Chemical Reactions and Equations (Chapter 10) PDF

CHEMICAL REACTIONS AND CHEMICAL EQUATIONS Chapter 10  Hydrogen + Oxygen → Water H2 and O2 are elements with unique properties But, when bonds break between the atoms and new ones are formed, we get a brand new substance, water 2 H2(g) + O2(g) → 2 H2O(l) Selected Prop...

CHEMICAL REACTIONS AND CHEMICAL EQUATIONS Chapter 10  Hydrogen + Oxygen → Water H2 and O2 are elements with unique properties But, when bonds break between the atoms and new ones are formed, we get a brand new substance, water 2 H2(g) + O2(g) → 2 H2O(l) Selected Properties Hydrogen Oxygen Water Melting Point –434°C –219°C 0°C Boiling Point –253°C –183°C 100°C State at Room Temp Gas Gas Liquid Density 0.08988 g/L 1.429 g/L 1.00 g/cm3 @ 0°C & 1 atm @ 0°C & 1 atm @ 4°C & 1 atm Chemical Reactions Chemical changes in matter result in new substances Rearrangements and/or exchanges of atoms produce new molecules Elements are not changed during a chemical reaction, only their arrangement Reactants  Products Reaction Evidence Color Bright shirt + UV from sun light = faded shirt over time Temperature sensitive materials can change color when exposed to heat Formation of solid Precipitate: liquid + liquid → solid Gas formation Bubbles in liquid (CO2) Flammable gases (H2, methane) Smelly gases (phosphorus, sulfur, ammonia) Hot!! Temperature change Combustion reactions cause your cars engine to get REALLY hot Light emission Light from a fire or glow sticks Don’t be fooled… While the changes are evidence of chemical reactions, the are not definite proof Chemical analysis is required for that The initial substances must change in some fashion in order for a chemical reaction to have taken place Remember, water boiling is a physical property, but what do you see here? Kind of look familiar to this? BUBBLES!! BUBBLES!!  Chemical Equations Shorthand way of describing a reaction Provides information about the reaction Formulas of reactants and products States of reactants and products Relative numbers of reactant and product molecules that are required Can be used to determine weights of reactants used and products that can be made Always goes from reactants (left side) to products (right side) CH4(g) + O2(g) → CO2(g) + H2O (g) Reactants → Products Some special notes… Atomic elements in their elemental form are represented by their elemental symbols (Mg, C, Ne) Do not forget about the diatomic elements – I will refer to these by name, but you have to remember that oxygen is O2, iodine is I2, and hydrogen is H2 There is a huge difference between atomic oxygen, elemental oxygen, an oxide ion, and a peroxide ion! This will have a major impact on the chemical equations IF you mess up on the chemical formula, everything else is just downhill from there! Something should look wrong Let’s take a quick look at the masses The Law of Conservation of Mass states that you must have the same mass of product formed as reactant you started with 16.05 g/mol 32.00 g/mol 44.01 g/mol 18.02 g/mol + + CH4(g) + O2(g) → CO2(g) + H2O(g) 48.05 g/mol 62.03 g/mol As we know, mass can’t simply just appear… Magic atoms? Let’s look at the atoms, starting with oxygen The Law of Conservation of Mass states that you must have the same atoms appear before and after the reaction 2O + + 3O CH4(g) + O2(g) → CO2(g) + H2O(g) Again, oxygen doesn’t just appear out of nowhere, so something is up… We’re not done yet… What about hydrogen? 