4. Chemical reactions

Transcription

1 4. Chemical reactions Chemical reactions types: Dissociation Precipitation Acid base or salification Redox Displacement Gas forming reaction Decomposition Synthesis Dr.ssa Rossana Galassi Prof.ssa Rossana Galassi

2 Why do reactions occur? For energetic reasons.energy can be considered as the measure of the capability to perform a work. In a reaction, work is needed to break chemical bonds of the reactants and to form others in the products

3 Energy is the capacity to do work. Potential Energy energy due to the position of the object or energy from a chemical reaction Kinetic Energy energy due to the motion of the object Potential and kinetic energy can be interconverted.

4 Energy is the capacity to do work. less stable change in potential energy EQUALS kinetic energy more stable A gravitational system. The potential energy gained when a lifted weight is converted to kinetic energy as the weight falls.

5 Energy is the capacity to do work. less stable change in potential energy EQUALS kinetic energy more stable A system of two balls attached by a spring. The potential energy gained by a stretched spring is converted to kinetic energy when the moving balls are released.

6 Energy is the capacity to do work. less stable change in potential energy EQUALS kinetic energy more stable A system of oppositely charged particles. The potential energy gained when the charges are separated is converted to kinetic energy as the attraction pulls these charges together.

7 Energy is the capacity to do work. less stable change in potential energy EQUALS kinetic energy more stable A system of fuel and exhaust. A fuel is higher in chemical potential energy than the exhaust. As the fuel burns, some of its potential energy is converted to the kinetic energy of the moving car.

8 Chemical reactions less stable reactants more stable products Reagents => products

9 Chemical reactions Double exchange or displacement Ba(NO 3 ) 2 + Na 2 SO 4 => BaSO 4 + 2NaNO 3 Single exchange or displacement Zn + 2HCl = ZnCl 2 + H 2 Redox 3Cu + 8HNO 3 => 3Cu(NO 3 ) 2 + 2NO + 4H 2 O Dissociation NaCl(s)=> Na+ (aq)+ Cl- (aq)

10 Chemical reactions Acid base or salification H 2 SO 4 + Na 2 O=> Na 2 SO 4 + H 2 O Decomposition Na 2 CO 3 (s) => Na 2 O (s) + CO 2 (g) Synthesis N 2 (g)+ 3H 2 (g) => 2NH 3 (g) When a gas is evolved from the reaction can be classified as gas forming reaction.

11 WATER SOLVENT In nature, reactions occur in watery media. Water is a molecule formed by two hydrogen and an oxygen atom.the molecule formula is: H 2 O Water is a polar molecule, that means that it has not a homogenous charge distribution in the surface of the molecule

12 Electron distribution in molecules of H 2 and H 2 O.

13 When ionic compounds are dissolved in water they dissociate in ions. Water molecules are able to separate charged moieties to form hydrated ions

14 The dissolution of an ionic compound in water. Prof.ssa Rossana Galassi

15 The electrical conductivity of ionic solutions.

16 Determining Moles of Ions in Aqueous Ionic Solutions PROBLEM: 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) 7.42x10 22 formula units of copper(ii) nitrate dissolved in water (d) 35 ml of 0.84 M zinc chloride PLAN: We have to relate the information given and the number of moles of ions present when the substance dissolves in water. SOLUTION: (a) (NH 4 ) 2 SO 4 (s) 2NH 4 + (aq) + SO 4 2- (aq) 5.0 mol (NH 4 ) 2 SO 4 2 mol NH mol (NH 4 ) 2 SO 4 = 10. mol NH mol SO 4 2-

17 Determining Moles of Ions in Aqueous Ionic Solutions continued (b) CsBr(s) 78.5 g CsBr Cs + (aq) + Br - (aq) mol CsBr = mol CsBr g CsBr = mol Cs + = mol Br - (c) Cu(NO 3 ) 2 (s) Cu 2+ (aq) + 2NO 3 - (aq) 7.42x10 22 formula units Cu(NO 3 ) 2 mol Cu(NO 3 ) x10 23 formula units = mol Cu(NO 3 ) 2 = mol Cu 2+ = mol NO 3 - (d) ZnCl 2 (aq) Zn 2+ (aq) + 2Cl - (aq) 0.84 mol ZnCl 2 35 ml ZnCl 2 = 2.9x110-2 mol ZnCl 2 L = 2.9x110-2 mol Zn 2+ = 5.8x10-2 mol Cl -

18 The hydrated proton.

19 Molarity Solution= solvent + solute To express the concentration in a solution of ions, substances such as acids, salts and so on (solute) the Molarity expression is commonly used. M = number of mole of solute in a liter of solution

20 Determining the Molarity of H + Ions in Aqueous Solutions of Acids PROBLEM: 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? PLAN: Use the formula to find the molarity of H +. SOLUTION: One mole of H + (aq) is released per mole of nitric acid (HNO 3 ) HNO 3 (l) H + (aq) + NO 3 - (aq) 1.4M HNO 3 (aq) should have 1.4M H + (aq).

