1 redox

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1 redox

  1. 1. OXIDATION AND REDUCTIONREDOX REACTIONSDefinition: Chemical reactions involving OXIDATION and REDUCTION occurring simultaneouslyOxidation and reduction in terms of:i. loss or gain of oxygenii. loss or gain of hydrogeniii. transfer of electronsiv. changes in oxidation numberOxidation: - combination of substance with oxygen - loss of hydrogen - loss of electrons - an increase in oxidation numberReduction: - removal of oxygen from substance - gain of hydrogen - gain of electrons - a decrease in oxidation number
  2. 2. Oxidant / Oxidizing agent:  The substance that causes oxidationReductant / Reducing agent:  The substance that causes reductionImportant Neutralization and Precipitation are NOT redox reactionsWHY? You tell me[wait until we discuss the ‘oxidation number’]
  3. 3. A. Loss or gains of oxygenOxidation: - combination of substance with oxygenReduction: - removal of oxygen from substanceExample:2CuO + C  2Cu + CO2CuO loses its oxygen to form copper. Carbon gains theoxygen to form carbon dioxide.CuO causes the oxidation of carbon. Carbon causes thereduction of CuO.Material undergoes oxidation: Carbon, CMaterial undergoes reduction: Copper(II) oxide, CuOOxidizing agent / oxidant : Copper(II) oxide, CuOReducing agent / reductant : Carbon, C
  4. 4. B. Loss or gains of hydrogenOxidation: - loss of hydrogenReduction: - gain of hydrogenExample:H2S + Cl2  S + 2HClH2S loses its hydrogen to form sulfur. Cl2 gains thehydrogen to form HCl.H2S causes the reduction of Cl2. Cl2 causes the oxidationof H2S.Material undergoes oxidation : Hydrogen sulphide, H2SMaterial undergoes reduction : Chlorine, Cl2Oxidizing agent / oxidant : Chlorine, Cl2Reducing agent / reductant : Hydrogen sulphide, H2S
  5. 5. C. Tranfer of electronsOxidation: - loss of electronsReduction: - gain of electronsExample 1: [Daniell cell]Zn + Cu2+  Zn2+ + CuElectrons transfer from zinc to copper(II) ions.Half reaction:One zinc atom loses 2 electrons to form one zinc ion.Zinc is oxidized to zinc ions.Oxidizing half-equation: Zn  Zn2+ + 2e-Half reaction:One copper(II) ion gains 2 electron to form one copperatom. Copper(II) ion is reduced to copper.Reduction half-equation: Cu2+ + 2e  Cu
  6. 6. Copper(II) ions act as oxidizing agent because itaccepts electrons.Zinc acts as reducing agent because it releaseselectrons.Oxidizing half-equation: Zn  Zn2+ + 2e-Reduction half-equation: Cu2+ + 2e-  Cu[balanced the number of electron please] Cancel the electronsZn + Cu2+ + 2e-  Cu + Zn2+ + 2e-Thus;Ionic Equation : Zn + Cu2+  Zn2+ + CuChemical equation: Zn + CuSO4  ZnSO4 + Cu[note: the sum of the two half-equations gives the ionic equation]
  7. 7. D. Changes in oxidation numberOxidation number?Definition: The oxidation number of an element is the charge that the atom of the element would have if complete transfer of electron occursRules: pg 107 figure 3.1 (read)Tips from the rules: • the oxidation number for atom and molecule is zero • the oxidation number for monoatomic ion is equal to its charge. • the sum of oxidation numbers of all elements in the compound is zero • the sum of oxidation number of all elements in polyatomic ions is equal to the charge of the ionsOxidation: - an increase in oxidation numberReduction:
  8. 8. - a decrease in oxidation numberExample: [ I use the same example but from different perspective]Zn + Cu2+  Zn2+ + Cu[easy way to detect which substance is oxidized or reduced] Oxidation number +2 Zn2+ Cu2+ Oxidation Reduction (loses electrons) (gains electrons) Zn Cu 0The oxidation number of zinc, Zn increases from0 to +2. Zn undergoes oxidation to zinc ions, Zn2+.The oxidation number of copper(II) ions, Cu2+ decreasesfrom +2 to 0. Cu2+ undergoes reduction to copper, Cu.Copper(II) ions, Cu2+ act as oxidizing agent.Zinc, Zn acts as reducing agent.Oxidizing half-equation: Zn  Zn2+ + 2e-Reduction half-equation: Cu2+ + 2e-  CuIonic Equation : Zn + Cu2+  Zn2+ + Cu[note: every redox reaction MUST have half equations
  9. 9. and ionic equation]Example:2Mg + O2  2MgO Oxidation number +2 Mg2+ Oxidation (loses electrons) 0 Mg O2 Reduction (gains electrons) -2 O2-Describe the process……..The oxidation number of magnesium, Mg increasesfrom 0 to +2. Mg undergoes oxidation to magnesiumions, Mg2+.Oxidizing half-equation: Mg  Mg2+ + 2e-1 magnesium atom loses 2 electrons to from 1mzgnesium ions. Mg undergoes oxidation tomagnesium ions, Mg2+.The oxidation number of oxygen, O2 decreases from0 to -2. O2 undergoes reduction to oxide ions, O2-.
  10. 10. Reduction half-equation: O2 + 4e-  2O2-1 molecule of oxygen gains 4 electron to form 2 oxideions. O2 undergoes reduction to oxide ions, O2-.Oxygen, O2 act as oxidizing agent.Magnesium, Mg acts as reducing agent.Oxidizing half-equation: Mg  Mg2+ + 2e- (×2) [it becomes: 2Mg  2Mg2+ + 4e]Reduction half-equation: O2 + 4e-  2O2-[balanced the number of electrons for both half equation] Cancel the electrons2Mg + O2 + 4e-  2Mg2+ + 2O2- + 4e-Thus;Ionic Equation: 2Mg + O2  2Mg2+ + 2O2- or 2Mg + O2  2MgOHW: pg 110 Learning task 3.1 Analyzing
  11. 11. [a and b] Explain each of the reactionSolution:(a) (i) 2H2 + O2  2H2O [rule: pg 107] Oxidation number +1 H+ Oxidation (loses electrons) 0 H2 O2 Reduction (gains electrons) 2- -2 OThe oxidation number of hydrogen, H2 increases from0 to +1. H2 undergoes oxidation to hydrogen ion, H+.Oxidizing half-equation: H2  2H+ + 2e-1 molecule of hydrogen loses 2 electrons to form2 hydrogen ions.The oxidation number of oxygen, O2 decreases from0 to -2. O2 undergoes reduction to oxide ions, O2-.Reduction half-equation: O2 + 4e-  2O2-1 molecule of oxygen gains 4 electrons to form
  12. 12. 2 oxide ions.Oxygen, O2 act as oxidizing agent.Hydrogen, H2 acts as reducing agent.Oxidizing half-equation: H2  2H+ + 2e- (×2)Reduction half-equation: O2 + 4e-  2O2-[balanced the number of electrons for both half equation] Cancel the electrons2H2 + O2 + 4e-  4H+ + 2O2- + 4e-Thus;The ionic equation 2H2 + O2  4H+ + 2O2- or 2H2 + O2  2H2OEasy lah!Prepared by;Kamal Ariffin Bin SaaimSMKDBL

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