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Oxidation and Reduction

Oxidation and Reduction. ………………………. Objectives. Oxidised, reduced Definition- oxidising/reducing agent, oxidant/reductant Rules of oxidation number Refer to oxidation number and decide-reduced or oxidised. Redox Reactions. = reactions involving RED uction and OX idation. Reduction.

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Oxidation and Reduction

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  1. Oxidation and Reduction ……………………….

  2. Objectives • Oxidised, reduced • Definition- oxidising/reducing agent, oxidant/reductant • Rules of oxidation number • Refer to oxidation number and decide-reduced or oxidised

  3. Redox Reactions. = reactions involving REDuctionandOXidation Reduction Oxidation

  4. Overview • Oxidation and reduction reactions always occur simultaneously (redox reactions)

  5. 3 ways of looking at oxidation and reduction • 1. oxidation is a gain of oxygen atoms, reduction is a loss of oxygen atoms • 2. oxidation is a loss of hydrogen atoms, reduction is a gain of hydrogen atoms • 3. oxidation is a loss of electrons, reduction is a gain of electrons • Most fundamental explanation, what we will be dealing with the most

  6. Oxygen Hydrogen e-

  7. REDOX REACTIONS Definitions: Remember “OILRIG” : Oxidation Is Loss ; Reduction Is Gain (of electrons) Oxidation states (also called oxidation numbers) are numbers assigned to EACH ATOM that takes part in a reaction. Oxidation states are assigned using a set of International rules.

  8. Redox-loss or gain of Oxygen • Remember that reduction is a loss of oxygen from a compound • We converted iron ore to metallic iron • We removed oxygen from the iron(III) oxide • 2Fe2O3 + 3C  4Fe + 3CO2 • Carbon was oxidized because it gained an oxygen

  9. Oxidizing and Reducing Agents • Now the confusing part… • CuO + H2 Cu + H2O • Cu goes from +2 to 0 • Cu is reduced, therefore it is called an oxidizing agent (OXIDANT) because it causes some other substance to be oxidized • H goes from 0 to +1 • H is oxidized, therefore it is called a reducing agent (REDUCTANT)because it causes some other substance to be reduced.

  10. Identifying Agents in an Equation • CuO + H2 Cu + H2O Reduction: CuO is the oxidizing agent Oxidation: H2 is the reducing agent

  11. Examples • Is the reactant oxidized or reduced? • Pb  PbO3 • SnO2  SnO • KClO3 KCl • C2H6O  C2H4O • C2H2  C2H6

  12. Leo the Lion! • LEO the lion says GER • Loss of electrons is oxidation, gain of electrons is reduction

  13. Pertaining to LEO… • Mg + S  MgS • Mg + S  Mg2+ + S2- • Magnesium is oxidized • Said to be the reducing agent • Substance in the reaction that loses electrons • Sulfide sulfur atom is reduced • Said to be the oxidizing agent • Substance in the reaction that gains electrons

  14. Oxidation Numbers • A count of the electrons transferred or shared in the formation or breaking of chemical bonds • To see if electron is lost or gained in the reaction an oxidation number is assigned to each element • Follow a set of rules to assign oxidation number…

  15. Rules for deciding Oxidation States (Numbers) : 1. In all UNCOMBINED ELEMENTS,atom’s ox. no. = 0 . LEARN and PRACTISE 2. In all COMPOUNDS,sum of ox. no.’s equals zero. 3. In all IONS,sum of ox. no.’s equals ion charge. + 1 • In all COMPOUNDS : Gp 1 elements Gp 2 elements + 2 Gp 3 elements + 3 Fluorine - 1 5. In a BINARY (2 elements) COMPOUND the more electronegative atom given NEGATIVE ox. no. and the less electronegative atom given POSITIVE ox. no. In most COMPOUNDS, except when bonded to a metal 6. H = + 1 - metal must have the positive ox. no. 7. O = - 2 except when bonded to F or in peroxides, e.g. Na2O2 - F must have the negative ox. no. )

  16. Oxidation Numbers • The oxidation numbers of atoms in a compound add up to zero. Find the oxidation state of C in CO2? ? – 4 = 0 ? = +4 Put the +!

  17. Oxidation Numbers • The oxidation numbers of atoms in an ion add up to the charge on the ion. Oxidation state of S in SO42-? ? – 8 = -2 S = +6

  18. Cl2 CO32- Ca2+ SO32- Al3+ ClO- H2O IO4- CO2 CH4 ClF MnO4- NO3- Na2S4O6 CuCl CuBr2 N2 C2O42- BrF5 Mn2O3 SF6 CO S2- BrF VCl2 Na2S NO2- BrO3- NH4+ H2SO4 SO42- I- S2O32- NH3 CCl4 Cr2O72- ASSIGN AN OXIDATION NUMBER / STATE TO EACH ATOM IN : O(-2)  C(+4) Cl(0) O(-2)  S(+4) Ca(+2) Al(+3) O(-2)  Cl(+1) H(+1)  O(-2) O(-2)  I(+7) H(+1)  C(-4) O(-2)  C(+4) F(-1)  Cl(+1) O(-2)  Mn(+7) Na(+1) & O(-2)  S(+2.5) O(-2)  N(+5) Br(-1)  Cu(+2) Cl(-1)  Cu(+1) N(0) O(-2)  C(+3) F(-1)  Br(+5) O(-2)  Mn(+3) F(-1)  S(+6) O(-2)  C(+2) S(-2) F(-1)  Br(+1) Cl(-1)  V(+2) Na(+1)  S(-2) O(-2)  N(+3) O(-2)  Br(+5) H(+1)  N(-3) O(-2) & H(+1)  S(+6) O(-2)  S(+6) I(-1) O(-2)  S(+2) H(+1)  N(-3) Cl(-1)  C(+4) O(-2)  Cr(+6)

  19. Problems • What is the oxidation number of each element? • I2 • Cr2O3 • AlCl3 • Na2SO4 • CaH2

  20. Identifying Redox Reactions using oxidation number 0 +3 -2 0 +3 -2 • 2 Al + Fe2O3 2 Fe + Al2O3 • Al increases from 0 to +3, it is Oxidized! • Fe decreases from +3 to 0, it is Reduced!

  21. Problems • Zn  Zn2+ • Fe3+  Fe2+ • CaCO3  CaO + CO2 • AgNO3  Ag • Do Practice Exercises

  22. Work out the oxidation number change for each of the following process and use it to decide whether it is an OXIDATION or a REDUCTION.  Cl(0)  (-1)  Ca(0)  (+2)  N(+4)  (+5)  Mn(+7)  (+2)  S(+4)  (+6)  I(+7)  (0)  S(+6)  (-2)  Br(0)  (+1) NONE N(-3)  (-3)  Cr(+6)  (+3)

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