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# Halfequations

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### Transcript of "Halfequations"

1. 1. AS Chemistry OXIDATION STATES, HALF EQUATIONS and REDOX REACTIONS
2. 2. REDOX REACTIONS = reactions involving REDuction and OXidation Definitions: OXIDATION REDUCTION = OXYGEN GAIN LOSS = HYDROGEN LOSS GAIN = ELECTRON LOSS GAIN = OXIDATION STATE INCREASE DECREASE 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. 3. Rules for deciding Oxidation States (Numbers) : 1. In all UNCOMBINED ELEMENTS, atom’s ox. no. = 0 . 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, 6. H = + 1 except when bonded to a metal - 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) 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 4. In all COMPOUNDS : Gp 1 elements Gp 2 elements +2 Gp 3 elements +3 Fluorine - 1
4. 4. ASSIGN AN OXIDATION NUMBER / STATE TO EACH ATOM IN : Cl2 Cl(0) CO32- O(-2)  C(+4) Ca2+ Ca(+2) SO32- O(-2)  S(+4) Al3+ Al(+3) ClO- O(-2)  Cl(+1) H2O IO4- CO2 H(+1)  O(-2) O(-2)  C(+4) CH4 O(-2)  I(+7) H(+1)  C(-4) ClF F(-1)  Cl(+1) MnO4- NO3- O(-2)  N(+5) Na2S4O6 CuCl BrF5 Cl(-1)  Cu(+1) N(0) F(-1)  Br(+5) SF6 F(-1)  S(+6) CO O(-2)  C(+2) S2- S(-2) Cl(-1)  V(+2) BrF F(-1)  Br(+1) Na2S NO2- O(-2)  N(+3) BrO3- Na(+1)  S(-2) O(-2)  Br(+5) NH4+ H(+1)  N(-3) H2SO4 O(-2) & H(+1)  S(+6) SO42- I- S2O32- O(-2)  S(+6) O(-2)  S(+2) NH3 I(-1) H(+1)  N(-3) CCl4 Cl(-1)  C(+4) Cr2O72- O(-2)  Cr(+6) N2 VCl2 O(-2)  Mn(+7) Na(+1) & O(-2)  S(+2.5) CuBr2 Br(-1)  Cu(+2) C2O42- O(-2)  C(+3) O(-2)  Mn(+3) Mn2O3
5. 5. Rem. OXIDATION REDUCTION = OXIDATION No. INCREASE DECREASE 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 Cl2  Cl- Cl(0)  (-1) Ca  Ca2+ NO2  NO3- Ca(0)  (+2) N(+4)  (+5) MnO4-  Mn2+ Oxidation Reduction SO2  SO42- Mn(+7)  (+2)  (+6) S(+4)      IO4-  I2 I(+7)  (0)  H2SO4  S2- S(+6)  (-2)  Br2  BrO- Br(0)  (+1) NH4+  NH3 Cr2O72-  Cr3+ N(-3)  (-3) Cr(+6)  (+3)  NONE 
6. 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) Ca atoms O2 mols + O2(g) -  2CaO(s) 0  +2 0  -2  Ca oxidised  O2 reduced HALF EQUATIONS : Ca  Ca2+ + 2eOxidation : Reduction : O2 + 4e-  2O2e.g. 2 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g) Na atoms H2O mols - 0  +1 H(+1)  H(0)  Na oxidised  H2O reduced HALF EQUATIONS : Na  Na+ + eOxidation : Reduction : 2H2O + 2e-  2OH- + H2
7. 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 Reduction : MnO4-  Mn2+ 2. Insert e- on left for reduction, right for oxidation Number of electrons CHANGE in oxidation = in half-equation number of “redox” atoms Reduction : MnO4- + 5e-  Mn2+ 3. Complete the balance (for atoms and charges) by inserting H2O and H+ or OH- as appropriate i.e. 4 O  4H2O  8H+ Reduction : MnO4- + 8H+ + 5e-  Mn2+ + 4H2O
8. 8. e.g.2 Cl2 ClO4-  Cl(0)  (+7) (NOT balanced ; occurs in alkali)  Cl2 is oxidised 1. Write formulas of “redox” particles and balance “changed” atoms Reduction : Cl2  2 ClO4- 2. Insert e- on left for reduction, right for oxidation Number of electrons CHANGE in oxidation = in half-equation number of “redox” atoms Reduction : Cl2  2 ClO4- + 14e- 3. Complete the balance (for atoms and charges) by inserting H2O and H+ or OH- as appropriate i.e. 8 O  8H2O  16OHReduction : Cl2 + 16OH-  2 ClO4- + 14e- + 8H2O
9. 9. e.g.3 Cu + HNO3  Cu2+ + NO2 Cu(0)  (+2) N(+5)  (+4) (NOT balanced ; occurs in acid)  Cu oxidised and HNO3 reduced 1. Write formulas of “redox” particles and balance “changed” atoms Reduction : HNO3  NO2 Oxidation : Cu  Cu2+ 2. Insert e- on left for reduction, right for oxidation Number of electrons CHANGE in oxidation = in half-equation number of “redox” atoms Oxidation : Cu  Cu2+ + 2e- Reduction : HNO3 + e-  NO2 3. Complete the balance (for atoms and charges) by inserting H2O and H+ or OH- as appropriate Oxidation : Reduction : Cu  Cu2+ + 2eHNO3 + H+ + e-  NO2 + H2O
10. 10. Write a half-equation for each of the following changes. 1. Cl2 to Cl- 6. Br- to Br2 2. Pb2+ to Pb 7. Al to Al3+ 3. H2SO4 to H2S 8. At- to At2 4. HNO3 to NO 9. Fe to Fe2+ 5. H2SO4 to SO2 10. Br- to Br2
11. 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 Oxidation : Cu  Cu2+ + 2e- Reduction : HNO3 + H+ + e-  NO2 + H2O X1 X2 Add : Cu + 2HNO3 + 2H+ + 2e-  Cu2+ + 2e- + 2NO2 + 2H2O
12. 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 Oxidation : Fe2+  Fe3+ + e- X5 Reduction : MnO4- + 8H+ + 5e-  Mn2+ + 4H2O X1 Add : 5Fe2+ + MnO4- + 8H+ + 5e-  5Fe3+ + 5e- + Mn2+ + 4H2O
13. 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 Oxidation : Cl2   +   2OH-    ClO- +  H2O  +  e- X 2 Reduction : Cl2  +  2e-    2Cl-  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. 14. Use the half-equations written earlier and combine  them to form the overall equation for.   1. Cl2 to Cl-  with      Br- to Br2 2. Pb2+ to Pb  with      Al to Al3+  3. H2SO4 to H2S  with      I- to I2 4. HNO3 to NO  with      Fe to Fe2+ 5. H2SO4 to SO2  with      Br- to Br2
15. 15. The End
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