4H + + 2H CH4(g) + O2(g) → CO2(g) + H2O(g)  Balanced equations A chemical equation in which the numbers of each type of atom on both sides of the equation are equal To balance an equation, we enter coefficients in front of the chemical formula as needed to balance the number of each atom on either side NOTICE: these are not the same as subscripts When adding coefficients: New atoms are not formed during the reaction, nor do they disappear It changes the number of molecules in the equation, but it does not change the kinds of molecules  Back to the previous equation To balance this equation, you’d insert a 2 before the O2 and the H2O 80.05 g/mol 80.05 g/mol Masses: 64.00 g/mol 36.04 g/mol 16.05 g/mol 44.01 g/mol + + CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) Atoms: 1C 4H 4O 1C 4H 4O Final result Balanced equation 1 CH4 molecule reacts with 2 O2 molecules to form 1 CO2 molecule and 2 H2O molecules Implied coefficient of ‘1’ CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) Writing a balanced reaction The first thing you should ever do when dealing with these types of problems Follow these guidelines every single time and you’ll be fine: 1. Write the unbalanced equation by writing the chemical formulas of each of the reactants and products. 2. If an element occurs in one compound on both sides of the equation, balance it first. 3. If an element occurs as a free element on either side of the equation, balance it last. 4. If the balanced equation contains coefficient fractions, change them into whole numbers by multiplying the entire equation by the appropriate factor. 5. CHECK your work!!! Example Solid silicon dioxide and solid carbon react to produce solid silicon carbide and carbon monoxide gas. Write a balanced equation for the reaction. Step 1 SiO (s) + C(s) → SiC(s) + CO(g) 2 2 Step 2 1 Si2 O 1 Si 2O 1 Start with Si: Si is balanced Balance O next: O is unbalanced Add a coefficient of 2 to the ‘O unit’ on the right side 2 O → 2 O : O is now balanced Example Solid silicon dioxide and solid carbon react to produce solid silicon carbide and carbon monoxide gas. Write a balanced equation for the reaction. Step 3 SiO2(s) + 3 C(s) → SiC(s) + 2 CO(g) 1 3C 1C 2C Balance C next:: C is unbalanced Add a coefficient of 3 to the free C on the left side 3 C → 3 C : C is now balanced Step 4: No fractions, so not necessary Example Solid silicon dioxide and solid carbon react to produce solid silicon carbide and carbon monoxide gas. Write a balanced equation for the reaction. SiO2(s) + 3 C(s) → SiC(s) + 2 CO(g) Step 5 Reactants Products P 1 Si atom 1 Si atom 2 O atoms 2 O atoms 3 C atoms 1 + 2 = 3 C atoms Try this one out Liquid octane (C8H18) and gaseous oxygen react to form gaseous carbon dioxide and gaseous water C8H18(l) + O2(g) → 8CO2(g) + H2O(g) Step 2: Start with C 8 on the left, 1 on the right Add a coefficient of 8 on the right 8C→8C P Try this one out Liquid octane (C8H18) and gaseous oxygen react to form gaseous carbon dioxide and gaseous water C8H18(l) + O2(g) → 8CO2(g) + 9H2O(g) Step 2: Balance H next 18 on the left, 2 on the right Add a coefficient of 9 on the right 18 H → 18 H P Try this one out Liquid octane (C8H18) and gaseous oxygen react to form gaseous carbon dioxide and gaseous water 25 C8H18(l) + 2O2(g) → 8CO2(g) + 9H2O(g) Don’t forget about the coefficients Step 3: Balance O you’ve already added 2 on the left, and [8(2) + 9] = 25 on the right Since you need 25 on the left, start with a coefficient of 25 /2 before the O2 on the left side (we will fix this in a second): 25/2 × O2 = 25 O 25 O → 25 O P  Try this one out Liquid octane (C8H18) and gaseous oxygen react to form gaseous carbon dioxide and gaseous water 2×C [ 8H18(l) + 25 O2(g) → 2 8CO2(g) + 9H2O(g) ] Step 4: Fix fraction coefficients Since we have a fraction with the denominator of 2, we must multiply everything by 2 Try this one out Liquid octane (C8H18) and gaseous oxygen react to form gaseous carbon dioxide and gaseous water 2 C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g) Step 5: Reactants Products 2 × C8 = 16 C atoms 16 C atoms 2 × H18 = 36 H atoms 18 × H2 = 36 H atoms 25 × O2 = 50 O atoms (16 × O2) + 18 O P = 50 O atoms Polyatomics make things a