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The Major Classes of Chemical Reactions Chapter 4 Learning Goals: Chapter 4 4.1 The Role of Water as a Solvent: Terms 4.2 Writing Equations for Aqueous Ionic Reactions 4.3 Precipitation Reactions 4.4 Acid-Base Reactions 4.5 Oxidation-Reduction (Redox) Reactions 4.6 Elements in Redox Reactions 4.7 Reversible Reactions: An Introduction to Chemical Equilibrium Water is a polar molecule due to oxygen’s ability to attract electrons (called electronegativity) more so than hydrogen. e- It’s just symbolic language eO H H O H H Ionic compounds dissolve when ion-water forces are larger than ion-ion forces holding the ionic crystal together. “Likes dissolve likes” Oxygen atom “sucks” electron density from hydrogen atoms leading to a “polar molecule”. Ionic substances conduct electricity, covalent substances do not. Strong Electrolyte – 100% dissociation into ions NaCl (s) H 2O Na+ (aq) + Cl- (aq) Weak Electrolyte – not completely dissociated CH3COOH CH3COO- (aq) + H+ (aq) Non-Electrolyte – no dissociation The dissolution of an ionic compound in water CH3OH An electrolyte is a substance that when dissolved in water results in a solution that can conduct electricity. A non-electrolyte is a substance that when dissolved, results in a solution that does not conduct electricity. Solubility (S) is the maximum amount of a solute that can dissolve in a fixed quantity of a solvent at a specified temperature. (Units of g solute/100 g water) Examples: Sucrose (sugar) - 203 g per 100 g H2O NaCl - 39.12 g per 100 g H2O (very soluble) AgCl - 0.0021 g per 100 g H2O (insoluble) Non-electrolyte No Dissociation Strong Electrolyte 100% Dissociation Weak Electrolyte Little dissociation Ionic compounds dissociate into ions thus its chemical formula tells us the number of moles of different ions in solution. How many moles of each ion are in the following solutions? (a) 5.0 mol of ammonium sulfate dissolved in water (b) 78.5 g of cesium bromide dissolved in water (c) 7.42 x 1022 formula units of copper(II) nitrate dissolved in water Covalent compounds do not dissociate! (e) 35 mL of 0.84 M glucose (C6H12O6) What does a 3.5 M FeCl3 mean? H 2O 2NH4+(aq) + SO42-(aq) (a) (NH4)2SO4(s) 5.0 mol (NH4)2SO4 2 mol NH4+ 1 mol (NH4)2SO4 = 10. mol NH4+ 5.0 mol SO42- H 2O (b) CsBr(s) Cs+(aq) + Br-(aq) mol CsBr 78.5 g CsBr 212.8 g CsBr (c) Cu(NO3)2(s) H 2O = 0.369 mol CsBr = 0.369 mol Cs+ = 0.369 mol Br- Cu2+(aq) + 2NO3-(aq) = 0.123 mol Cu2+ 7.42x1022 mol Cu(NO3)2 formula = 0.123 mol Cu(NO3)2 units Cu(NO3)2 6.022x1023 formula units = 0.246 mol NO 3 H 2O (d) ZnCl2(aq) Zn2+(aq) + 2Cl-(aq) 1L 0.84 mol ZnCl2 35 mL ZnCl2 = 2.9x110-2 mol ZnCl2 L 103mL = 2.9x110-2 mol Zn2+ = 5.8x10-2 mol Cl- What does 3.5 M FeCl3 mean? 3.5 M FeCl3 = 3.5 moles FeCl3 1 Liter solution (1) a homogeneous solution of 3.5 moles of dry 100% pure FeCl3 dissolved in 1.00 Liter total solution volume (not 1 L of liquid!). (3) Note: It does not mean 3.5 moles of FeCl3 is dissolved in 1.00 liter of water! (4) [Fe3+] = 3.5M and [Cl-] = 3 x 3.5 M (5) It can be used as a conversion factor - There are 3-classes of chemical reactions that occur in aqueous solution. 1. Precipitation Reaction-– an insoluble solid is formed from specific cation-anion combinations. 2. Acid-Base Reaction-– a protons donor substance reacts with a hydroxide donor substance forming a salt and water. Precipitation Involves Substances Cations Anions Involves Substances Involves Substances H+ ions OH- ions Combine to Form Insoluble Precipitate Salt and H2O Predicted by In a precipitation reactions a metal cation and nonmetal anion combine & form an insoluble solid that “precipitates”. Oxidation Reduction Combine to Form 3. Oxidation-Reduction Reaction-electron donor substances react with react with substances that accept electrons. Acid-Base Neutralization Solubility Rules Oxidation Reduction Which is the Loss of e- Gain of e- Which is called Reducing Agent Oxidizing Agent How Do We Predict If A Precipitate Rx Will Occur? We memorize solubility rules and apply the rule “if it can happen, it will happen”. AgNO3(aq) + Na2CrO4(aq) => Ag2CrO4 (s) + NaNO3(aq) Precipitation occurs when ionic attractive forces overcome water’s tendency to hydrate and dissolve. Spectator ions: any species that remains soluble. Solubility Rules For Ionic Compounds Silberberg The “driving force” of a precipitation reaction is the act of PRECIPITATION. Soluble Ionic Compounds 1. Salts of Group 1A and ammonium ion (NH4+) 2. NO3-, CH3COO- or C2H3O2- , ClO43. Cl-, Br- I- except those of Ag+, Pb2+, Cu+, and Hg22+. Insoluble Ionic Compounds 1. All OH, except those of Group 1A and the larger members of Group 2A (beginning with Ca2+). 