1. AS Chemistry OXIDATION STATES, HALF EQUATIONS and REDOX REACTIONS

2. OXIDATIONREDUCTION = OXYGEN = HYDROGEN = ELECTRON = OXIDATION STATE REDOX REACTIONS = reactions involving REDuction and OXidation 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.

3. 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 + 1 + 2 - 1 + 3 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. (Rule 5) 7. O = - 2except when bonded to F or in peroxides, e.g. Na 2 O 2 - F must have the negative ox. no. (Rule 4) LEARN and PRACTISE Gp 2 elements Gp 3 elements Fluorine

4. 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)

5. 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

6. Half Equations = equations showing the SEPARATE oxidation (loss of e - ) and reduction (gain of e - ) processes in any redox reaction e.g. 1 2Ca(s) + O 2 (g)  2CaO(s) Ca atoms -  Ca oxidised O 2 mols -  O 2 reduced Oxidation: Reduction: HALF EQUATIONS : 0  +2 0  -2 Ca  Ca 2+ + 2e - O 2 + 4e -  2O 2- e.g. 2 2Na(s) + 2H 2 O(l)  2NaOH(aq) + H 2 (g) Na atoms -  Na oxidised H 2 O mols -  H 2 O reduced Oxidation: Reduction: HALF EQUATIONS : 0  +1 H(+1)  H(0) Na  Na + + e - 2H 2 O + 2e -  2OH - + H 2

7. General Method for Writing Half Equations e.g.1 MnO 4 -  Mn 2+ (NOT balanced ; occurs in acid) Mn(+7)  (+2)  MnO 4 - is reduced Number of electrons in half-equation 3. Complete the balance (for atoms and charges) by inserting H 2 O and H + or OH - as appropriate 1. Write formulas of “redox” particles and balance “changed” atoms Reduction : MnO 4 -  Mn 2+ CHANGE in oxidation number of “redox” atoms 2. Insert e- on left for reduction, right for oxidation Reduction : MnO 4 - + 5e -  Mn 2+ = Reduction : MnO 4 - + 8H + + 5e -  Mn 2+ + 4H 2 O i.e. 4 O  4H 2 O  8H +

8. e.g.2 Cl 2  ClO 4 - (NOT balanced ; occurs in alkali) Cl(0)  (+7)  Cl 2 is oxidised Number of electrons in half-equation 3. Complete the balance (for atoms and charges) by inserting H 2 O and H + or OH - as appropriate 1. Write formulas of “redox” particles and balance “changed” atoms Reduction : Cl 2  2 ClO 4 - CHANGE in oxidation number of “redox” atoms 2. Insert e- on left for reduction, right for oxidation Reduction : Cl 2  2 ClO 4 - + 14e - = Reduction : Cl 2 + 16OH -  2 ClO 4 - + 14e - + 8H 2 O i.e. 8 O  8H 2 O  16OH -

9. e.g.3 Cu + HNO 3  Cu 2+ + NO 2 (NOT balanced ; occurs in acid) Cu(0)  (+2) N(+5)  (+4)  Cu oxidised and HNO 3 reduced Number of electrons in half-equation 3. Complete the balance (for atoms and charges) by inserting H 2 O and H + or OH - as appropriate 1. Write formulas of “redox” particles and balance “changed” atoms Oxidation : Reduction : Cu  Cu 2+ HNO 3  NO 2 CHANGE in oxidation number of “redox” atoms 2. Insert e- on left for reduction, right for oxidation Oxidation : Reduction : Cu  Cu 2+ + 2e - HNO 3 + e -  NO 2 = Oxidation : Reduction : Cu  Cu 2+ + 2e - HNO 3 + H + + e -  NO 2 + H 2 O

10. Write a half-equation for each of the following changes. 1.Cl 2 to Cl - 6. Br - to Br 2 2.Pb 2+ to Pb 7. Al to Al 3+ 3.H 2 SO 4 to H 2 S 8. At - to At 2 4.HNO 3 to NO 9. Fe to Fe 2+ 5.H 2 SO 4 to SO 2 10. Br - to Br 2

11. Combining half-equations to produce the full equation (a) An oxidation half-equation must be combined with a reduction half-equation (b)Combine in the ratio which balances out the electrons lost during oxidation with those gained during reduction. i.e. oxidation number changes must balance Oxidation : Reduction : Cu  Cu 2+ + 2e - HNO 3 + H + + e -  NO 2 + H 2 O X 1 X 2 Add : Cu + 2HNO 3 + 2H + + 2e -  Cu 2+ + 2e - + 2NO 2 + 2H 2 O Example 1

12. Remember (a) An oxidation half-equation must be combined with a reduction half-equation (b)Combine in the ratio which balances out the electrons lost during oxidation with those gained during reduction. Oxidation : Reduction : Fe 2+  Fe 3+ + e - MnO 4 - + 8H + + 5e -  Mn 2+ + 4H 2 O X 5 Add : 5Fe 2+ + MnO 4 - + 8H + + 5e -  5Fe 3+ + 5e - + Mn 2+ + 4H 2 O Example 2 X 1

13. Remember (a) An oxidation half-equation must be combined with a reduction half-equation (b)Combine in the ratio which balances out the electrons lost during oxidation with those gained during reduction. Oxidation : Reduction : Cl 2 + 2OH -  ClO - + H 2 O + e - Cl 2 + 2e -  2Cl - X 2 Add : 2Cl 2 + 4OH - + 2e -  2ClO - + 2e - + 2Cl - + 2H 2 O Example 3 X 1 Cl 2 + 2OH -  ClO - + Cl - + H 2 O Cancel by 2 Note : Chlorine is BOTH oxidised and reduced. Such a reaction is called a DISPROPORTIONATION

14. Use the half-equations written earlier and combine them to form the overall equation for. 1.Cl 2 to Cl - with Br - to Br 2 2.Pb 2+ to Pb with Al to Al 3+ 3.H 2 SO 4 to H 2 S with I - to I 2 4.HNO 3 to NO with Fe to Fe 2+ 5.H 2 SO 4 to SO 2 with Br - to Br 2

15. The End