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. Oxidation Reduction = reactions involving REDuction and OXidation

Overview Oxidation and reduction reactions always occur simultaneously (redox reactions)

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

OxidationReduction Gains oxygenLose Oxygen Lose HydrogenGains hydrogen Lose ElectronsGains electrons Oxygen Hydrogen e-e-

OXIDATIONREDUCTION = OXYGEN = HYDROGEN = ELECTRON = OXIDATION STATE REDOX REACTIONS Definitions:GAINLOSS LOSSGAIN LOSSGAIN INCREASEDECREASE 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.

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 –2Fe 2 O 3 + 3C  4Fe + 3CO 2 –Carbon was oxidized because it gained an oxygen

Oxidizing and Reducing Agents Now the confusing part… CuO + H 2  Cu + H 2 O 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.

Identifying Agents in an Equation CuO + H 2  Cu + H 2 O Reduction: CuO is the oxidizing agent Oxidation: H 2 is the reducing agent

Examples Is the reactant oxidized or reduced? Pb  PbO 3 SnO 2  SnO KClO 3  KCl C 2 H 6 O  C 2 H 4 O C 2 H 2  C 2 H 6

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

Pertaining to LEO… Mg + S  MgS Mg + S  Mg 2+ + S 2- 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

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…

Rules for deciding Oxidation States (Numbers) : 5. In a BINARY (2 elements) COMPOUND 1. In all UNCOMBINED ELEMENTS, atom’s ox. no. = 0. 2.In all COMPOUNDS, sum of ox. no.’s equals zero. 3. In all IONS, sum of ox. no.’s equals ion charge. 4.In all COMPOUNDS : Gp 1 elements the more electronegative atom given NEGATIVE ox. no. and the less electronegative atom given POSITIVE ox. no. In most COMPOUNDS, 6. H = + 1 except when bonded to a metal - metal must have the positive ox. no. 7. O = - 2except when bonded to F or in peroxides, e.g. Na 2 O 2 - F must have the negative ox. no. ) LEARN and PRACTISE Gp 2 elements Gp 3 elements Fluorine

Oxidation Numbers The oxidation numbers of atoms in a compound add up to zero. F O-2 H+1 Cl Find the oxidation state of C in CO 2 ? ? – 4 = 0 ? = +4 Put the +!

Oxidation Numbers The oxidation numbers of atoms in an ion add up to the charge on the ion. F O-2 H+1 Cl Oxidation state of S in SO 4 2- ? ? – 8 = -2 S = +6

ASSIGN AN OXIDATION NUMBER / STATE TO EACH ATOM IN : Cl 2 CO 3 2- Ca 2+ SO 3 2- Al 3+ ClO - H2OH2O IO 4 - CO 2 CH 4 ClF MnO 4 - NO 3 - Na 2 S 4 O 6 CuCl CuBr 2 N2N2 C 2 O 4 2- BrF 5 Mn 2 O 3 SF 6 CO S 2- BrF VCl 2 Na 2 S NO 2 - BrO 3 - NH 4 + H 2 SO 4 SO 4 2- I-I- S 2 O 3 2- NH 3 CCl 4 Cr 2 O 7 2- Cl(0) Ca(+2) Al(+3) H(+1)  O(-2) O(-2)  C(+4) F(-1)  Cl(+1) O(-2)  N(+5) Cl(-1)  Cu(+1) N(0) F(-1)  Br(+5) F(-1)  S(+6) S(-2) Cl(-1)  V(+2) O(-2)  N(+3) H(+1)  N(-3) O(-2)  S(+6) O(-2)  S(+2) Cl(-1)  C(+4) O(-2)  C(+4) O(-2)  S(+4) O(-2)  Cl(+1) O(-2)  I(+7) H(+1)  C(-4) O(-2)  Mn(+7) Na(+1) & O(-2)  S(+2.5) Br(-1)  Cu(+2) O(-2)  C(+3) O(-2)  Mn(+3) O(-2)  C(+2) F(-1)  Br(+1) Na(+1)  S(-2) O(-2)  Br(+5) O(-2) & H(+1)  S(+6) I(-1) H(+1)  N(-3) O(-2)  Cr(+6)

Problems What is the oxidation number of each element? I 2 Cr 2 O 3 AlCl 3 Na 2 SO 4 CaH 2

Identifying Redox Reactions using oxidation number Al + Fe 2 O 3  2 Fe + Al 2 O 3 Al increases from 0 to +3, it is Oxidized! Fe decreases from +3 to 0, it is Reduced!

Problems Zn  Zn 2+ Fe 3+  Fe 2+ CaCO 3  CaO + CO 2 AgNO 3  Ag Do Practice Exercises

Rem.OXIDATIONREDUCTION = OXIDATION No. INCREASEDECREASE 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.PROCESS Ox. No.’s OxidationReduction Cl 2  Cl - Ca  Ca 2+ NO 2  NO 3 - MnO 4 -  Mn 2+ SO 2  SO 4 2- IO 4 -  I 2 H 2 SO 4  S 2- Br 2  BrO - NH 4 +  NH 3 Cr 2 O 7 2-  Cr 3+ Cl(0)  (-1) Ca(0)  (+2) N(+4)  (+5) Mn(+7)  (+2) S(+4)  (+6) I(+7)  (0) S(+6)  (-2) Br(0)  (+1) N(-3)  (-3) Cr(+6)  (+3)          NONE