2. CO32- and PO43-) are insoluble, except those of Group 1A(1) and NH4+. 3. All S2- except those of Group 1A, Group 2A and NH4+. If it can happen, it will. How To Predict Whether A Precipitation Reaction Example: Pb(NO3)2(aq) + NaI(aq) ==> ? 2. Consult the Solubility table to see if PbI2 or NaNO3 are insoluble. 3. PbI2 is insoluble so that reaction will occur. 1. Note the ions present in the reactants. 2. Consider the possible cation-anion combinations. 3. Refer to the table of solubility rules and decide whether any of the ion combinations is insoluble. 4. If a combination is insoluble, that RXN will occur. 5. Write the molecular, ionic and net ionic equation for the reaction. There are 3-types of equations that are written for precipitation, acid-base and redox reactions 1. The molecular equation shows all reactants and products as undissociated compounds. precipitate PbI2(s) + 2NaNO3(aq) 2. The total ionic equation shows all of the soluble ionic substances dissociated into ions. Pb2+ + 2NO3- + 2Na+ + 2I- Pb(NO3)2(aq) + NaI(aq) ==> PbI2(s) 1. Recognize => IONIC => Solubility Rules Apply Look at ions that are possible from the reactants, then use “Solubility Rules” ! 2NaI(aq) + Pb(NO3)2(aq) Example: The reaction of the salts: PbI2 (s) + 2Na+ + 2NO3- Na+ and NO3- are spectator ions---they don’t do much but sit there! 3. The net ionic equation eliminates the spectator ions and shows the actual chemical change taking place. Pb2+ + 2IPbI2 (s) Learning Check: Precipitation Reactions Predict whether a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions. (a) sodium sulfate(aq) + strontium nitrate(aq) (b) ammonium perchlorate(aq) + sodium bromide(aq) (c) silver nitrate(aq) + sodium chromate(aq) 4. Now write 3-equations 1)molecular, 2) ionic and 3) net equation as: 2NaI(aq) + Pb(NO3)2(aq) 2Na+ + 2I- + Pb2+ + 2I-(aq) + Pb2+(aq) 2NO3 PbI2(s) + 2NaNO3(aq) - PbI2 + 2Na+ + 2NO3PbI2(s) Putting It All Together: Predicting Precipitation 0. Know how to read a Table of Solubility Rules 1. Know ionic nomenclature so you can write the correct ionic formula of reactants and products. 2. Write the molecular equation by writing the chemical formula for reactants and products. 3. Break the compounds into their ions and write the ionic equation for the reaction. 3. Refer to the table of solubility rules and decide whether any of the ion combinations is insoluble. 4. If a candidate is insoluble, that reaction will occur. 5. Remove the spectator ions and write the net ionic equation that summarized the reaction. Solubility Rules For Ionic Compounds in Water 1. Convert names formulas, write a balanced equation showing intact compounds (not dissociated). Molecular Equation Na2SO4(aq) + Sr(NO3)2 (aq) Molecular Equation NH4ClO4(aq) + NaBr (aq) NH4Br (aq) + NaClO4(aq) 2NaNO3(aq) + SrSO4(s) 2. Dissociate the formula equation to an ionic equation showing cations and anions with charge and phase. Ionic Equation Examing the solubility table shows that none of the possible combinations of ions would result in an insoluble precipitate (NH4ClO4 is soluble as is NH4Br, NaBr and NaClO4) In this case all ions are spectator ions. 2Na+(aq) +SO42-(aq)+ Sr2+(aq)+2NO3-(aq) 2Na+(aq) +2NO3-(aq)+ SrSO4(s) Ionic Equation 3. Cancel the spectator ions to obtain the net ionic equation Net Ionic Equation SO42-(aq)+ Sr2+(aq) NH4+ + ClO4+(aq) + Na+ + Br- (aq) NH4+ + Br -(aq) + Na+ + ClO4 -(aq) SrSO4(s) Does a precipitate form when silver nitrate is mixed with a solution of sodium chromate? Write the molecular, ionic and net ionic equations for the reaction. Write the net ionic equation for the reaction of silver nitrate with sodium chromate. 1. Write the balanced molecular equation. 2. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions. 3. Cancel the spectator ions on both sides of the ionic equation to obtain the net ionic equation 2. Acid-Base Reactions There are 3-classes of chemical reactions that occur in aqueous solution. 1. Precipitation Reaction-– an insoluble solid is formed from specific cation-anion combinations. 2. Acid-Base Reaction-– a protons donor substance reacts with a hydroxide donor substance forming a salt and water. 