Oxidation and Reduction

1. Oxidation and Reduction

2. Oxidation of Food: What a Waste! • Fruits and Vegetables oxidised when left in open air • Solution: Seal in plastic wrap • More radical: Add lemon juice to the cut fruit

3. People! Oxidation of… • Oxidation of nutrients causes increased activity of cells, leading to aging skin • Solution: Beauty products?

4. What is a redox reaction? • Redox – reduction + oxidation • Both processes occur simultaneously • Hence, one species is oxidised, another is reduced • So, what is oxidation, and what is reduction? • 3 different versions of the definition:

5. Redox Oxidation Reduction gain in oxygen loss of oxygen loss of hydrogen gain in hydrogen loss of electrons gain of electrons

6. Oxidation • Gain of oxygen in a species • E.g. Mg is oxidized to MgO 2Mg(s) + O2(g)  2MgO(s) • Loss of hydrogen in a species • E.g. HCl is reduced to Cl2 MnO2(s) + 4HCl(aq)  MnCl2(aq) + 2H2O(g) + Cl2(g)

7. Reduction • Loss of oxygen in a species • E.g. CuO is reduced to Cu CuO(s) + H2(g)  Cu(s) + H2O(l) • Gain of hydrogen in a species • E.g. C2 H4 is reduced to C2 H6 in the hydrogenation process C2 H4(g) + H2(g)  C2 H6(g)

8. Oxidation and Reduction • In terms of Electrons (OIL RIG: Oxidation Is Loss, Reduction Is Gain): 2Mg(s) + O2(g)  2MgO(s) Oxidation : Mg  Mg2+ + 2e- (Loss of electrons by Mg) Reduction : O2 + 4e-  2O2- (Gain of electrons by O2) • The two equations above are known as half-equations.

9. Oxidation numbers For the redox reaction such as S(s) + O2(g)  SO2(g), would the definition of electron transfer be applicable? A new definition is developed to overcome the problem • In terms of Oxidation numbers: • Oxidation: Increase in oxidation number in a species • Reduction: Decrease in oxidation number in a species The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred.

10. Simple rules on oxidation numbers • Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 • In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 • The oxidation number of oxygen is usually–2. • In H2O2 and O22- it is –1.

11. Deduce the oxidation number of (i) sulfur in sulfur dioxide (ii) nitrogen in NH4+ (iii)carbon in HCO3- (iii) chromium in Cr2O72- • The oxidation number of hydrogen is +1except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. 5. 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.

12. Naming inorganic compounds • The oxidation number is inserted immediately after the name of ion. Example • Iron(II)chloride, FeCl2 and iron(III)chloride, FeCl3 • Potassium chromate(VI), K2CrO7 and potassium manganate(VII), KMnO4

13. Identifying redox reactions Consider the following reactions • 2FeCl2(s) + Cl2(g)  2FeCl3(s) • Mn(NO3)2(s)  MnO2(s) + 2NO2(g) • (NH4) 2Cr2O7(s)  Cr2O3(s) + 4H2O(g) + N2(g) What if there’s no change in the oxidation numbers during the chemical reaction?

14. Non-redox reactions • The oxidation states of the elements remained unchanged in the following reactions: • Neutralisation reactions:

15. Precipitation reactions: • Complex formation: • Another reaction: ligand Tetraammine copper(II) complex(deep blue solution)

16. Disproportionation • A disproportionation reaction is a redox reaction in which one species is simultaneously oxidised and reduced. Example 2H2O2(aq)  2H2O(l) + O2(g) • Reactant : • Oxidation number of O in H2O2 is -1 • Product : • Oxidation number of O in H2O is -2 • Oxidation number of O in O2 is 0 Reduction Oxidation

