1. Oxidation Numbers & Balancing equation

2. Oxidation Number Oxidation number is defined as  The charge an atom has  Or appears to have  When electrons are distributed  according to certain rules

3. Oxidation Number Rules The oxidation number of an Element is 0 group One elements is +1 group Two elements is +2 in compounds

4. The oxidation number of an ion is equal to the charge on the ion halogens is -1 (in binary compounds) (except ……????)

5. The oxidation number of H in a compound is +1 – except in metal hydrides when it is -1

6. The oxidation number of O in a compound is -2 – except (x2) in peroxides when it is -1 (H 2 O 2 ) in OF 2 when it is +2 (why?)

7. Oxidation numbers add up to zero in a compound add up to the charge of a complex ion

8. What is the oxidation number of each element in :- (write down before you go on) H 2 0 MnO 4 ¯ I 2 KBrO 3 Na 2 S 2 O 3 H 2 O 2 NaOCl

9. The oxidation number of each element is :- H 2 0 MnO 4 ¯ I 2 KBrO 3 Na 2 S 2 O 3 H 2 O 2 NaOCl +1-2 +7-2 0 +1+5-2 +1+2-2 +1-2+1

10. Learning Check Can I give the oxidation number RULE for a)Oxygen b)Hydrogen c)free element d)Neutral atom (sum) e)Ion (simple and complex) f)Group 1 element g)Group 2 element h)HALOGEN STILL NOT The End - click to go on

11. Balancing Equations with oxidation numbers STEPS 1.Assign oxidation numbers 2.Identify what is oxidised and reduced 3.Write half equation SIDE by SIDE for each (showing number of electrons on the move for one atom of each) 4.Rewrite for the number of atoms given e.g. Cr 2 5.Balance the electrons 6.REWRITE the original equation using these “prefixes” 7.Balance remainder by inspection 8.CHECK – do the charges on each side cancel out??

12. Example Assign & Identify Cr 2 O 7 2- + Fe 2+ + H +  Cr 3+ + Fe 3+ + H 2 0 +6 -2 +2 +1 +3 +3 +1 -2 Reduced x3 Oxidised x1

13. 2 Identify oxidised or reduced & number of electrons lost or gained per atom & as given ReducedOxidised Cr + 3e -  CrFe – e -  Fe ATOM Cr 2 + 6e -  2CrFe – e -  Fe GIVEN Balance Electrons Cr 2 + 6e -  2CrFe – e -  Fe Rewrite and sub back Cr 2 O 7 2 - + 6 Fe 2+ + H+  2 Cr 3+ + 6 Fe 3+ + H 2 0 Balance remainder by inspection Cr 2 O 7 2 - + 6 Fe 2+ + H+  2 Cr 3+ + 6 Fe 3+ + H 2 0 6 6 6 714

14. Check Charges on each side should balance Cr 2 O 7 2 - + 6 Fe 2+ + 14 H +  2 Cr 3+ + 6 Fe 3+ + 7 H 2 0 LEFT RIGHT 2-6+ 12+18+ 14+___24+