3. Oxidation-Reduction Reaction-electron donor The effects of acid rain on a statue of George Washington taken in 1935 (left) and 2001 (right) marble. substances react with react with substances that accept electrons. Precipitation Involves Substances Cations Anions Acid-Base Neutralization Oxidation Reduction Involves Substances Involves Substances H+ ions OH- ions Combine to Form Combine to Form Insoluble Precipitate Salt and H2O Oxidation Reduction Which is the Loss of e- Predicted by Which is called Solubility Rules Chemistry uses 3-definitions of acids and bases: 1) Arrehenius 2) Bronsted-Lowry 3) Lewis Gain of e- Reducing Agent Oxidizing Agent Acid: substance that has a covalent H atom in its formula, and releases a proton H+ when dissolved in water. H 2O Chem 7 HA(g) ==> H+(aq) + A-(aq) Chem 11 Name Acid Definition Base Definition Arrhenius Substance that increases H+ Substances that increase OH- BrønstedLowry Substances that donate H+ Substances that accept H+ Lewis Electron-pair acceptor Electron-pair donar HCl(g), HBr(g), HI (g), HNO3(l), H2SO4(l), HClO4(l) HCl(aq), HBr(aq), HI (aq), HNO3(aq), H2SO4(aq), HClO4(aq) Chemists symbolize the reactions of acids with water two confusing ways. Don’t let it bug you. H+ representation H 2O H3O+ Representation HCl(g) ===> H+ + Cl- HCl(g) + H2O => H3O+ + Cl- H 2O HNO3(l) ===> H+ + NO3- HNO3(g) + H2O => H3O+ + NO3- Base: a substance that contains OH in its formula, and releases hydroxide ions (OH-) when dissolved in water. H 2O MOH(s) ==> OH- + M+ NaOH(s), KOH (s), LiOH (s), Mg(OH)2(s), Ca(OH)2(s) NaOH(aq), KOH (aq), LiOH (aq), Mg(OH)2(aq), Ca(OH)2(aq) Acids and bases are classified as either strong or weak. We use arrows to symbolize the difference. 100% ionized = strong electrolyte = ==> arrow Strong Acid H2SO4(aq) ==> 2H+(aq) + SO42-(aq) Strong Base KOH(aq) ==> OH-(aq) + K+ (aq) <20% ionized = weak electrolyte = <==> arrow H 2O HA(g) ===> H+ + AGeneralized Acid HA HA(g) + H2O => H3O+ + AGeneralized Acid HA Weak Acid HNO2(aq) <==> H+(aq) + NO2-(aq) Weak Base NH4OH(aq) <==> OH-(aq) + NH4+ (aq) Strong acids and strong bases dissociate completely, conduct electricity and are strong electrolytes. Strong acids and strong bases completely dissociate or completely ionize in water: HA H+ + A- HA(g) + H2O(l) Before dissociation Uni-directional arrow used to heavily product-favored H3O+(aq) + A-(aq) After dissociation AH3O+ HCl(aq) H+(aq) + Cl-(aq) HNO3(aq) H+ (aq) + NO3-(aq) H2SO4(aq) Ca(OH)2(aq) Na+(aq) + OH-(aq) Ca2+(aq) + 2OH-(aq) HA H3O+ A- We must memorize common strong acids and strong bases. All are strong electrolytes that dissociate completely in solution. Strong Bases Strong Acids hydrochloric acid HCl hydrobromic acid HBr lithium hydroxide: LiOH sodium hydroxide: NaOH hydroiodic acid HI nitric acid HNO3 sulfuric acid H2SO4 perchloric acid HClO4 potassium hydroxide: KOH calcium hydroxide Ca(OH)2 strontium hydroxide Sr(OH)2 barium hydroxide H-X acids Oxide containing acids Ba(OH)2 Group I and II Hydroxides Weak acids and weak bases dissociate only to a slight extent in water Ka << 1. HA HA(aq) + H2O(l) H+ + AH3O+(aq) + A-(aq) Remember anything not strong is weak! Common Weak Acids and Their Anions Acid HF CH3COOH HCN HNO2 H2CO3 H2SO3 H3PO4 (COOH)2 Anion FCH3COOCNNO2CO32SO32PO43(COO)22- All not strong is weak! After dissociation HNO2 (aq) <=> NO2- (aq) + H+(aq) CH3CO2H(aq) <=> CH3CO2- (aq) + H+ HA HA HA HA HA HA HA HA H3O+ A- Anion Name fluoride ion acetate ion cyanide ion nitrite ion carbonate ion sulfite ion phosphate ion oxalate ion Weak acids and weak bases do not dissociate or ionize to a large extent in solution, and so are weak electrolytes (small value of Ka). reactant-favored Before dissociation Strong H+ (aq) + HSO4-(aq) NaOH(aq) HA Strong H2SO3 (aq) <=> HSO3- (aq) + H+(aq) Acids can have one, two or three acidic protons depending on their structure. Monoprotic acids--only one H+ available HCl H+ + ClHNO3 H+ + NO3CH3COOH H+ + CH3COO- Acids and bases have distinct properties. Acids: • • Strong electrolyte, strong acid • Strong electrolyte, strong acid • Weak electrolyte, weak acid Diprotic acids--two acidic H+ available for reaction H2SO4 H+ + HSO4- Strong electrolyte, strong acid HSO4- H+ + SO42- Weak electrolyte, weak acid Acid Acrid sour taste React with metals (Group I,II) to yield H2 gas Changes plant dye litmus from blue to red React with carbonates and bicarbonates to produce CO2 gas 200 Million MT H2SO4 Base Bases: Triprotic acids--three acidic H+ H3PO4 