21 Writing Equations for Aqueous Ionic Reactions The molecular equation shows all of the reactants and products as intact, undissociated compounds. The total ionic equation shows all of the soluble ionic substances dissociated into ions. The net ionic equation eliminates the spectator ions and shows the actual chemical change taking place.

22 A precipitation reaction and its equation.

23 The reaction of Pb(NO 3 ) 2 and NaI. NaI(aq) + Pb(NO 3 ) 2 (aq) PbI 2 (s) + NaNO 3 (aq) 2NaI(aq) + Pb(NO 3 ) 2 (aq) PbI 2 (s) + 2NaNO 3 (aq) 2Na + (aq) + 2I - (aq) + Pb 2+ (aq) + 2NO 3- (aq) PbI 2 (s) + 2Na + (aq) + 2NO 3- (aq) 2NaI(aq) + Pb(NO 3 ) 2 (aq) PbI 2 (s) + 2NaNO 3 (aq) double displacement reaction (metathesis)

24 Predicting Whether a Precipitate Will Form 1. Note the ions present in the reactants. 2. Consider the possible cation-anion combinations. 3. Decide whether any of the ion combinations is insoluble. See Table 4.1 (next slide) for solubility rules.

25 Table 4.1 Solubility Rules For Ionic Compounds in Water Soluble Ionic Compounds 1. All common compounds of Group 1A(1) ions (Li +, Na +, K +, etc.) and ammonium ion (NH 4+ ) are soluble. 2. All common nitrates (NO 3- ), acetates (CH 3 COO - or C 2 H 3 O 2- ) and most perchlorates (ClO 4- ) are soluble. 3. All common chlorides (Cl - ), bromides (Br - ) and iodides (I - ) are soluble, except those of Ag +, Pb 2+, Cu +, and Hg Insoluble Ionic Compounds 1. All common metal hydroxides are insoluble, except those of Group 1A(1) and the larger members of Group 2A(2)(beginning with Ca 2+ ). 2. All common carbonates (CO 3 2- ) and phosphates (PO 4 3- ) are insoluble, except those of Group 1A(1) and NH All common sulfides are insoluble except those of Group 1A(1), Group 2A(2) and NH 4+.

26 PROBLEM: Predicting Whether a Precipitation Reaction Occurs; Writing Ionic Equations 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) PLAN: write ions combine anions & cations SOLUTION: (a) Na 2 SO 4 (aq) + Sr(NO 3 ) 2 (aq) => 2NaNO 3 (aq) + SrSO 4 (s) 2Na + (aq) +SO 4 2- (aq)+ Sr 2+ (aq)+2no 3 - (aq)=> 2Na + (aq) +2NO 3 - (aq)+ SrSO 4 (s) check for insolubility Table 4.1 eliminate spectator ions for net ionic equation SO 4 2- (aq)+ Sr 2+ (aq) => SrSO 4 (s) (b) NH 4 ClO 4 (aq) + NaBr (aq) => NH 4 Br (aq) + NaClO 4 (aq) All reactants and products are soluble so no reaction occurs.

27 Table 4.2 Selected Acids and Bases Acids Strong hydrochloric acid, HCl hydrobromic acid, HBr hydroiodic acid, HI nitric acid, HNO 3 sulfuric acid, H 2 SO 4 Bases Strong sodium hydroxide, NaOH potassium hydroxide, KOH calcium hydroxide, Ca(OH) 2 strontium hydroxide, Sr(OH) 2 barium hydroxide, Ba(OH) 2 perchloric acid, HClO 4 Weak hydrofluoric acid, HF Weak ammonia, NH 3 phosphoric acid, H 3 PO 4 acetic acid, CH 3 COOH (or HC 2 H 3 O 2 )

28 Strong acid and strong base an acid or a base are defined strong when in solution they are completely dissociated (ionized) H 2 O HCl (g) => H + (aq) + Cl - (aq) H 2 O Na(OH) (s)=> Na + (aq) + OH - (aq)