bit easier… Al(s) + H2SO4(aq) → Al2(SO4)3(aq) + H2(g) Al and H exist as “free” elements, so do those last Remember when I was talking about polyatomics and I told you to keep them as a single entity? This is one major reason why! Don’t split up the atoms that make up the polyatomic if they stay the same on both sides Since there are 3 SO42– on the right side, you want 3 on the left, so add a 3 for a coefficient Polyatomics make things a bit easier… Al(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + H2(g) Then balance the rest from there 2 Al on the right, so put a 2 in front of the Al on the left Polyatomics make things a bit easier… 2 Al(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + H2(g) Then balance the rest from there 2 Al on the right, so put a 2 in front of the Al on the left Since there are now 6 H on the left (due to the 3 coefficient you added to the H2SO4 already), you want 6 on the right, so add a coefficient of 3 on the right Polyatomics make things a bit easier… 2 Al(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g) Now check the “units” Reactants Products 2 × Al = 2 Al atoms 1 × Al2 = 2 Al atoms 3 × H2 = 6 H atoms 3 × H2 = 6 H atoms 3 × S = 3 S atoms 1 × (S)3 = 3 S atoms 3 × O4 = 12 O atoms (O4)3 = 12 O atoms What happens when a solute dissolves? Attractive forces hold the particles of the solid solute together There are also attractive forces between the solvent molecules When we mix the solute with the solvent, there are attractive forces everywhere! Solvent-Solvent, Solute-Solute, Solvent-Solute… If the attractions between solute and solvent are strong enough, the solute will dissolve into its individual ions Charge distribution in water The unequal sharing of electrons in H2O causes the oxygen side of the molecule to have a partial negative charge (d–) and the hydrogen side to have a partial positive charge (d+) These different partial charges on different parts of the molecule cause ions of a particular charge (Na+ or Cl–) to be drawn to different regions of the molecule Salt and water For NaCl, the solvent-solute attractions overcome the solute-solute and solvent-solvent attractions Na+: the positive charge is attracted to the negative side of water’s dipole moment Cl–: the negative charge is attracted to the positive side of water’s dipole moment Salt water Since the attraction between Na+ and Cl– is weaker than the attraction of the Na+ to water and Cl– to water, NaCl will dissolve when placed in water H2O will strip the crystal of its ions, and then five other H2Os will completely surround the ions This is known as hydration NaCl dissolving in water Each ion is attracted to the surrounding water molecules and pulled off and away from the crystal When it enters the solution, the ion is surrounded by water molecules, insulating it from other ions The result is a solution with free moving charged particles able to conduct electricity Solubility of Ionic Compounds When most ionic compounds dissolve in water, the resulting solution does not contain any of the intact ionic compound itself The compound dissociates completely into its component ions in the water However, not all ionic compounds dissolve in water If we add AgCl to water it remains solid and appears as a white powder at the bottom of the water In general, a compound is termed soluble if it dissolves in water and insoluble if it does not Soluble NaCl NaCl fully dissolves in water A sodium chloride solution, NaCl(aq), does not contain any NaCl units Only dissolved Na+ ions and Cl− ions are present Substances (such as NaCl) that completely dissociate into ions in solution are strong electrolytes AgNO3 A silver nitrate solution, AgNO3(aq), does not contain any AgNO3 units Only dissolved Ag+ ions and NO3− ions are present AgNO (aq) is also a strong 3 electrolyte solution When