H2PO4HPO42- H+ + H2PO4 H+ + HPO42H+ + PO43- • 50 Million MT NaOH/yr Bitter taste 3 million containers Slippery feel Changes plant dye litmus from red to blue • React and neutralizes the effects of acids • - • Weak electrolyte, weak acid Acids and bases are everywhere. Common Acids , soft drinks Most anti-perspirants, water treatment plants, paper Determining the Molarity of H+ Ions in Aqueous Solutions of Acids Nitric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H+ ion. What is the molarity of H+(aq) in 1.4M nitric acid? What is the H+ molarity of 0.70 M H2SO4? Common Bases Of H+ .466 M H3PO4? Of 2.5 M NaOH ? Determining the Molarity of H+ Ions in Aqueous Solutions of Strong Acids or Bases Nitric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H+ ion. What is the molarity of H+(aq) in 1.4M nitric acid? One mole of H+(aq) is released per mole of nitric acid (HNO3) HNO3(l) H 2O H+(aq) + NO3-(aq) Acids react with bases in a chemical reaction called a “neutralization reaction” forming a salt and water. NaOH (aq) + HCl (aq) ==> H2O + Na+ + Clbase + acid ==> Water + salt 1.4M HNO3(aq) is 1.4M H+(aq). What is the molarity of H+ in a 0.70 M H2SO4? What is the molarity of H+ in a 0.466 M H3PO4? A salt’s cation comes from a parent base and anion comes from a parent acid! Chemists use general symbols to represent the “neutralization reaction” to form a salt and water. NaOH (aq) + HCl (aq) ==> H2O + Na+ + Clbase + acid ==> Water + salt Learning check: write the molecular, ionic and net equation for the neutralization between calcium hydroxide and sulfuric acid. 1. The molecular equation shows all reactants and products as undissociated compounds. Ca(OH)2 (aq) + H2SO4 (aq) ===> 2H2O + CaSO4 2. The total ionic equation MOH (aq) + HX (aq) ==> H2O + Alkali Metal Cation M+ + X- Halide Anion Shows all soluble ionic substances dissociated into ions. Ca2+ + 2OH- + 2H+ + SO42- ===> 2H2O + Ca2+ + SO42Ca2+ and SO42- are spectator ions---they just watch! 3. The net ionic equation Eliminate spectator ions and show actual chemical change H+(aq) + OH- (aq) ===> H2O(l) Learning check: Write molecular, ionic and net equations for the following acid base neutralization reactions. Writing Ionic Equations for Acid-Base Reactions strontium hydroxide(aq) + perchloric acid(aq) (a) Sr(OH)2(aq)+2HClO4(aq) strontium hydroxide(aq) + perchloric acid(aq) 2H2O(l)+Sr(ClO4)2(aq) (b) Sr2+(aq)+2OH-(aq)+ 2H+(aq)+2ClO4-(aq) 2H2O(l)+Sr2+(aq)+2ClO4-(aq) barium hydroxide(aq) + sulfuric acid(aq) Nitric acid(aq) + barium hydroxide(aq) (c) 2OH-(aq)+ 2H+(aq) barium hydroxide(aq) + sulfuric acid(aq) (b) Ba(OH)2(aq) + H2SO4(aq) Acetic acid(aq) + potassium 2H2O(l) 2H2O(l) + BaSO4(aq) Ba2+(aq) + 2OH-(aq)+ 2H+(aq)+ SO42-(aq) 2H2O(l)+Ba2+(aq)+SO42-(aq) (c) 2OH-(aq)+ 2H+(aq) Weak acids dissociate to a very small extent and this fact is reflected in their equations as well using a double-arrow (<==>). 2H2O(l) Learning Check: Acids and Bases 1. Name and characterize each as a base, acid or salt? HF(g), HI(aq), LiOH(aq), Mg(OH)2, Na2SO4 CH3COONH4 Molecular equation 2. Name and classify the following as strong, weak acid or base? NaOH(aq) + CH3COOH(aq) CH3COONa(aq) + H2O Weak acid is not dissociated! Total ionic equation Na+(aq)+ OH-(aq) + CH3COOH(aq) CH3COO-(aq) + Na+(aq) + H2O(l) Net ionic equation OH-(aq) + CH3COOH(aq) CH3COO-(aq) + H2O(l) HClO4, Sr(OH)2, HClO2, NH3(g), H3PO4(aq), H2SO4(aq), HNO3(aq) 3. Write the M/I/NI equation for the reactions of a) hydrochloric acid with calcium hydroxide and b) phosphoric acid with sodium hydroxide 4. What acids and bases were reacted to form the following salts? Show using balanced equations. 1) NaNO2 2) CaSO4 3) Mg(PO4)2 Learning check: 1. Identify the following as a strong or weak acid or base a salt. If a salt what is the parent acid and base? HF, HI, LiOH, Ca(OH)2, Na2SO4 CH3COO-, NH4+ 2. Classify the following as strong, weak acid or base? HClO4, Sr(OH)2, HClO2, NH3, H3PO4, H2SO4 3. What is the correct formula of the salt formed in the neutralization reaction of hydrochloric acid with calcium hydroxide? Name the type of reaction and write M/I/NI ionic equations for the following: 1. 2. 3. 4. 5. 6. 