29 Writing Ionic Equations for Acid-Base Reactions PROBLEM: Write balanced molecular, total ionic, and net ionic equations for each of the following acid-base reactions and identify the spectator ions. (a) strontium hydroxide(aq) + perchloric acid(aq) (b) barium hydroxide(aq) + sulfuric acid(aq) PLAN: SOLUTION: reactants are strong acids and bases and therefore completely ionized in water products are water spectator ions (a) Sr(OH) 2 (aq)+2hclo 4 (aq) 2H 2 O(l)+Sr(ClO 4 ) 2 (aq) Sr 2+ (aq) + 2OH - (aq)+ 2H + (aq)+ 2ClO 4 - (aq) 2H 2 O(l)+Sr 2+ (aq)+2clo 4 - (aq) 2OH - (aq)+ 2H + (aq) (b) Ba(OH) 2 (aq) + H 2 SO 4 (aq) 2H 2 O(l) 2H 2 O(l) + BaSO 4 (aq) Ba 2+ (aq) + 2OH - (aq)+ 2H + (aq)+ SO 4 2- (aq) 2H 2 O(l)+BaSO 4 (s)

30 Molarity When a solid (Solute) is dissolved in a liquid (solvent), commonly water, an homogeneous mixture is formed. This mixture is called SOLUTION. The ammount of the solute in the solution or in the solvent is called CONCENTRATION. An expression of the concentration is the Molarity. Molarity: number of mole of solute in 1L of solution Prof.ssa Rossana Galassi

31 A 1M solution means that a mole of solute is dissolved in water till a 1L of solution is formed. 1L of 0.1M HCl solution means that in 1L of solution there are 0.1 mole of HCl or g of HCl (MM HCl = g/mol) n= g / MM g = mol x MM

32 How many grams of Na 2 CO 3 are contained in 100 ml of a 0.2M solution? MM Na2CO3 = 106 g/mol n = 100 x 0.2/1000 = 0.02 mol 0.02 mol x 106 g/mol = 2.12 g of Na 2 CO 3 The solution can be prepared weighting 2.12 g of sodium carbonate and adding water till a volume of 1 Liter.

33 250 ml of a 0.1M solution of HCl contains the following number of mole 0.1 : 1000= n : ml x 0.1 mol/l n = = mol 1000 ml/l n = V (ml) x M (mol/l)/1000 (ml/l)

34 Titration This is a method for quantitative analysis. In a titration, the reaction between the titrant and the titolate occurs. The equivalent point is reached when a volume of titrant reacts completely with the volume of titolate according to the titration reaction. Ca(OH) HCl => CaCl 2 + 2H 2 O

35 An acid-base titration. The colour at the stoichiometric point is given by the presence of an indicator The indicator changes its color when the solution does not contain acid anymore Prof.ssa Rossana Galassi

36 PROBLEM: PLAN: volume(l) of base mol of base mol of acid M of acid Finding the Concentration of Acid from an Acid-Base Titration You perform an acid-base titration to standardize an HCl solution by placing ml of HCl in a flask with a few drops of indicator solution. You put M NaOH into the buret, and the initial reading is 0.55 ml. At the end point, the buret reading is ml. What is the concentration of the HCl solution? multiply by M of base molar ratio divide by L of acid SOLUTION: NaOH(aq) + HCl(aq) ( ) ml x 1L 10 3 ml NaCl(aq) + H 2 O(l) = L L X M = 5.078x10-3 mol NaOH Molar ratio is 1: x10-3 mol HCl L = M HCl

37 An aqueous strong acid-strong base reaction on the atomic scale. Prof.ssa Rossana Galassi

38 An acid-base reaction that forms a gaseous product. Molecular equation NaHCO 3 (aq) + CH 3 COOH(aq) CH 3 COONa(aq) + CO 2 (g) + H 2 O(l) Total ionic equation Na + (aq)+ HCO 3- (aq) + CH 3 COOH(aq) CH 3 COO - (aq) + Na + (aq) + CO 2 (g) + H 2 O(l) Net ionic equation HCO 3- (aq) + CH 3 COOH(aq) CH 3 COO - (aq) + CO 2 (g) + H 2 O(l)