compounds containing polyatomic ions such as NO3− dissolve, the polyatomic ions dissolve as intact units AgCl Not all ionic compounds dissolve in water AgCl is considered insoluble in water It stays intact as AgCl even in water It does not break up into Ag+ and Cl– AgCl appears as a white solid in water Soluble vs Insoluble NaCl and AgNO3 are both soluble and will break up into their respective ions If you were to mix these two solutions, solid AgCl would form and drop of solution because it is insoluble When will a salt dissolve? Whether a particular compound is soluble or insoluble depends on several factors Predicting whether a compound will dissolve in water is not easy The best way to do it is to do some experiments to test whether a compound will dissolve in water, and then develop some rules based on those experimental results We call this method the empirical method Lucky for you, a lot of other people have already done these experiments and have determined all the rules for you Unfortunately, you still have to LEARN these rules…  Solubility rules (in H2O) Know this chart!!! *this follows your book, but is slightly different than other sources Water-soluble Compounds Insoluble Exceptions All compounds of Group I (Li+, Na+, K+, etc.) None All compounds containing an ammonium ion (NH4+) None All common nitrates (NO3–), acetates (C2H3O2–) and None perchlorates (ClO4–) All common chloride (Cl ), bromides (Br ), and – – Ag+, Hg22+, Pb2+ iodides (I–) All common sulfates (SO42–) Ag+, Hg22+, Pb2+, Ca2+, Sr2+, Ba2+ Water-insoluble Compounds Soluble Exceptions All common carbonates (CO32–), phosphates (PO42–), Group I metal ions, NH4 + chromates (CrO42–), and sulfides (S2–) All common hydroxides (OH–) Group I metal ions, NH4+, Ba2+  Water-soluble Water-soluble Compounds Insoluble Exceptions All compounds of Group I (Li+, Na+, K+, etc.) None All compounds containing an ammonium ion (NH4+) None All common nitrates (NO3–), acetates (C2H3O2–) and None perchlorates (ClO4–) All common chloride (Cl ), bromides (Br ), and – – Ag+, Hg22+, Pb2+ iodides (I–) All common sulfates (SO42–) Ag+, Hg22+, Pb2+, Ca2+, Sr2+, Ba2+ Everything that contains these are always going to be soluble, no exceptions If it contains Cl–, Br–, or I–, it will normally be soluble; unless it contains Ag+, Hg22+, or Pb2+, in which case, it will be insoluble If it contains SO42–, it will normally be soluble; unless it contains Ag+, Hg22+, Pb2+, Ca2+, Sr2+, Ba2+, in which case, it will be insoluble  Water-insoluble Water-insoluble Compounds Soluble Exceptions All common carbonates (CO32–), phosphates (PO42–), Group I metal ions, NH4 + chromates (CrO42–), and sulfides (S2–) All common hydroxides (OH–) Group I metal ions, NH4+, Ba2+ If it contains CO32–, PO42–, CrO42–, or S2–, it will normally be insoluble; unless it contains any ion from Group I or NH4+, in which case, it will be soluble If it contains OH–, it will normally be insoluble; unless it contains any ion from Group I, NH4+, or Ba2+, in which case, it will be soluble Is this compound soluble? AgBr Compounds containing Br– are normally soluble, but Ag+ is an exception; insoluble Is this compound soluble? AgBr Insoluble CaCl2 Compounds containing Cl– are normally soluble, and Ca2+ is NOT an exception; soluble Is this compound soluble? AgBr Insoluble CaCl2 Soluble Pb(NO3)2 Compounds containing NO3– are always soluble Precipitation reactions Precipitation reactions are reactions in which a solid forms when we mix two solutions Some reactions between aqueous solutions of ionic compounds produce an ionic compound that is insoluble in water The insoluble product is called a precipitate The sodium carbonate in laundry detergent reacts with dissolved Mg2+ and Ca2+ ions to form solids that precipitate from solution Predicting precipitation reactions 1. Determine