7. hydrochloric acid and potassium hydroxide sodium carbonate and hydrochloric acid aluminum nitrate and sodium phosphate potassium chloride + iron(II) nitrate Iron sulfide and hydrochloric acid acetic acid and magnesium hydroxide sodium sulfite and hydrochloric acid 4. What is the chemical formula of the salt produced by the neutralization of sodium hydroxide with sulfurous acid? Reactions between acids and carbonate containing compounds yield gaseous CO2 and H2O + salt. Oxidation Reduction Involves simultaneous Molecular equation Oxidation NaHCO3(aq) + CH3COOH(aq) Reduction Which is the CH3COONa(aq) + CO2(g) + H2O(l) Loss of eTotal ionic equation Gain of e- Which is called Na+(aq)+ HCO3-(aq) + CH3COOH(aq) Reducing Agent CH3COO-(aq) + Na+(aq) + CO2(g) + H2O(l) Oxidizing Agent Which Net ionic equation Increases oxidation number HCO3-(aq) + CH3COOH(aq) CH3COO-(aq) + CO2(g) + H2O(l) Redox reactions trends can be understood by applying the properties of elements in the periodic table. Non-metals tend to gain electrons (be reduced) forming anions which are oxidizing agents. An oxidation-reduction reaction occurs when there is a net movement of electrons from one reactant to another. One reactant gives the other receives! 2Mg(s) + O2 (g) 2MgO (s) Oxidized Reduced Oxidizing Perspective Mg loses electron(s) Metals tend to loose electrons (oxidized) forming cations Metals are reducing agents Decreases oxidation number Mg(s) is oxidized Mg is the reducing agent Mg increases its oxidation number We learn rules to figure out what is oxidized and what is reduced! Reduction Perspective O gains electron(s) O is reduced O is the oxidizing agent O decreases its oxidation number Oxidation and reduction occur simultaneously. 2Mg(s) + O2 (g) 2MgO (s) A set of rules called “assigning oxidation numbers” is used to identify what species is being oxidized and reduced in a chemical reaction. 1. The oxidation state of a free elements in atomic or molecular for are assigned an oxidation number of 0. Oxidized Reduced Na, Be, K, Pb, H2, O2, P4 = 0 2Mg2+ + 4e- Oxidation half-reaction (lose e-) 2Mg O2 + 4e- 2O2- Reduction half-reaction (gain e-) 2. The oxidation number of a an monoatomic ion is equal to the charge on the ion (Group Number Rule) Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 2Mg + O2 + 4e2Mg + O2 2Mg2+ + 2O2- + 4e2MgO Ionic Equation Net Equation 4. Fluorine has an oxidation state of –1 in its compounds 3. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. H2O (H = +1 and O = 2) Sum = 0 Quick Table Assigning Oxidation Numbers 5. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. 1. The oxidation state of any free element = 0 6. The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1 (exceptions). 3. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. 7. When there is a conflict the lower numbered rules take priority over the higher numbered rules. Oxidation numbers of all the elements in HCO3- ? HCO3O = -2 H = +1 3x(-2) + 1 + ? = -1 C = +4 Assign oxidation numbers to the following ions and the elements within a compound. IF7 HCO3O = -2 H = +1 3x(-2) + 1 + ? = -1 C = +4 7x(-1) + ? = 0 I = +7 K2Cr2O7 O = -2 F = -1 K = +1 7x(-2) + 2x(+1) + 2x(?) = 0 Cr = +6 NaIO3 Na = +1 O = -2 3x(-2) + 1 + ? = 0 I = +5 2. In monatomic ions, the oxidation number is equal to the charge on the ion (Use Group Number) 4. Fluorine has an oxidation state of –1 in its compounds 5. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds (–1). 6. The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1 (exceptions). 7. When there is a conflict the lower numbered rules take priority over the higher numbered rules. Assign oxidation numbers to the following ions and the elements within a compound. 1.! Na2SO3 ! 1.!Na +1 ! 2.!S +4 ! 3.!O -2 2.! PF3 ! 1.!P +3 ! 2.!F -1 3.! CrO32! 1.!Cr +4 ! 2.!O -2 4.! Cr2O72! 1.!Cr +6 ! 2.!O -2 5.! (NH4)3PO4 ! 1.!N -3 ! 2.!H +1 ! 3.!P +5 ! 4.!O -2 Elements that make up compounds can take on a range of “oxidation states”. Example: N Chemists use oxidation rules to identify what is being oxidized and what is being reduced in a chemical reaction. The reactions below occur as written. Use the rules for assigning oxidation numbers to the oxidizing and reducing agents in the reactions below? (increase in ON = reduced, decrease = oxidized) 0 -4 +1 CH4 + 2O2 +2 -2 2HgO -4 -2 +1 -2 CO2 + 2H2O 0 0 2Hg + O2 +2 0 +2 0 Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Identify the oxidizing agent and reducing agent in each of the following: Zn (s) + CuSO4 (aq) Identify the oxidizing agent and reducing agent in each of the following: ZnSO4 (aq) + Cu (s) +2 +6 -2 0 Zn (s) + CuSO4 (aq) +2 +6 -2 0 ZnSO4 (aq) + Cu (s) Oxidized 1. Assign oxidation numbers 2. Identify the species being oxidized (reactant that increases in oxidation number) and the specie being reduced (reactant that decrease in oxidation number). 