Unit 5: Principles and Applications of Science II PDF

Unit 5: Principles and Applications of Science II Learning Aim A: Properties and Uses of Substances Relating Properties to Uses and Production of Substances Learning Objectives Be able to write the formulae for metal oxides/hydroxides Write equations for the reactions of soluble metal oxid...

Unit 5: Principles and Applications of Science II Learning Aim A: Properties and Uses of Substances Relating Properties to Uses and Production of Substances Learning Objectives Be able to write the formulae for metal oxides/hydroxides Write equations for the reactions of soluble metal oxides with water Write equations for the reactions of metal oxides and hydroxides with common acids State the uses of some common metal oxides and hydroxides Describe, using equations, the amphoteric nature of alumina, Al2O3 Acid vs. Base/Alkali Match the terms to their definitions Acid A compound formed by an acid-base reaction A compound that dissociates in water to form hydrogen Base ions (H+) Alkali A compound that reacts with an acid to form a salt and water Salt: A base that dissolves in water to form hydroxide ions (OH-) Acid vs. Base/Alkali  Group 1 → X+ Group 6 → X2- Group 2 → X2+ Group 7 → X- Metal Oxides Group 3 → X3+ Metal + Oxygen → Metal oxide Work out the formulae of the following metal oxides: a) Sodium oxide Na2O d) Magnesium oxide MgO b) Calcium oxide CaO e) Potassium oxide K2O c) Lithium oxide Li2O f) Caesium oxide Cs2O Group 1 and 2 metal oxides are basic Metal Hydroxides Group 1 and 2 metal oxides are basic Metal oxides dissolve in water to form alkaline solutions Na2O + H2O → 2NaOH CaO + H2O → Ca(OH)2 Hydroxide ion → OH- Group 1 → X+ Group 6 → X2- Group 2 → X2+ Group 7 → X- Metal Hydroxides Group 3 → X3+ Metal oxide + Water → Metal hydroxide Write balanced symbol equations for the following reactions: a) Potassium oxide + water → Potassium hydroxide K2O + H2O → 2KOH b) Magnesium oxide + water → Magnesium hydroxide MgO + H2O → Mg(OH)2 c) Lithium oxide + water → Lithium hydroxide Li2O + H2O → 2LiOH Checking Understanding Copper oxide Sodium hydroxide Can it neutralise Yes Yes acids? Is it a base? Yes Yes Can it dissolve in No Yes water? Is it an alkali? No Yes  Neutralisation Reaction Acid + Base → Salt + Water Write balanced symbol equations for the following reactions: a) Calcium hydroxide + Hydrochloric acid → Calcium chloride + Water Ca(OH)2 + 2HCl → CaCl2 + H2O b) Sodium oxide + Sulfuric acid → Sodium sulfate + Water Na2O + H2SO4 → Na2SO4 + H2O c) Barium oxide + Nitric acid → Barium nitrate + Water BaO + 2HNO3 → Ba(NO3)2 + H2O Monday, 03 October 2022 Acid + Base → Salt + Water Starter: Write balanced symbol equations for the following reactions: a) sodium hydroxide + Hydrochloric acid → sodium chloride + Water NaOH + HCl → NaCl + H2O b) berrylium oxide + Nitric acid → Berrylium nitrate + Water BeO + HNO3 → Be(NO3)2 + H2O c) Lithium oxide + sulfuric acid → Lithium sulfate + Water LiO + H2SO4 → Li2SO4 + H2O Uses of Metal Oxides & Hydroxides Substance Uses Calcium oxide Dissolved in water to form calcium hydroxide which is then (Quicklime) used by farmers to raise the pH of acidic soil Used as a desiccant when preserving books in a library: Magnesium oxide neutralises acidic sulfur oxides produced when papers are oxidised in air Sodium hydroxide Used in making plastics and soaps; found in drain cleaner Magnesium Used for treating acid indigestion by raising the pH of stomach hydroxide Used in the treatment of acidic effluent produced by factories: Calcium hydroxide H2SO4 + Ca(OH)2 → CaSO4 + 2H2O  Alumina / Aluminium Oxide / Al2O3 Amphoteric: Can act as both an acid and a base Acting as a base: Aluminium oxide + Hydrochloric acid → Aluminium chloride + Water Al2O3 + 6HCl → 2AlCl3 + 3H2O Acting as an acid: Aluminium oxide + Sodium hydroxide + Water → Sodium aluminate Al2O3 + 6NaOH + 3H2O → 2Na2Al(OH)6 Alumina / Aluminium Oxide / Al2O3 Match the use to the property that makes it suitable Paint Inert Sunscreen Corrosion resistant Glass Conductor Wiring Alumina / Aluminium Oxide / Al2O3 Alumina is a refractory material – material that is physically and chemically stable at very high temperatures (over 3000 °C) Where might this be useful? Linings for furnaces, kilns and reactors What property do you think it must have to be a refractory material? High melting point Alumina / Aluminium Oxide / Al2O3 Redox Reactions Electrolysis The decomposition of an ionic compound using electricity Ions must be free to move: The compound must either be molten or in solution  Products of molten electrolysis Q1. What would be the products of the molten electrolysis for: Positive Mg2+ ions (cations) Negative Cl- ions (anions) attract to negative electrode (cathode). attract to positive electrode (anode). Cl2 Magnesium chloride Here they are reduced by gaining 2 electrons. Here they are oxidised by each losing 1 electron. Mg2+ + 2e- → Mg - + This must happen to two Cl- ions as halogens Mg2+ Cl- must form diatomic molecules. As Mg is a metal, solid metal deposits form around the negative electrode (cathode). 