what ions each aqueous reactant has 2. Determine formulas of possible products Exchange ions Positive (+) ion from one reactant with the negative (–) ion from the other Balance charges of combined ions to get the formula of each product 3. Determine solubility of each product in water Use the solubility rules If a product is insoluble (or slightly soluble), it will precipitate 4. If no product will precipitate, write No Reaction after the arrow 5. If any of the possible products are insoluble, write the formula as the product of the reaction using (s) to signify it’s a solid 6. Write any soluble products with (aq) after the formula to indicate aqueous 7. Balance the equation - Only change coefficients, not subscripts KI and Pb(NO3)2 KI(aq) + Pb(NO3)2(aq) soluble soluble Before mixing, KI and Pb(NO3)2 are both fully dissociated in their respective solutions + KI and Pb(NO3)2 KI(aq) + Pb(NO3)2(aq) soluble soluble As soon as they are mixed, before any reaction takes place, all four ions are present KI and Pb(NO3)2 KI(aq) + Pb(NO3)2(aq) → ? soluble soluble Potentially insoluble products are now possible You just need to figure out which ones, if any would form Original Compounds Products KI and Pb(NO3)2 KI(aq) + Pb(NO3)2(aq) → KNO3(?) + PbI2(?) soluble soluble Determine if the products are: Both soluble  NO REACTION occurs Either is insoluble  a precipitate will form Original Compounds Products soluble insoluble KI and Pb(NO3)2 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) soluble soluble soluble insoluble KNO3 is soluble, and will remain ions PbI2 is insoluble and will precipitate out of solution as a solid Lead(II) Iodide Notice the (s) Potassium iodide and lead(II) nitrate 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) Potassium iodide is soluble Lead(II) nitrate is soluble Lead iodide is insoluble Bright yellow solid “No Reaction” is possible Precipitation reactions do not KI(aq) + NaCl(aq) → NO REACTION always occur when mixing two aqueous solutions If none of the products are insoluble, no solid will form When a potassium iodide solution is mixed with a sodium chloride solution, no reaction occurs KI is soluble NaCl is soluble NaI is soluble KCl is soluble Precipitation example Write the equation for the reaction that occurs (if any) when solutions of sodium carbonate and copper(II) chloride are mixed. Products Solubility Rules NaCl Anything that contains Na+ is always soluble CuCO3 Compounds containing CO32– are usually insoluble and Cu2+ is not an exception; insoluble Precipitation example Write the equation for the reaction that occurs (if any) when solutions of sodium carbonate and copper(II) chloride are mixed. Na2CO3(aq) + CuCl2(aq) → NaCl(aq) + CuCO3(s) Balance: Na2CO3(aq) + CuCl2(aq) → 2 NaCl(aq) + CuCO3(s)  Another precipitation example Write the equation for the reaction that occurs (if any) when solutions of lithium nitrate and sodium sulfate are mixed. NO REACTION Potential Products Solubility Rules Li2SO4 Anything that contains Li+ is always soluble NaNO3 Anything that contains Na+ is always soluble Representing Aqueous Reactions An equation showing the complete neutral formulas for each compound in the aqueous reaction as if they existed as molecules is called a molecular equation 2 KOH(aq) + Mg(NO3)2(aq) ® 2 KNO3(aq) + Mg(OH)2(s) In actual solutions of soluble ionic compounds, dissolved substances are present as ions Equations that describe the material’s structure when dissolved are called complete ionic equations Complete Ionic Equation Rules of writing the complete ionic equation: Aqueous, strong electrolytes are written as ions Soluble salts, strong acids, strong bases Insoluble substances, weak electrolytes, and nonelectrolytes are written in molecule form Solids, liquids, and gases are not dissolved, hence are written in molecule form 