3. The species that is oxidized is the reducing agent, the species that is reduced is called the oxidizing agent. Reduced Zn Zn2+ + 2e- Zn is oxidized Cu2+ + 2e- Copper wire reacts with silver nitrate to form silver metal and soluble cupric nitrate. What is the oxidizing agent and the reducing agent in this reaction? Zn is the reducing agent Cu Cu2+ is reduced Cu2+ is the oxidizing agent Copper wire reacts with silver nitrate to form silver metal and soluble cupric nitrate. What is the oxidizing agent and the reducing agent in this reaction? (Translate words to equations) 0 +1 Cu (s) + 2AgNO3 (aq) Oxidized Cu is oxidized Ag+ is reduced +2 0 Cu(NO3)2 (aq) + 2Ag (s) Reduced Cu is the reducing agent Ag+ is the oxidizing agent Identify the oxidizing agent and reducing agent in each of the following: Identify the oxidizing agent and reducing agent in each of the following: +2 -2 (a) 2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g) (b) PbO(s) + CO(g) (c) 2H2(g) + O2(g) Pb(s) + CO2(g) 2H2O(g) +2 -2 (b) PbO(s) + CO(g) 0 +4 -2 Pb(s) + CO2(g) The O.N. of C increases; it is oxidized; CO is the reducing agent. The O.N. of Pb decreases; it is reduced; PbO is the oxidizing agent. 0 0 +1 -2 1. Know the rules for assigning oxidation numbers 2. Assign oxidation numbers to all reactants and products 3. Increasing ON = Reactant Oxidized (reducing agent) Decreasing ON = Reactant Reduced (oxidizing agent) Using oxidation numbers (c) 2H2(g) + O2(g) 2H2O(g) The O.N. of H increases; it is oxidized; it is the reducing agent. The O.N. of O decreases; it is reduced; it is the oxidizing agent. Balancing Redox Reactions • Few redox reactions are easy to balance by inspection. • Most redox reactions involve H+ or OH- as sometimes they are reactants or products. 1. When an element combines with O2 the element has been oxidized and O2 reduced. 2. When an element combines with a halogen the element is oxidized and the halogen reduced. 3. When a metal combines with something, the metal is been oxidized. Whatever the metal combined with, is reduced. Ion-Electron Method To Balance Redox Rxtns 1. Divide the equation into two half-reactions-oxidation and reduction 2. Balance all atoms other than H and O 3. Balance O by adding H2O 4. Balance H by adding H+ 5. Balance net charge by adding e6. Balance e- gain to e- loss and add the half-reactions 7. Add the half reactions and cancel common factors. 8. Add to both sides of the equation the same number of OHas there are H+ 9. Combine H+ and OH- to form H2O and cancel common H2O 10. Cancel common H2O • We normally use H+ or OH- in excess and we use them with water to balance a redox reaction. • We use the “Half-Reaction (Ion-Electron) Method” Balancing Redox Reactions In Acid Balancing the Equation SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq) Sulfite ion + permanganate ion → sulfate ion + managenese Note that the equation is not balanced in mass (atoms) or in charge! Ion-Electron (Half-reaction) Method 1. Determine the oxidation states of the various species. Example 5-6 4. Balance O by adding H2O to the side that needs O: 4+ 6+ 7+ 2+ SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq) 2. Identify the oxidizing and reducing agents. Write the halfreactions 4+ 6+ SO32-(aq) → SO42-(aq) 7+ 2+ MnO4-(aq) → Mn2+(aq) H2O(l) + SO32-(aq) → SO42-(aq) MnO4-(aq) → Mn2+(aq) + 4 H2O(l) 5. Balance hydrogen by adding H+: H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq) 8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) 3. Balance atoms other than H and O: Already balanced for elements S and Mn Balancing Redox Equations 6. Add sides. e- to balance charge. Charge should be equal on both H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq) + 2 e-(aq) 5 e-(aq) + 8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) Balancing Redox Reactions 8. Add both equations and simplify: 5 SO32-(aq) + 2 MnO4-(aq) + 6H+(aq) → 9. Check the result 5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l) 10. If a basic solution is desired then add OH- to neutralize the H+ on both sides of the equation. Cancel common water. 