39 The redox process in compound formation.

40 Table 4.3 Rules for Assigning an Oxidation Number (O.N.) General rules 1. For an atom in its elemental form (Na, O 2, Cl 2, etc.): O.N. = 0 2. For a monoatomic ion: O.N. = ion charge 3. The sum of O.N. values for the atoms in a compound equals zero. The sum of O.N. values for the atoms in a polyatomic ion equals the ion s charge. Rules for specific atoms or periodic table groups 1. For Group 1A(1): O.N. = +1 in all compounds 2. For Group 2A(2): O.N. = +2 in all compounds 3. For hydrogen: O.N. = +1 in combination with nonmetals 4. For fluorine: O.N. = -1 in combination with metals and boron 5. For oxygen: O.N. = -1 in peroxides O.N. = -2 in all other compounds(except with F) 6. For Group 7A(17): O.N. = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group

41 Determining the Oxidation Number of an Element PROBLEM: Determine the oxidation number (O.N.) of each element in these compounds: (a) zinc chloride (b) sulfur trioxide (c) nitric acid PLAN: The O.N.s of the ions in a polyatomic ion add up to the charge of the ion and the O.N.s of the ions in the compound add up to zero. SOLUTION: (a) ZnCl 2. The O.N. for zinc is +2 and that for chloride is -1. (b) SO 3. Each oxygen is an oxide with an O.N. of -2. Therefore the O.N. of sulfur must be +6. (c) HNO 3. H has an O.N. of +1 and each oxygen is -2. Therefore the N must have an O.N. of +5.

42 Highest and lowest oxidation numbers of reactive main-group elements. Prof.ssa Rossana Galassi

43 A summary of terminology for oxidation-reduction (redox) reactions. e - X Y transfer or shift of electrons X loses electron(s) X is oxidized X is the reducing agent X increases its oxidation number Y gains electron(s) Y is reduced Y is the oxidizing agent Y decreases its oxidation number

44 Recognizing Oxidizing and Reducing Agents PROBLEM: Identify the oxidizing agent and reducing agent in each of the following: (a) 2Al(s) + 3H 2 SO 4 (aq) (b) PbO(s) + CO(g) (c) 2H 2 (g) + O 2 (g) Al 2 (SO 4 ) 3 (aq) + 3H 2 (g) Pb(s) + CO 2 (g) 2H 2 O(g) PLAN: Assign an O.N. for each atom and see which atom gained and which atom lost electrons in going from reactants to products. An increase in O.N. means the species was oxidized (and is the reducing agent) and a decrease in O.N. means the species was reduced (is the oxidizing agent). SOLUTION: (a) 2Al(s) + 3H 2 SO 4 (aq) Al 2 (SO 4 ) 3 (aq) + 3H 2 (g) The O.N. of Al increases; it is oxidized; it is the reducing agent. The O.N. of H decreases; it is reduced; H 2 SO 4 is the oxidizing agent.

45 Recognizing Oxidizing and Reducing Agents continued (b) PbO(s) + CO(g) Pb(s) + CO 2 (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 (c) 2H 2 (g) + O 2 (g) 2H 2 O(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.

46 Balancing Redox Equations by the Oxidation Number Method PROBLEM: Use the oxidation number method to balance the following equations: (a) Cu(s) + HNO 3 (aq) Cu(NO 3 ) 2 (aq) + NO 2 (g) + H 2 O(l) (b) PbS(s) + O 2 (g) PbO(s) + SO 2 (g) SOLUTION: (a) Cu(s) + HNO 3 (aq) Cu(NO 3 ) 2 (aq) + NO 2 (g) + H 2 O(l) O.N. of Cu increases because it loses 2e - ; it is oxidized and is the reducing agent. O.N. of N decreases because it gains1e - ; it is reduced and is the oxidizing agent. loses 2e - Cu(s) + HNO 3 (aq) balance other ions Cu(NO 3 ) 2 (aq) + NO 2 (g) + H 2 O(l) balance unchanged polyatomic ions x2 to balance e -

47 Balancing Redox Equations by the Oxidation Number Method continued Cu(s) + 4 HNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2NO 2 (g) + 2H 2 O(l) (b) PbS(s) + O 2 (g) PbO(s) + SO 2 (g) loses 6e - PbS(s) + O 2 (g) PbO(s) + SO 2 (g) gains 2e - per O; need 3/2 O 2 to make 3O 2- total of a 6e - exchange Multiply by 2 to have whole number coefficients. 2PbS(s) + 3O 2 (g) 2PbO(s) + 2SO 2 (g)