2 Cl- → Cl2 + 2e- As Cl2 is a gas, we will see effervescence around the positive electrode (anode). electrolysis Magnesium chloride → Magnesium + Chlorine  Products of molten electrolysis Q2. What would be the products of the molten electrolysis for: Positive Na+ ions (cations) Negative Cl- ions (anions) attract to negative electrode (cathode). attract to positive electrode (anode). Cl2 Sodium chloride Here they are reduced by gaining 1 electron. Here they are oxidised by each losing 1 electron. Na+ + e- → Na - + This must happen to two Cl- ions as halogens Na+ Cl- must form diatomic molecules. As Na is a metal, solid metal deposits form around the negative electrode (cathode). 2 Cl- → Cl2 + 2e- As Cl2 is a gas, we will see effervescence around the positive electrode (anode). electrolysis Sodium chloride → Sodium + Chlorine  Products of molten electrolysis Q3. What would be the products of the molten electrolysis for: Positive Zn2+ ions (cations) Negative Br- ions (anions) attract to negative electrode (cathode). attract to positive electrode (anode). Br2 Zinc Bromide Here they are reduced by gaining 2 electrons. Here they are oxidised by each losing 1 electron. Zn2+ + 2e- → Zn - + This must happen to two Br- ions as halogens Zn2+ Br- must form diatomic molecules. As Zn is a metal, solid metal deposits form around the negative electrode (cathode). 2 Br- → Br2 + 2e- Br2 is a liquid under standard conditions, however it will evaporate due to the high temperature so we will see effervescence around the positive electrode (anode). electrolysis Zinc bromide → Zinc + Bromine  Products of molten electrolysis The products of molten electrolysis are the products of reducing or oxidising the ions! Magnesium chloride Sodium bromide Copper oxide Mg2+ Cl- Na+ Br- Cu2+ O2- Solid deposits of Chlorine (Cl2) Solid deposits of Bromine (Br2) Solid deposits of Oxygen (O2) Magnesium Gas Sodium Gas Copper Gas (Mg) Metal (Na) Metal (Cu) Metal  Pb2+ Lead Bromide - PbBr2 Br- Lead bromide contains lead Pb2+ and bromide Br- ions. 1. To which electrode does each ion go to? 2. What happens to each ion at each electrode? 1. Negative bromide anions Br- 1. Positive lead cations Pb2+ attract attract towards the positive electrode towards the negative electrode (anode). 2. Br- ions are oxidised by losing + - (cathode). 2. Pb2+ ions are reduced by gaining 1 electrons. ___ 2 electrons. ___ Pb2+ Br- Pb2+ Br- Br- Pb2+ Pb2+ Br- Pb2+ Br- Br- Pb2+ Pb2+ How can we improve these answers? At the negative electrode: Lead ions are reduced by each gaining 2 electrons. - Pb2+ + 2e- → Pb Pb2+ This forms atoms of lead which form metallic bonds. Pb2+ Pb2+ Pb2+ We will see solid deposits of this lead around the negative electrode (cathode). Rule: Whenever metal ions are reduced, we see solid deposits around the negative electrode (cathode). At the positive electrode: Bromide ions are oxidised by each losing 1 electron. + Br - → Br + e- Br2 However, bromine cannot exist as a lone atom so it Br- Br- covalently bonds another atom of itself to form Br2 Br- Br- Br + Br → Br2 Overall we write this as: Rule: Whenever a group 7 2 Br- → Br2 + 2e- element is oxidised, it forms a diatomic molecule X2 Monday, 03 October 2022 Electrolysis Electrolysis of Molten Sodium Chloride, NaCl(l)  Electrolysis of Brine (Sodium Chloride Solution, NaCl(aq)) Task: Complete the diagram Concentrated sodium 2H+ + 2e- → H2 chloride solution (NaCl (aq) ) turns into H+ and Na+ sodium hydroxide 2Cl- → Cl2 + 2e- (NaOH (aq) ) solution OH- left in solution Hydrogen gas Na+ left in solution Cl- and OH- Anode Cathode Chlorine gas Electrolysis Electrolysis – Equilibrium Process When a metal is placed in water, it loses electrons and becomes a positive ion: Mg(s) → Mg2+(aq) + 2e- This in turn attracts the negative electrons