7. Multiply the half-reactions to balance all e- in both equations: 5 SO32-(aq) + 2 MnO4-(aq) + 6H+ + 6OH- (aq) → 5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l) + 6OH- 5 H2O(l) + 5 SO32-(aq) → 5 SO42-(aq) + 10 e-(aq) + 10 H+(aq) 16 H+(aq) + 10 e-(aq) + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l) 5 SO32-(aq) + 2 MnO4-(aq) + 3H2O (aq) → 5 SO42-(aq) + 2 Mn2+(aq) + 6OH- Balancing Redox Reactions x2 4H+ + MnO4- +3e- x3 C2O42- + 2H2O • Method used to balance redox reactions 3+ 2+ 2+ 2CO32- + 4H+ + 2e- 4+ Fe3+(aq) + Sn2+ → Fe2+ + Sn4+ Oxidation: Sn2+ → Sn4+(aq) + 2 e- Reduction: Fe3+(aq) + 1 e- → Fe2+ Atoms and mass are balanced but charge is not. Let’s multiply by 2 to get charge to cancel Sn2+ → Sn4+(aq) + 2 e2 Fe3+(aq) + 2 e- → 2Fe2+ Overall: MnO2+ 2H2O Fe3+(aq) + Sn2+ → Fe2+ + Sn4+ 8H+ + 2MnO4- + 6e3C2O42- + 6H2O 2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l) + 4OH- 2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq) 2MnO2+ 4H2O 6CO32- + 12H+ + 6e- 2MnO2(s) + 6CO32-(aq) + 4H+(aq) + 4OH- 2MnO2(s) + 6CO32-(aq) + 2H2O(l) The redox reaction between dichromate and iodide anions Cr2O72- Cr3+ + I2 I- Balancing redox reactions by the half-reaction method PROBLEM: Permanganate ion is a strong oxidizing agent and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate anion to form carbonate anion and solid manganese dioxide. Balance the skeleton ionic reaction that occurs between KMnO4 and Na2C2O4 in basic solution. MnO4-(aq) + C2O42-(aq) PLAN: MnO2(s) + CO32-(aq) Proceed in acidic solution and then neutralize with base. SOLUTION: +7 MnO44H+ + MnO4+ 3e- MnO4- R C2O42- MnO2 O +4 MnO2 +3 C2O42- MnO2 + 2H2O C2O42- + 2H2O CO32- +4 2 CO 23 + 2CO32- + 4H + 2e- Balancing redox reactions in acidic solution Cr2O72-(aq) + I-(aq) → Cr3+(aq) + I2(aq) 1. Divide the equation into two half-reactions---an oxidation and a reduction use oxidation numbers if necessary. 2. Balance all atoms other than H and O 14H+(aq) + Cr2O72-(aq) + 6I-(aq) 3. Balance O by adding H2O (count) 2Cr3+(aq) + 3I2(aq) + 7H2O(l) 4. Balance H by adding H+ 5. Balance net charge by adding e6. Balance e- gain to e- loss and add the half-reactions 7. Add the half reactions and cancel common factors. 8. For basic reactions neutralize the H+ with OH- but add it on both sides. Cancel common water. Balancing redox reactions in basic solution ❐ Add OH- to neutralize the H+ ions. 14H+(aq) + Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l) + 14OH-(aq) + 14OH-(aq) 2- 14H2O + Cr2O7 + 6 I- 2Cr3+ + 3I2 + 7H2O + 14OH- Show a balanced equation for the oxidation of Fe2+ to Fe3+ by Cr2O72- (to Cr3+) in acid solution? 1. Write the unbalanced equation for the reaction ion ionic form. Fe2+ + Cr2O72- 2. Identify what is being oxidized and reduced using oxidation numbers. Separate the equation into two half-reactions. +2 Oxidation: ❐ Reconcile the number of water molecules. 7H2O + Cr2O72- + 6 I- 2Cr3+ + 3I2 + 14OH- Do a final check on atoms and charges. Fe3+ + Cr3+ Reduction: +3 Fe2+ Fe3+ Cr2O72- Cr3+ +6 +3 3. Balance the atoms other than O and H in each half-reaction. Cr2O72- 2Cr3+ 4. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms. Cr2O722Cr3+ + 7H2O + 214H + Cr2O7 2Cr3+ + 7H2O 5. Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe2+ Fe3+ + 2Cr3+ + 7H2O 6e- + 14H+ + Cr2O72- How Can We Predict Redox Reactions? Show the oxidation and reduction that occur and write an overall ionic equation for the reaction of hydrochloric acid with iron to produce hydrogen and iron(II). 1. Translate words to chemical formula HCl + Fe(s). 2. What will be oxidized and reduced by looking at the activity series 3. Write the half-reactions maintaining mass and charge balance 4. Sum the two half-reactions 2H+ (s) + 2e- (g) Fe(s) + 2H+(aq) Fe2+(aq) + 2 e+ H2(g) Binary Ionic Compound 2NaCl 2. Non-Metal + Non-metal 0 S + O2 3. Compound + Element +3 0 PCl3+ Cl2 4. Compound + Compound +2 -2 +1 -2 CaO(s)+ H2O Reduction: 6e- + 14H+ + Cr2O7214H+ + Cr2O72- + 6Fe2+ 2Cr3+ + 7H2O 6Fe3+ + 2Cr3+ + 7H2O 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 9. For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation. There are 5-Main Types of Redox Reactions 1. Combination A + B => AB -produce fewer products than reactants. i. Element + Element => Compound ii. Element + Compound => New Compound iii. Compound + Compound => New Compound 2. Decomposition AB => A + B 3. Displacement AB + C => AC + B -produce more products than reactants. -methathesis (switching places) element in compound is both oxidized and reduced Fe2+(aq) Combination reactions produce fewer products than reactants. 