48 Dismutation redox equation gains 1e - per Cl Cl 2 (g) + NaOH(aq) NaCl (aq) + NaClO 3 (aq) + H 2 O(l) loses 5e - per Cl N ox. 0 N ox. -1 N ox. +5 Cl 2 + 2e - 2Cl - X 5 Cl 2-10e - 2Cl 5+ X 1 5Cl + 10e Cl - Cl - 10e - 2 2Cl 5+- 6Cl 2 10Cl - + 2Cl 5+ 6Cl 2 (g) + 12 NaOH(aq) Prof.ssa Rossana Galassi 10 NaCl (aq) + 2NaClO 3 (aq) + 6H 2 O(l)

49 Watery media redox equation (acid media) loses 5e - per I I 2 (g) + HNO 3 (aq) NO (aq) + HIO 3 (aq) + H 2 O(l) N ox. 0 gains 3e - per N N ox.+2 N ox.+5 I 2 + H + + NO 3 - NO + H + + IO H 2 O(l) I 2 2IO - + 6H 2 O H e - X 3 NO H + + 3e - NO + 2H 2 O X 10 3I H 2 O +40H NO e - 6IO H e NO + 20H 2 O 3I 2 + 4H NO 3-6IO NO + 2H 2 O 3I 2 (g) + 10 HNO 3 (aq) Prof.ssa Rossana Galassi 10NO (aq) + 6HIO 3 (aq) + 2 H 2 O(l)

50 Organic Compounds Redox Equation (basic media) loses 4e - per C CH 3 CH 2 OH(aq) + KMnO 4 (aq) CH 3 COOK (aq) + MnO 2 (aq) + KOH(aq) H 2 O(l) gains 3e - per Mn N ox.-1 N ox.+7 N ox.+3 N ox.+4 CH 3 CH 2 OH + K+ + MnO 4 - CH 3 COO - + MnO 2 + K + + OH - + H 2 O CH 3 CH 2 OH + 5OH - CH 3 COO - + 4H + 4e - 2 O X 3 MnO - MnO 4 + 2H + 3e O + 4OH - X 4 3CH 3 CH 2 OH + 15OH - + 4MnO H + 12e - 2 O 3CH 3 COO - +12H + 12e - 2 O + 4 MnO OH - 3CH 3 CH 2 OH + 4MnO 4-3CH 3 COO - +4H 2 O + OH MnO 2 3CH 3 CH 2 OH(aq) + 4 KMnO 4 (aq) Prof.ssa Rossana Galassi 3CH 3 COOK (aq) + 4 MnO 2 (aq) + KOH(aq)+4 H 2 O(l)

51 Common Oxidizing agents Oxygen, ozone, O 2 and O 3 Nitric acid, concentrated or diluted Peroxides, H 2 O 2, Na 2 O 2 Peroxi- compounds, Permanganate, MnO 4 - Cromate or bicromate, Cr 2 O 7 2-, CrO 4 - Halogens, Cl 2, Br 2, I 2.

52 Common Reducing agents Hydrogen, H 2 Metal hydrides, CaH 2, NaH, LiAlH 4. Oxalates, CaC 2 O 4 Covalent hydrides, such as H 2 S, BH 3, PH 3 Thiocompounds Metals, Na, Ca, Fe etc

53 Oxidizing agents Reaction product Reducing agent Reaction product Oxygen, O 2 Oxide ion Hydrogen, H 2 Hydronium ion (aq) Halogens, F 2, Cl 2, Halide ion Metals Metal ions Br 2, I 2 Nitric acid NO or NO 2 depending on the HNO 3 concentration Cr 2 O 7 2- Cr 3+ ion C or CO CO or CO 2 MnO 4 - Mn 2+ or MnO 2 or MnO 4 2- depending on the ph

54 A redox titration. In this case the indicator is not necessary as the KMnO 4 is colored

55 PROBLEM: Finding an Unknown Concentration by a Redox Titration Calcium ion (Ca 2+ ) is required for blood to clot and for many other cell processes. An abnormal Ca 2+ concentration is indicative of disease. To measure the Ca 2+ concentration, 1.00mL of human blood was treated with Na 2 C 2 O 4 solution. The resulting CaC 2 O 4 precipitate was filtered and dissolved in dilute H 2 SO 4. This solution required 2.05mL of 4.88x10-4 M KMnO 4 to reach the end point. The unbalanced equation is KMnO 4 (aq) + CaC 2 O 4 (s) + H 2 SO 4 (aq) MnSO 4 (aq) + K 2 SO 4 (aq) + CaSO 4 (s) + CO 2 (g) + H 2 O(l) (a) Calculate the amount (mol) of Ca 2+. (b) Calculate the amount (mol) of Ca 2+ ion concentration expressed in units of mg Ca 2+ /100mL blood. PLAN: (a) volume of KMnO 4 soln multiply by M molar ratio mol of Ca 2+ ratio of elements in formula mol of CaC 2 O 4