back to the metal ion, forming a metal atom: Mg2+(aq) + 2e- → Mg(s) The following equilibrium is established: Mg2+(aq) + 2e- ⇌ Mg(s) Electrolysis – Equilibrium Process Mg2+(aq) + 2e- ⇌ Mg(s) Electrochemical Series There is a potential difference between the negative charge on the metal and the positive charge of the solution around it This is measured as a standard electrode potential (E°, volts) Electrochemical Series The lower the metal is in the electrochemical series, the more likely it is to be discharged CuSO4(aq): Copper atoms form instead of hydrogen gas (H2) +0.34 V vs 0 V  Electrolysis of Brine (Sodium Chloride Solution, NaCl(aq)) Task: Complete the diagram Concentrated sodium 2H+ + 2e- → H2 chloride solution (NaCl (aq) ) turns into H+ and Na+ sodium hydroxide 2Cl- → Cl2 + 2e- (NaOH (aq) ) solution OH- left in solution Hydrogen gas Na+ left in solution Cl- and OH- Anode Cathode Chlorine gas  Monday, 03 October 2022 Transition Metals Give the full electronic configurations of: a) Cobalt atom 1s2 2s2 2p6 3s2 3p6 3d7 4s2 b) Manganese atom 1s2 2s2 2p6 3s2 3p6 3d5 4s2 c) Zinc atom 1s2 2s2 2p6 3s2 3p6 3d10 4s2 Transition Metals Q: in which block are transition metals found? d block elements A transition metal is an element that forms one or more stable ions which have incompletely filled d orbitals 4s-sub-shell fills before 3d-sub-shell 4s electrons are lost before 3d electrons Form ions with more than one stable oxidation state Used as catalysts Transition Metals Give the full electronic configurations of: a) Co2+ 1s2 2s2 2p6 3s2 3p6 3d7 b) Mn5+ 1s2 2s2 2p6 3s2 3p6 3d3 c) Zinc3+ 1s2 2s2 2p6 3s2 3p6 3d9 Water Complex Ions In solution, transition metal compounds form complex ions ligand A complex ion consists of a central metal ion surrounded by ligands A ligand is a molecule or an ion that donates a pair of electrons to the central transition metal ion to form a dative covalent bond dative covalent bond [Fe(H2O)6]2+ Contact Process Process of making sulfuric acid, H2SO4 Catalyst: Vanadium (V) oxide, V2O5 Q: What is a catalyst? Increases the rate of reaction by lowering the activation energy. Task: Complete the flowchart by adding the equations for the reaction taking place at each stage. Contact Process 1. Sulfur reacts with oxygen to produce sulfur dioxide: S(s) + O2(g) → SO2(g) 2. Sulfur dioxide is mixed with excess air to be converted into sulfur trioxide: 2SO2(g) + O2(g) ⇌ 2SO3(g) Vanadium (V) oxide, V2O5, catalyses the reaction 3. Sulfur trioxide is dissolved in concentrated sulfuric acid to form disulfuric acid: H2SO4(l) + SO3(g) → H2S2O7(l) 4. This is then added to water to produce more concentrated sulfuric acid: H2S2O7(l) + H2O(l) → 2H2SO4(l) Task: Complete the flowchart by adding the equations for the reaction taking place at each stage. S(s) + O2(g) → SO2(g) 2SO2(g) + O2(g) ⇌ 2SO3(g) Vanadium (V) oxide, V2O5, catalyses the reaction H2S2O7(l) + H2O(l) → 2H2SO4(l) H2SO4(l) + SO3(g) → H2S2O7(l) Extension: What is the oxidation state of ? Vanadium oxide, V2O5 ? Contact Process - Details Sulfur dioxide can also be produced by reacting sulfide ores with excess air: 4FeS2(g) + 11O2(g) → 2Fe2O2(s) + 8SO2(g) Vanadium (V) oxide has the ability to change its oxidation state Sulfur dioxide is oxidised to sulfur trioxide by the vanadium (V) oxide So the vanadium (V) oxide is reduced to vanadium (IV) oxide: 2SO2 + 2V2O5 → 2SO3 + 2V2O4 Vanadium (IV) oxide is then oxidised with the oxygen: 2V2O4 + O2 → 2V2O5 Vanadium (V) oxide catalyst takes part in the reaction and is changed in the reaction, but remains chemically unchanged by the end of the reaction Adding sulfur trioxide to water is too uncontrollable; hence, we first dissolve in concentrated sulfuric acid Haber Process Process of making ammonia, NH3 N2(g) + 3H2(g) ⇌ 2NH3(g) Catalyst: Iron N2 and H2 molecules absorb onto the surface of the iron The covalent bonds within the molecules are weakened This weakening of the bond lowers the activation energy for the reaction Reaction between N2 and H2 takes places at the surface of the iron After the reaction, NH3 molecules desorb from the surface of the iron Haber Process https://www.sciencephoto.com/media/727620/view/haber-process-catalysis-animation Checking Understanding Complete the following. Write in full sentences. 1. Explain the products of the electrolysis of brine. 2. Describe how vanadium (V) oxide acts as a catalyst in the contact process. Include any relevant equations.