2 KOH(aq) + Mg(NO3)2(aq) ® 2 KNO3(aq) + Mg(OH)2(s) 2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq) → 2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s) You MUST include the charges for each ion Molecular and ionic equations Molecular equation Shows the complete neutral formulas for every compound in the reaction AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s) Solid ppt will not break into their Complete ionic equation constituent parts Shows the reactants and products as they are actually present in the solution AgAg(aq)(aq) +3NO (aq) Na++(aq) Na++(aq) + Cl–(aq) + – + + NO – (aq)3 + Cl–(aq) → Na++(aq) (aq)++NO NO – –(aq) + AgCl(s) → Na 3 (aq)+ 3 AgCl Spectator ions Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) → AgCl(s) + Na+(aq) + NO3–(aq) Spectator ions Notice that some ions appear on both sides of the reaction This means that they really don’t participate in the overall reaction, so they are just spectators Net ionic equations Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) → AgCl(s) + Na+(aq) + NO3–(aq) Ag+(aq) + Cl–(aq) → AgCl(s) If you get rid of the spectator ions, you have only the species that participate in the reaction This is referred to as the net ionic reaction Equations summary Molecular: shows the complete, neutral formulas HCl(?) + NaOH(?) → H2O(?) + NaCl(?) Complete ionic: shows all of the species as they are actually present in solution H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) Identify any spectator ions → H2O(l) + Na+(aq) + Cl–(aq) Cancel out all spectator ions Net ionic: shows only the species that actually participate in the reaction H+(aq) + OH–(aq) → H2O(l) Examples Write the ionic and net ionic equation for the following: K2SO4(?) + 2 AgNO3(?) ® 2 KNO3(?) + Ag2SO4(?) Complete Ionic 2 K+(aq) + SO42−(aq) + 2 Ag+(aq) + 2 NO3−(aq) → 2 K+(aq) + 2 NO3−(aq) + Ag2SO4(s) Net ionic 2 Ag+(aq) + SO42−(aq) ® Ag2SO4(s) I only wrote the Ag+ first because the solid formed is silver sulfate Examples Write the ionic and net ionic equation for the following: Na2CO3(aq) + 2 HCl(aq) ® 2 NaCl(aq) + CO2(g) + H2O(l) Gases and liquids do not dissociate in water, so they Complete Ionic stay in their molecular form 2 Na+(aq) + CO32−(aq) + 2 H+(aq) + 2 Cl−(aq) → 2 Na+(aq) + 2 Cl−(aq) + CO2(g) + H2O(l) Net ionic CO32−(aq) + 2 H+(aq) ® CO2(g) + H2O(l) Acids and Bases Acids Sour taste Able to dissolve some metals Tend to form H+ ions in solution Bases Bitter taste Slippery feel Tend to form OH– ions in solution  Acid-Base Reactions Reactions that form water upon mixing of an acid and a base Called neutralization reactions because the acid and base neutralize each other’s properties and result in water Like precipitate reactions, the cation of one reactant combines with the anion of the other, and vice-versa Net ionic: HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) H+(aq) + OH–(aq) → H2O(l) Acid-Base Reactions Acids ionize in water to form H + ions More precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H3O+ Most chemists use H+ and H3O+ interchangeably Bases dissociate in water to form OH  ions Bases, such as NH3, that do not contain OH ions, produce OH by pulling H off water molecules In the reaction of an acid with a base, H + from the acid combines with OH– from the base to make water The cation from the base combines with the anion from the acid to make a salt The net ionic equation for all strong acid-base reactions is: H+(aq) + OH–(aq) → H2O(l) Acid-Base Reaction HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)  Know these common acids Acid Formula Hydrochloric acid HCl Hydrobromic acid HBr Nonmetal or Hydroiodic acid HI polyatomic Nitric acid Hydrogen is HNO3 Sulfuric acid a MUST! H2SO4 Perchloric acid HClO4 Weak Acetic acid HC2H3O2 acids Hydrofluoric acid HF Binary Acids Acids are molecular