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Topic 9 Oxidation and Reduction

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  1. Topic 9 Oxidation and Reduction • Introduction • Oxidation numbers • Redox equations • Reactivity • Voltaic cells • Electrolytic cells

  2. 9.1 Introduction to oxidation and reduction Definition: Oxidation: Particle that loses electron(s). Na  Na+ + e- Reduction: Particle that gains electron(s). Cl2+ 2e- 2 Cl-or • If something is oxidised, something else must be reduced in a chemical reaction.

  3. OILRIG Oxidation Is Loss, Reduction Is Gain … ofelectrons

  4. Half reactions • Consider the redox reaction: 2 Na + Cl2 2 NaCl • This reaction can be split into two “half reactions”: • Oxidation reaction: 2 Na  2 Na+ + 2 e- • Reduction reaction: Cl2 + 2e- 2 Cl-

  5. Oxidation number • It’s sometimes difficult to see if it is a redox reaction or if a particle has been oxidised/reduced. • Then you have to find out the oxidation numbers or oxidation state of the atoms in the compound. • The oxidation number is placed above the atom symbol often in roman figures but not in IB =>Use normal figures e.g +2 (Charge: 2+)

  6. Rules for determining oxidation number: • Elements in their element state: 0 • The total oxidation state = the charge of the compound H2O total oxidation state = 0 H3O+ total oxidation state = 1 • “Atomic ions”: the same oxidation number as its charge Cl- oxidation number = -1 element Na+ oxidation number = +1 Al3+ oxidation number = +3

  7. Some elements often have the same oxidation number • Fluorine: -1 • Hydrogen: +1 (except in hydrides = -1) • Oxygen: -2 (except in peroxides = -1, bonded to F= pos.) • The halogens: -1 (except when bonded to oxygen or a more electronegative halogen.)

  8. Find the oxidation number in following compounds • Cl2 0 => element • Cu2+ +2=> atomic ions: the same oxidation number as its charge • CH4 H: +1 • C: -4=> total oxidation state = the charge of the compound • H2SO3 H: +1 => total +2 O: -2=> total –6 S: +4( 2 + 4 - 6 = 0) • NO3- O: -2=> total -6 N: 5 (5 – 6 = -1)

  9. 9.2 In a chemical reaction • Oxidation: the atom that has an increase in oxidation number. • Reduction: the atom that has a decrease in oxidation number. • No change in oxidation number for any atom = no redox reaction.

  10. Is following compounds being oxidised or reduced? • NO  NO3- N: +2  N: +5 => oxidation • N2O  NH3 N: +1  N: -3 => reduction • HNO2 NO2- N: +3  N: +3 => no redox

  11. Naming compounds: • Oxidation number is also often used in naming compounds: • FeCl2: Iron(II)chloride (2 = Ox. no. of Fe) • CuO Copper(II)oxide • Cu2O Copper(I)oxide

  12. Redox reactions • Balance and put together half-reactions: by calculating electrons Silver ions react with magnesium metal: • Oxidation reaction: Mg  Mg2++ 2 e- • Reduction reaction: 2 Ag+ + 2 e-2 Ag Total rection: Mg + 2 Ag+ Mg2+ + 2 Ag

  13. Reducing agent • An element like sodium, Na, is eager to become an ion through oxidation: Na  Na+ + e- • Then some other particle, X, must be reduced (X + e- X-) • Sodium is then said to be a reducing agent • A reducing agent reduces a compound by self being oxidised

  14. Oxidising agent • An element like chlorine, Cl, is eager to become an ion through reduction: Cl+ e-Cl- • Then some other particle must become oxidised • Chlorine is then said to be a oxidizing agent • An oxidising agent oxidises a compound by self being reduced

  15. Balance redox reactions • An easy example to start with: Cu + Ag+ Cu2+ +Ag Divide into Half equations 1. Balance one of the reactants; Cu  Cu2+ + 2e- 2. Balance the other; Ag+ + e-  Ag Have to bemultiplied by 2 to get the same number of electrons. 3. Add the reactions; Cu + 2 Ag+ Cu2+ + 2 Ag 4. Check atoms and charge.

