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AS Chemistry OXIDATION STATES, HALF EQUATIONS and REDOX REACTIONS

AS Chemistry OXIDATION STATES, HALF EQUATIONS and REDOX REACTIONS. REDOX REACTIONS. = reactions involving RED uction and OX idation. Definitions :. Remember “ OILRIG ” : O xidation I s L oss ; R eduction I s G ain (of electrons).

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AS Chemistry OXIDATION STATES, HALF EQUATIONS and REDOX REACTIONS

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

  2. REDOX REACTIONS = reactions involving REDuctionandOXidation 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.

  3. 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. (Rule 5) 7. O = - 2 except when bonded to F or in peroxides, e.g. Na2O2 - F must have the negative ox. no. (Rule 4)

  4. 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) O(-2)  C(+4) H(+1)  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)

  5. 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)

  6. Half Equations = equations showing the SEPARATEoxidation (loss of e-) and reduction (gain of e-) processes in any redox reaction e.g. 1 2Ca(s) + O2(g)  2CaO(s) 0  +2 Caoxidised Ca atoms - O2 mols -  O2reduced 0  -2 HALF EQUATIONS : Ca  Ca2+ + 2e- Oxidation : Reduction : O2 + 4e-  2O2- e.g. 2 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g) 0  +1 Naoxidised Na atoms - H2O mols - H(+1)  H(0)  H2O reduced HALF EQUATIONS : Na  Na+ + e- Oxidation : Reduction : 2H2O + 2e-  2OH- + H2

  7. General Method for Writing Half Equations e.g.1 MnO4- Mn2+ (NOT balanced ; occurs in acid)  MnO4- is reduced Mn(+7)  (+2) 1. Write formulas of “redox” particles and balance “changed” atoms MnO4- Mn2+ Reduction : 2. Insert e- on left for reduction, right for oxidation Number of electrons in half-equation CHANGE in oxidation number of “redox” atoms = MnO4-+ 5e-  Mn2+ Reduction : 3. Complete the balance(for atoms and charges) by inserting H2O and H+ or OH- as appropriate i.e. 4 O  4H2O  8H+ MnO4-+ 8H++ 5e- Mn2+ + 4H2O Reduction :

  8. e.g.2 Cl2 ClO4- (NOT balanced ; occurs in alkali)  Cl2 is oxidised Cl(0)  (+7) 1. Write formulas of “redox” particles and balance “changed” atoms Cl2 2 ClO4- Reduction : 2. Insert e- on left for reduction, right for oxidation Number of electrons in half-equation CHANGE in oxidation number of “redox” atoms = Cl2 2 ClO4- + 14e- Reduction : 3. Complete the balance(for atoms and charges) by inserting H2O and H+ or OH- as appropriate i.e. 8 O  8H2O  16OH- Cl2+ 16OH-  2 ClO4- + 14e- + 8H2O Reduction :

  9. e.g.3 Cu + HNO3 Cu2+ + NO2 (NOT balanced ; occurs in acid)  Cu oxidised Cu(0)  (+2) N(+5)  (+4) and HNO3 reduced 1. Write formulas of “redox” particles and balance “changed” atoms HNO3 NO2 Reduction : Cu  Cu2+ Oxidation : 2. Insert e- on left for reduction, right for oxidation Number of electrons in half-equation CHANGE in oxidation number of “redox” atoms = Reduction : Oxidation : HNO3+ e- NO2 Cu  Cu2+ + 2e- 3. Complete the balance(for atoms and charges) by inserting H2O and H+ or OH- as appropriate Oxidation : Cu  Cu2++ 2e- HNO3+ H++ e- NO2 + H2O Reduction :

  10. Write a half-equation for each of the following changes. • Cl2 to Cl- 6. Br- to Br2 • Pb2+ to Pb 7. Al to Al3+ • H2SO4 to H2S 8. At- to At2 • HNO3 to NO 9. Fe to Fe2+ • H2SO4 to SO2 10. Br- to Br2

  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 Example 1 X 1 Oxidation : Cu  Cu2++ 2e- X 2 HNO3+ H++ e- NO2 + H2O Reduction : Add : Cu + 2HNO3+ 2H++ 2e- Cu2+ + 2e-+2NO2 + 2H2O

  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. Example 2 X 5 Oxidation : Fe2+ Fe3++ e- MnO4-+ 8H++ 5e- Mn2+ + 4H2O Reduction : X 1 Add : 5Fe2+ + MnO4-+ 8H++ 5e- 5Fe3++ 5e- + Mn2+ + 4H2O

  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. Example 3 X 2 Oxidation : Cl2+ 2OH- ClO-+ H2O+ e- Cl2+ 2e- 2Cl- Reduction : X 1 Add : 2Cl2+ 4OH-+ 2e- 2ClO-+ 2e- + 2Cl- + 2H2O Cancel by 2 Cl2+ 2OH- ClO-+ Cl- + H2O 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. • Cl2 to Cl-with Br- to Br2 • Pb2+ to Pb with Al to Al3+ • H2SO4 to H2S with I- to I2 • HNO3 to NO with Fe to Fe2+ • H2SO4 to SO2with Br- to Br2

  15. The End

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