0 6Fe3+ + 6e- 4. Disproportionation--common element or H2 1. Combination Reactions: A + B => C 1. Metal + Non-metal 2Na(s)+ Cl2 6Fe2+ 8. Verify that the number of atoms and the charges are balanced. 2Cr3+ + 7H2O Fe(s) Oxidation: 1e- 6. If necessary, equalize the number of electrons in the two halfreactions by multiplying the half-reactions by appropriate coefficients. 6Fe2+ 6Fe3+ + 6e6e- + 14H+ + Cr2O72- 7. Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. Binary Molec Cmpd 5. Combustion CH4 + O2 => CO2 + H2O 2. Decomposition Reactions --produces more products than reactants through the action of heat (thermal decomposition) or electricity (electrolytic decomposition). 1. Compound +4 -2 Element A + Element B electricity SO2 2H2O New Compound +5 -1 PCl5 New Compound +2 -2 +1 Ca(OH)2 2HgO heat 2H2 + O2 2Hg + O2 2. Compound +1 +5 -2 2KClO3 Compound A + Element B heat +1 -1 0 2KCl + 3O2 Activity Series For Metals 3. Displacement Reactions Pure metals higher in the series will be oxidized by a dissolved elemental ion below it in the series which is reduced. A more “reactive metal” will displace a “less reactive metal” or H2 within their compounds to form the oxidized metal and the reduced form of the less active metal or H2. double A + BC AC + B Methathesis = displacement +1 -2 0 H2 DISPLACEMENT Metals above H2 react with acids (H+), or H2O to displace H2(g). H+ reduced to H2. Al(s) + H+ => Al3+ + H2(g) Ag(s) + H+ => No reaction +2 -2 +1 0 Ca(s) + 2H2O(aq) → Ca(OH)2 + H2 H2 Displacement Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) +4 -1 0 0 +2 -1 0 TiCl4(aq) + 2Mg(s) Ti + 2MgCl2 +2 +2 0 Zn(s) + CuSO4 (aq) => ZnSO4 (aq) + Cu (s) METAL DISPLACEMENT More active metal + Salt of less active metal => Reduced Less active metal + Salt of more active metal. Al(s) + Fe3+(aq) => Al3+ + Fe(s) Zn(s) + CuSO4 => ZnSO4 + Cu(s) Metal Displacement Metal Displacement (Activity Series) Zn Rod Predict whether the following reactions will occur as written. 1. Fe(s) + 3Cu2+ => 2Fe3+ + Cu(s) 2. Sn(s) + Ca2+ => Sn2+ + Ca(s) CuSO4 3. Au(s) + 3AgNO3(aq) => Au(NO3)3 CuSO4 Solid Zn rod dunked A solid Zn rod into a solution dunked into a of solution4.of copper CuSO sulfate (CuSO4) + 3 Ag(s) Use activity series: Zn metal is higher than Cu in the series. It will displace Cu2+ from solution. The Zn is oxidized and the Cu2+ is reduced to solid Cu(s) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Strength as oxidizing agent A more active halogen will displace a less active (heavier) halide from their binary salts. 0 F2 +1 -1 0 +1 -1 Cl2 + 2NaI ===> I2 + 2NaCl Br2 + 2NaI ===> I2 + 2NaBr I2(s) + 2F- ===> No reaction Cl2 Br2 I2 Cl2 and Br2 both are higher and will displace Ifrom its compounds, forming I2 and the chloride and bromide salt respectively. A less active halogen will not displace a more active halide from their salts. Halogen + Salt of Less Active Halide gives Less Active Halide + salt of more active halide. 4. 2 Al(s) + Cr2O3(s) ! Al2O3(s) + 2 Cr(s) 5. Pt(s) + 4 HCl(aq) ! PtCl4(aq) + 2 H2(g) More active metal + Salt of less active metal => Reduced Less active metal + Salt of more active metal. 4. Disproportionation Reaction A single element in a compound is both oxidized and reduced (both oxidizing and reducing agent in one) +1 -1 -2 +1 +1 -2 H2O2 + 2OH- 0 -2 +1 Cl2 + 2OH- 0 2H2O + O2 +1 -2 -1 +1 -2 ClO- + Cl- + H2O Identifying the Type of Redox Reaction 5. Combustion Reaction In a combustion reaction a fuel or typically a hydrocarbon reacts with oxygen forming carbon dioxide and water. The oxygen is increases in O.N. and is therefore oxidized (reducing agent). Classify each of the following redox reactions as a combination, decomposition, or displacement reaction, write a balanced molecular equation for each, as well as total and net ionic equations for part (c), and identify the oxidizing and reducing agents: (a) magnesium(s) + nitrogen(g) +4 -1 0 +4 -2 CH4 + 2O2 (b) hydrogen peroxide(l) +1 -2 CO2 + 2H2O 2C4H10 + 13O2 (c) aluminum(s) + lead(II) nitrate(aq) 8CO2 + 10H2O Identifying the Type of Redox Reaction 0 0 +2 -3 3Mg(s) + N2(g) Mg3N2 (aq) (a) Combination Mg is the reducing agent; N2 is the oxidizing agent. +1 -1 +1 -2 H2O2(l) 2 H2O2(l) 0 H2O(l) + 1/2 O2(g) or (b)Decomposition 2 H2O(l) + O2(g) H2O2 is the oxidizing and reducing agent. 0 +2 +5 -2 Al(s) + Pb(NO3)2(aq) 2Al(s) + 3Pb(NO3)2(aq) +3 +5 -2 0 (c)Displacement Al(NO3)3(aq) + Pb(s) 2Al(NO3)3(aq) + 3Pb(s) Pb(NO3)2 is the oxidizing and Al is the reducing agent. magnesium nitride (aq) water(l) + oxygen gas aluminum nitrate(aq) + lead(s)