  16. Balance redox reactions in acidic solutions Cu + HNO3 +….  Cu(NO3)2 + NO + ….. 1. Half reactions: (1) Cu  Cu2+ + 2e- (2) HNO3 NO 2. Check and balance half reactions for O and H O not balanced in (2) => add H2O so oxygen balances (2) HNO3 NO + 2 H2O ; H not balanced in (2) => add H as H+ (acidic solution, remember…) (2) HNO3+ 3 H+ NO + 2 H2O

  17. Balance redox reactions in acidic solutions (II) (1) Cu  Cu2+ + 2e- (2) HNO3+ 3 H+ NO + 2 H2O 3. Check and balance half reactions for charges If charges not balanced =>add e- (2) HNO3 + 3 H+ + 3 e- NO + 2 H2O 4. Check and balance half reactions so they use the same no of e- Multiply (1) with 3 and (2) with 2 => 6 e- produced and consumed in each half-reaction: (1) 3Cu 3 Cu2+ + 6 e- (2)2 HNO3+6 H+ + 6 e-2 NO + 4H2O 5. Add the reactions: 3 Cu +2 HNO3 + 6 H+ + 6 e- 3 Cu2+ + 6 e- + 2 NO + 4 H2O Check atoms and e- : Equation balanced

  18. Balance redox reactions in acidic solutions (III) 6. Remove electrons. 3 Cu +2 HNO3 + 6 H+ 3 Cu2+ + 2 NO + 4 H2O 7. Fix the reaction (if needed). Add 6 NO3- on both side: 3 Cu +2 HNO3 + 6 H+ + 6 NO3- 3 Cu2+ + 2 NO + 4 H2O + 6 NO3- 8. Simplify: 3 Cu +8 HNO3 3 Cu(NO3)2 + 2 NO + 4 H2O Check atoms and charges.

  19. 9.3 Reactivity • Redox couple: A species that gain or lose electron(s), e.g. Na++ e-→ Na • If a compound is a good reducing agent (easily oxidized/lose e-), the “other form” will be a bad oxidising agentand vice versa • Redox couples can be arranged in a reactivity series

  20. Redox reactivity series Redox couple Oxidised formReduced form Na+ + e- Na Goodred. agent Mg2+ + 2e-Mg Fe2+ + 2e- Fe 2 H+ + 2e-  H2 Cu2+ + 2e- Cu I2+ 2e- 2 I- Br2+ 2e-2 Br- Good ox. agentF2 + 2e- 2 F -

  21. In the upper part of the redox reactivity seriesthe redox couple prefer to be on the left side (oxidised form) In the lower part of the table the redox couple prefer to be on the right side (reduced form) Eg. sodium is rather Na+ than Na, flourine is rather F2 than F-

  22. You can use the redox reactivity series to predict if a reaction is possible or not by comparing redox couples • If the redox couple standing above in the table is going to the left and the redox couple standing below is going to the right, then a reaction will occur. • Mg2+ + 2e-  Mg • Br2 + 2e- 2 Br- => Mg + Br2 MgBr2 is possible • If the above couple going to the right and the pair below to the left no reaction will occur. • E.g. F2 + 2I- 2F- + I2 OK but I2 + F- no reaction

  23. 9.4 Voltaic cell • The power of oxidising and reduction can be given in volts (V) • You can measure the potential between two metals and their ions in a galvanic cell • In a galvanic/voltaic cell chemical energy is converted into electrical energy • It’s a spontaneous reaction

  24. The less noble metal (Zn) will oxidise: Zn  Zn2+ +2e-. This will be the negative pole. The electrons will pass the voltmeter and reach the copper metal and copper ions will be reduced: Cu2+ +2e- Cu. This will be the positive pole. In the salt bridge ions will go from Copper half cell to Zinc half cell. It balances the charges and will give you a closed circuit.

  25. Cell diagram Cu2+ (aq) Cu (s) + - Zn (s) Zn2+ (aq) Positive electrode, cathode Negative electrode, anode

  26. Standard electrode potentials • In the table “standard electrode potential” you find potentials of different redox couples compared to a “standard hydrogen electrode”. • If you want to calculate the potential, E, you take the difference between the positive half cell and the negative half cell: E = E+ -E- • Use of Voltaic cells: Batteries

  27. 9.5 Electrolytic cell • In electrolysis electrical energy is converted to chemical energy. It’s a non-spontaneous reaction. • Electrolysis can be done in ionic aqueous solution (= electrolyte) or in molten salt (= electrolyte).

  28. Electrolysis of CuBr2 + pole/electrode = Anode: 2 Br- Br2 + 2e- Oxidation Reduction -pole/electrod= Cathode: Cu2+ + 2e-  Cu

  29. Electrolysis ofmoltensalt: e.g. NaCl(l) Use of electrolysis: Electroplating, analysis, chargeable batteries, purification of metals

  30. Differences between Voltaic cells and Electrolysis Voltaic cell Electrolysis Non-spontaneous Electrical energy Chemical energy + electrode: Oxidation (Anode) - electrode: Reduction (Cathode) • Spontaneous • Chemical energy Electrical energy • + electrode: Reduction (Cathode) • - electrode: Oxidation (Anode)