17. Consider another example: Cl2(g) + H2O(l)  HOCl(aq) + HCl(aq)

18. Oxidising and Reducing agent • An oxidising agent is a substance that brings about oxidation by accepting electrons from the substance it oxidises. It is always reduced in the process. • A reducing agent is a substance that brings about reduction by donating electrons to the substance it reduces. It is always oxidised in the process. 2Mg + O2 → 2MgO • Mg is oxidised, and thus is the reducing agent • O2 is reduced, and thus is the oxidising agent

19. List of common Oxidising and Reducing Agents • Realise something? • H2O2 is both an oxidising and a reducing agent! • If a stronger oxidising agent is present, H2O2 is reducing

20. Identifying Redox reactions (a) Consider the reaction S8 + 12O2 8SO3 (i) Does it involve ionic or covalent compounds? (ii) Is this a redox reaction? Explain (iii) What is the oxidising agent and reducing agent?

21. Consider each of the 2 reactions below • 2AgNO3 + Na2S  Ag2S + 2NaNO3 • 5As4O6 + 8MnO4- + 18H2O  20 AsO43- + 8 Mn2+ + 36H+ (i) Is this a redox reaction? Explain (ii) What is the oxidising agent and reducing agent?

22. Balancing redox equations The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution. (1) Write the unbalanced equation for the reaction in ionic form. Fe2+ + Cr2O72-  Fe3+ + Cr3+ (2) Separate the equation into two half equations Oxidation : Fe2+ Fe3+ Reduction: Cr2O72- - Cr3+ (3) Balance the atoms other than O and H in each half eqn Cr2O72- 2Cr3+ (4) Add H2O to balance O atoms and H+ to balance H atoms. Cr2O72- 2Cr3+ + 7H2O 14H+ + Cr2O72- 2Cr3+ + 7H2O

23. (5) Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe2+ Fe3+ + e- 14H+ + Cr2O72- + 6e-  2Cr3+ + 7H2O (6) If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 6Fe2+ 6Fe3+ + 6e- 14H+ + Cr2O72- + 6e-  2Cr3+ + 7H2O

24. (7) Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. You should also cancel like species. Oxidation : 6Fe2+ 6Fe3+ + 6e- Reduction : 14H+ + Cr2O72- + 6e-  2Cr3+ + 7H2O Overall eqn: 14H+ + Cr2O72- + 6Fe2+  6Fe3+ + 2Cr3+ + 7H2O (8) Verify that the number of atoms and the charges are balanced. 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. You should combine H+ and OH- to make H2O.

25. Example Balance the following reaction:

26. Example Write a redox equation for the reduction of acidified manganate(VII) ions and the oxidation of methanol using the balanced half-equations below: • 2 H2O(l) + CH3OH(l)  CO2(g) + H2O(l) + 6H+(aq) + 6e- • MnO4-(aq)+ 8H+(aq) + 5e- Mn2+ (aq)+ + 4H2O(l)

27. Reactivity Series and cell voltage Negative electrode • Potassium • Sodium • Calcium • Magnesium • Aluminium • Carbon • Zinc • Iron • Tin • Lead • Copper • Mercury • Silver • Gold • Magnesium • Zinc • Iron • Tin • Lead • Copper • Silver Reactivity decrease Lower Voltage Higher Voltage Positive electrode

28. 2Al + Fe2O3 2Fe + Al2O3 THERMITE REACTION

29. Uncontrolled spontaneous reaction . How can this be harnessed to do work?

30. . Controlled redox reaction Zn  Zn2+ + 2e− Cu2+ + 2e−  Cu Zn + Cu 2+ Cu + Zn2++ overall reaction but electrons are forced outside of the cells

31. Voltaic Cell ΔG = neg Galvanic Battery

32. Voltaic Cells anode oxidation cathode reduction - + spontaneous redox reaction 19.2

33. Explain what happens in this system. Follow the circulation of one negative charge in the cell.

34. An electric current flows between the anode and cathode because there is a difference in electrical potential between the 2 electrodes, measured by a voltmeter (in volts) : CELL POTENTIAL. • The voltaic cell depends on • the nature of the electrodes and the ions • the concentrations of ions and • the temperature at which the celll is operated.