compounds that ionize when they dissolve in water The molecules are “ripped” apart by their attraction for the water When acids ionize, they form H+ cations The percentage of molecules that ionize varies from one acid to another Acids that ionize virtually 100% are called strong acids. “” means all HCl is converted to HCl(aq) → H+(aq) + Cl−(aq) H+ and Cl− Acids that only ionize a small percentage are called weak “” means some acids HF still remains HF(aq)  H+(aq) + F−(aq) in solution along with H+ and F–  Know these common bases Base Formula Sodium hydroxide Usually a NaOH metal Lithium hydroxide LiOH Hydroxide Potassium hydroxide KOH is a MUST! Calcium hydroxide Ca(OH)2 Barium hydroxide Ba(OH)2 Ammonia* NH3 (weak base) * Ammonia does not contain OH–, but it produces it in a reaction with water that only occurs to a small extent: NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq) Try one out… What is the chemical equation for aqueous sulfuric acid mixed with aqueous potassium hydroxide? ACID? Sulfuric acid = H2SO4 BASE? Potassium hydroxide = KOH Both are in water, so don’t forget (aq) H2SO4(aq) + KOH(aq) What salt will form along with water? K2SO4(aq) + H2O(l) Write a balanced equation H2Acid SO4(aq) + 2 KOH(aq) Base → 2 H2Water O(l) + K2Ionic SO4(aq) Salt Oxidation-Reduction Reactions Reactions that involve a transfer of electrons are called oxidation-reduction reactions More commonly known as redox reactions Just because it’s shorter than saying oxidation-reduction If you’ve ever used a battery, you’ve preformed a redox reaction Many (NOT ALL) redox reactions involve the reaction of a substance with oxygen Oxidation and Reduction To convert a free element into an ion, the atoms must gain or lose electrons Of course, if one atom loses electrons, another must accept them Atoms that lose electrons are being oxidized, while atoms that gain electrons are being reduced 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na → Na+ + 1 e– (oxidation) Cl2 + 2 e– → 2 Cl– (reduction)  Transfer of electrons 2 Mg(s) + O2(g) → 2 MgO(s) All solid elements have a charge of Elemental diatomics Determine the charge ‘0’ always have a charge on components of of ‘0’ compounds as we have in the past I discuss charges and oxidation numbers in a few slides, so don’t panic yet… Transfer of electrons reduced 0 –2 O gains 2 e– 2 Mg(s) + O2(g) → 2 MgO(s) 0 +2 Mg loses 2 e– oxidized Oxidation – atom loses e–s (increase in charge) Reduction – atom gains e–s (decrease in charge) Na and Cl2 are both reduced pure elements 0 –1 2 Na(s) + Cl2(g) → 2 NaCl(s) 0 +1 oxidized Oxidizing and Reducing Agents Oxidation = loss of electrons The reducing agent is the reactant that loses electrons The reactant that reduces an element in another reactant The reactant that contains the element that is oxidized Reduction = gain of electrons The oxidizing agent is the reactant that gains electrons The reactant that oxidizes an element in another reactant The reactant that contains the element that is reduced Cl2(aq) + 2 KBr(aq) → 2 KCl(aq) + Br2(aq) Br loses an electron; Br is oxidized from –1 to 0 → KBr is the reducing agent Cl gains an electron; Cl is reduced from 0 to –1 → Cl2 is the oxidizing agent Oxidation Numbers We need a method for determining how the electrons are transferred Chemists assign a number to each element in a reaction called an oxidation state (oxidation number) that allows them to determine the electron flow in the reaction Even though they look like them, oxidation states are not ion charges! Oxidation states are imaginary charges assigned based on a set of rules Ion charges are real, measurable charges Oxidation Numbers General Rules 1. Elements in their natural state [Na(s), Br2(l), N2(g), Fe(s), etc.]: O. N. = 0 2. Monoatomic ions (Na+, Ca2+, O2–, Cl–, etc.): O. N. = ion’s charge 3. The