56 KMnO 4 (aq) + CaC 2 O 4 (s) + H 2 SO 4 (aq) MnSO 4 (aq) + K 2 SO 4 (aq) + CaSO 4 (s) + CO 2 (g) + H 2 O(l) K + + MnO Ca 2+ + C 2 O H + + SO 4 2- Mn 2+ + SO K + +Ca 2+ + CO 2 + H 2 O MnO H + + 5e Mn H 2 O C 2 O 4 2-2CO 2 + 2e X 2 X 5 2MnO H e + 5C 2 O 4 2-2Mn H 2 O +10CO e 2MnO H + + 5C 2 O 4 2-2Mn H 2 O +10CO 2 2KMnO 4 (aq) + 5CaC 2 O 4 (s) + 8H 2 SO 4 (aq) 2MnSO 4 (aq) + K 2 SO 4 (aq) + 5CaSO 4 (s) + 10CO 2 (g) + 8 H 2 O(l)

57 Finding an Unknown Concentration by a Redox Titration continued SOLUTION: PLAN: (b) 2.05mL soln L 4.88x10-4 mol KMnO = 1.00x mol KMnO ml L 1.00x10-6 mol KMnO 4 5mol CaC 2 O 4 2mol KMnO 4 = 2.50x10-6 mol CaC 2 O x10-6 mol CaC 2 O 4 1mol Ca 2+ mol Ca 2+ /1mL blood multiply by 100 mol Ca 2+ /100mL blood multiply by M g Ca 2+ /100mL blood 10-3 g = 1mg mg Ca 2+ /100mL blood 1mol CaC 2 O 4 = 2.50x10-6 mol Ca 2+ SOLUTION: 2.50x10-6 mol Ca 2+ 1mL blood 2.50x10-4 mol Ca mL blood =2.50x10-4 mol Ca mL blood 40.08g Ca 2+ mol Ca 2+ =10.0mg Ca 2+ /100mL blood mg 10-3 g

58 Combining elements to form an ionic compound. A redox reaction causing the KCl synthesis Prof.ssa Rossana Galassi

59 Decomposing a compound to its elements. A redox reaction causing the decomposition of mercury(ii)oxide Prof.ssa Rossana Galassi

60 An active metal displacing hydrogen from water. A redox reaction causing the hydrogen displacement Prof.ssa Rossana Galassi

61 The displacement of H from acid by nickel. O.N. increasing oxidation occurring reducing agent O.N. decreasing reduction occurring oxidizing agent Ni(s) + 2H + (aq) Ni 2+ (aq) + H 2 (g)

62 Displacing one metal with another.

63 strength as reducing agents The activity series of the metals. Li K Ba Ca Na Mg Al Mn An Cr Fe Cd Co Ni Sn Pb can displace H from water can displace H from steam can displace H from acid H 2 Cu Hg Ag Au cannot displace H from any source Prof.ssa Rossana Galassi

64 Identifying the Type of Redox Reaction PROBLEM: 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) magnesium nitride (aq) (b) hydrogen peroxide(l) water(l) + oxygen gas (c) aluminum(s) + lead(ii) nitrate(aq) aluminum nitrate(aq) + lead(s) PLAN: Combination reactions produce fewer products than reactants. Decomposition reactions produce more products than reactants. Displacement reactions have the same number of products and reactants.

65 Identifying the Type of Redox Reaction continued (a) Combination Mg(s) + N 2 (g) Mg 3 N 2 (aq) Mg is the reducing agent; N 2 is the oxidizing agent. (b) Decomposition H 2 O 2 is the oxidizing and reducing agent. H 2 O 2 (l) H 2 O(l) + O 2 (g) 2 H 2 O 2 (l) 2 H 2 O(l) + O 2 (g) (c) Displacement Al(s) + Pb(NO 3 ) 2( aq) 2Al(s) + 3Pb(NO 3 ) 2( aq) Al(NO 3 ) 3 (aq) + Pb(s) 2Al(NO 3 ) 3 (aq) + 3Pb(s) Pb(NO 3 ) 2 is the oxidizing and Al is the reducing agent.