35. 1.10 V 1.0 M 1.0 M CELL POTENTIAL, E • Electrons are “driven” from anode to cathode by an electromotive force or emf. • For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 ˚C and when [Zn2+] and [Cu2+] = 1.0 M. Zn and Zn2+, anode Cu and Cu2+, cathode

36. CELL POTENTIAL, E • For Zn/Cu cell, potential is +1.10 V at 25 ˚C and when [Zn2+] and [Cu2+] = 1.0 M. • is the STANDARD CELL POTENTIAL, Eo • a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 ˚C.

37. Cell potentials of selected v cells The further apart the two metals are in the reactivity series, the higher the cell potential.

38. Reactivity Series and cell voltage Negative electrode • Potassium • Sodium • Calcium • Magnesium • Aluminium • Carbon • Zinc • Iron • Tin • Lead • Copper • Mercury • Silver • Gold • Magnesium • Zinc • Iron • Tin • Lead • Copper • Silver Reactivity decrease Lower Voltage Higher Voltage Positive electrode

39. Electrolysis • Electricity is passed from a battery through a liquid which may be a solution/molten solid. • The plates which carry the electricity into the liquid are called electrodes • Molten ionic compounds or aqueous solution of ionic compounds that allows electricity to pass through are called electrolytes

40. Electrodes • Metal plates or graphite rods that conduct electricity into the electrolyte • Eg. Platinum, copper • Cathode: • Electrode that is connected to the negative terminal of the battery. Postively charged ions, cations, moved towards the cathode • Anode: • Electrode that is connected to the positive terminal of the battery. Negatively charged ions, anions, moved towards the anode

41. Electrolytic cell

42. Conduction of electricity • Conductor is a substance which conducts electricity but is not chemically changed during the conduction • Presence of freely moving valence electrons • Eg. All metals and graphite • Insulator does not allow the passage of electricity. • Valence electrons are held in fixed positions • Eg. Sulphur, phosphorus, diamond, solid state crystalline salts, wood and glass

43. Electrolytes and non-Electrolytes • Electrolytes: • Molten ionic compounds or aqueous solution of ionic compounds that allows electricity to pass through and are decomposed in the process • Eg. Acids, Alkali, Salts dissolved in water, molten salts • Non-electrolytes: • Does not allow passage of electricity • Eg. Distilled water, alcohol, turpentine, oil, paraffin, organic solvents

44. When electricity is passed through an electrolyte, chemical decomposition occurs • This involves the ‘splitting up’ of the electrolyte • Since all electrolytes are ionic, composed of positively and negatively charged ions • The process: When an electric current pass through the electrolyte, ions in the solution migrate towards the oppositely charged electrode • This discharge of ions at the electrodes results in the chemical decomposition of the electrolyte to form its elements.

45. At the anode, negatively charged ions lose their electron(s) to the anode (connected to positive terminal of battery) to form neutral atoms. • The negatively charged ions are said to be oxidised and discharged at the anode. • Oxidation occured at the anode. • At the cathode, positively charged ions gain electron(s) from the cathode (connected to negative terminal of battery) to form neutral atoms. • The positively charged ions are said to be reduced and discharged at the cathode. • Reduction occured at the cathode.

46. Rules for Predicting Selective Discharge of Cations

47. Rules for Predicting Selective Discharge of Anions

48. Electrolysis of Concentrated Hydrochloric Acid Carbon rods as electrodes

49. Electrolysis of Molten Compounds • Many ionic compounds are binary compounds. • A binary compound is a compound containing only 2 elements. It contains a metal cation and a non-metal anion. • The electrolysis of a molten binary compound will yield a metal and a non-metal as products.

50. Electrolysis of Molten Lead(II) Bromide Carbon rods as electrodes