sum of O.N. values for all atoms in a neutral compound = 0 The sum of O.N. values for all atoms in an ion = ion’s charge 4. For covalent compounds, the element closest to F gets the negative charge Rules for Specific Atoms / Periodic Table Groups 5. Group 1A: O. N. = +1 in all compounds 6. Group 2A: O. N. = +2 in all compounds 7. Hydrogen: O. N. = +1 in combination with nonmetals O. N. = –1 in combination with metals and boron (RARE!) 4. Fluorine: O. N. = –1 in all compounds 5. Oxygen: O. N. = –2 in almost all compounds (except with F) O. N. = –1 in peroxides 6. Group 7A: O. N. = –1 in combination with metals, nonmetals (except O), and other halogens lower in the group Oxidation Numbers Covalent compounds (not ionic) The element closest to fluorine has the negative charge The other(s) have a positive charge Examples: NO : oxygen = –2; nitrogen = +2 SO2 : oxygen = –2; sulfur = +4 ICl2 : chlorine = –1; iodine = +2 ClF3 : fluorine = –1; chlorine = +3 Oxidation Numbers Polyatomic ions and charged complexes The sum of the oxidation numbers must equal the overall charge of your complex For neutral compounds, the sum equals 0 Examples: NO3– : oxygen = –2 ; nitrogen = +5 CO32– : oxygen = –2; carbon = +4 NH4+ : hydrogen = +1; nitrogen = –3 Identifying Redox Components Determine what is being reduced and oxidized, and what the agents are reduced 0 –2 2 Ca(s) + O2(g) → 2 CaO(s) 0 +2 oxidized Oxygen is gaining 2 e–s → O is reduced O2 is the oxidizing agent Calcium is losing 2 e–s → Ca is oxidized Ca is the reducing agent  Another one… Determine what is being reduced and oxidized, and what the agents are reduced +3 –2 0 –2 Al(s) + Fe2O3(s) → Fe(s) + Al2O3(s) 0 +3 oxidized Iron is gaining 3 e–s → Fe is reduced Fe2O3 is the oxidizing agent Aluminum is losing 3 e–s → Al is oxidized Al is the reducing agent One more… Determine what is being reduced and oxidized, and what the agents are reduced reduced +1 0 –2 +1 –2 CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) –4 +4 oxidized Oxygen is gaining 2 e–s in two different cases → O is reduced O2 is the oxidizing agent Carbon is losing 8 e–s → C is oxidized CH4 is the reducing agent Redox recap Oxidation = loss of electrons Loss of e– (negative unit) will increase your charge Reduction = gain of electrons To reduce means to “make less”; Gaining e– s → make your charge “more negative” Reducing agent: the reactant that contains the element being oxidized Oxidizing agent: the reactant that contains the element being reduced Helpful mnemonic: OIL RIG : Oxidation Is Loss; Reduction Is Gain LEO GER : Lose Electrons – Oxidation; Gain Electrons – Reduction Oxidation and Reduction MUST occur together If one substance loses electrons, another must gain electrons You cannot have a gain of electrons if they weren’t lost from somewhere else, and vice-versa Combination reactions X+Y→Z An element and another element may react to form ionic or covalent compounds A metal and a nonmetal form an ionic compound Two nonmetals form a covalent compound Decomposition reactions Z→X+Y Basically, the opposite of combination reactions A compound decomposes to form two or more products Displacement / replacement reactions Single-displacement X + YZ → XZ + Y Double-displacement (swapping partners) AB + CD → AD + CB Combustion Reactions Products are oxides of the fuel Carbon based fuels (hydrocarbons) Products are usually CO2 and H2O (and heat) CxHy + O2 → CO2 + H2O Redox combustion reactions Always a redox reaction Reaction of a substance (referred to as a fuel) with O2 to form one or more oxygen-containing compounds, often including water Products are oxides of the fuel Carbon base fuels (hydrocarbons) CO2, H2O CxHy + O2 → CO2 + H2O Carbon is always oxidized; the fuel is always the reducing agent Oxygen is always reduced; the oxygen gas is always the oxidizing agent