. 2 Na + Br → 2 NaBr 2 Solution Both reactants are the elemental forms of their atoms, so the Na and Br atoms have oxidation numbers of 0. In the ionic product, the Na+ ions have an oxidation number of +1, while the Br− ions have an oxidation number of −1. 2 N a + Br2 → 2 N aBr 0 0 +1−1 Sodium is increasing its oxidation number from 0 to +1, so it is being oxidized; bromine is decreasing its oxidation number from 0 to −1, so it is being reduced: Equation of sodium and bromine reacting, labeled with sodium's oxidation number change from 0 to +1 and bromine's oxidation number change from 0 to -1 Because oxidation numbers are changing, this is a redox reaction. The total number of electrons being lost by sodium (two, one lost from each Na atom) is gained by bromine (two, one gained for each Br atom). Exercise 7.9.2: Oxidation of Carbon Identify what is being oxidized and reduced in this redox reaction. C + O2 → C O2 Answer C is being oxidized from 0 to +4; O is being reduced from 0 to −2 Oxidation reactions can become quite complex, as attested by the following redox reaction: + − 2+ 6H (aq) + 2 M nO (aq) + 5 H2 O2 (l) → 2 M n (aq) + 5 O2 (g) + 8 H2 O(l) 4 −1 +2 0 +7 To demonstrate that this is a redox reaction, the oxidation numbers of the species being oxidized and reduced are listed; can you determine what is being oxidized and what is being reduced? This is also an example of a net ionic reaction; spectator ions that do not change oxidation numbers are not displayed in the equation. Eventually, we will need to learn techniques for writing correct (i.e., balanced) redox reactions. Combustion Reactions A combustion reaction is a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O as one reactant. The combustion of hydrogen gas produces water vapor. 2 2H (g) + O (g) → 2 H O (g) (7.9.6) 2 2 2 7.9.3 https://chem.libretexts.org/@go/page/47511 Notice that this reaction also qualifies as a combination reaction. Figure 7.9.2 : Explosion of the Hindenberg. The Hindenberg was a hydrogen-filled airship that suffered an accident upon its attempted landing in New Jersey