1. Oxidation and Reduction Or, “Do you know where your electrons are?”

2. Definitions  Oxidation is the process of losing electrons (oxidation state becomes more positive)  Na  Na + + 1e -  Reduction is the process of gaining electrons (oxidation state becomes more negative)  Cl + 1e -  Cl -

3. Definitions L osing E lectrons O xidation goes G aining E lectrons R eduction

4. Definitions O xidation I s L osing R eduction I s G aining

5. Oxidation state  Charge on an ion  Na +, Ca +2, O -2  The number of electrons unequally shared in a covalent bond.  H 2 O : H is +1, O is -2

6. Oxidation state assignment rules  Any element has oxidation number of zero  Oxygen has an oxidation number of -2, except in peroxides where it is -1  Hydrogen is +1 except in hydrides, where it is -1 – in HCl the H is +1, but in NaH it is -1  Nitrogen is -3 except with oxygen

7. Oxidation state assignment rules  Halogens are -1 except with oxygen or each other  All other oxidation numbers are assigned so that the sum of all the oxidation numbers equals the charge on the particle.  In examples not covered here the atom with greater electronegativity gets the negative charge.

8. Oxidation state assignment rules  NH 3  H= +1, N= -3  NI 3  N= -3, I = +1

9. Oxidation state assignment rules  NF 3  N= +3, F= -1 H3O+H3O+  H= +1, O= -2

10. Oxidation state assignment rules  NO 3 -  O= -2, N= +5  Cr 2 O 7 -2  O= -2, Cr= +6

11. Redox reaction  Any reaction that results in a change of oxidation state for any reactant. N 2 + 3H 2  2NH 3 0 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O 0 0-3,+1 +5+2

12. Redox Reaction  2Fe + 3CuSO 4  3Cu + Fe 2 (SO 4 ) 3 0 +2 0 +3  Oxidizing agent – the reactant that is reduced C + O 2  CO 2  Oxygen is reduced (0 to -2), so it is the oxidizing agent

13. Oxidizing and reducing agents RR educing agent – the reactant that is oxidized 33 H 2 + 2Cr +3  6H + + 2Cr HH ydrogen is oxidized (0 to +1), so it is the reducing agent EE xample: Identify the oxidizing and reducing agents in the following reaction: 22 HCl + Zn  ZnCl 2 + H 2 ZZ n – reducing agent HH + – oxidizing agent

14. Redox and electronegativity  C + O 2  CO 2  Carbon is oxidized because it has lost some electron density to oxygen, which has greater electronegativity.  Oxygen is reduced because it gained some electron density from carbon

15. Balancing redox equations  Charge Balance  Redox is a transfer of electrons, so the number of electrons lost by the reducing agent = number of electrons gained by oxidizing agent  Total charge of reactants must = total charge of products Cr +6 + Fe +2  Cr +3 + Fe +3  Even though the atoms are balanced, the charge is not.

16. Balancing redox equations  Oxidation number method:  Identify all changes in oxidation number  Cr +6 + Fe +2  Cr +3 + Fe +3 -3 +1

17. Balancing redox equations  Use coefficients to make the changes cancel Cr +6 + Fe +2  Cr +3 + Fe +3 -3+1x3 = +3 33

18. Balancing redox equations  Check charge balance Cr +6 + 3Fe +2  Cr +3 + 3Fe +3 +12  +12 +5 +3 +2 +5 HNO 3 + H 3 AsO 3  NO + H 3 AsO 4 + H 2 O -3 +2 Use least common multiple – 6 2HNO 3 + 3H 3 AsO 3  2NO + 3H 3 AsO 4 + H 2 O

19. Balancing Redox Equations  Half reactions method  Every redox reaction consists of two half reactions Fe + Cu +2  Fe +3 + Cu oxidation Fe  Fe +3 + 3e - reduction Cu +2 + 2e -  Cu Oxidation and reduction reactions always happen in pairs

20. Balancing Redox Equations  Sum of appropriate numbers of half reactions yields a balanced equation – use coefficients to make # electrons lost = # electrons gained 2(Fe  Fe +3 + 3e - ) + 2(Cu +2 + 2e -  Cu) = 2Fe + 3Cu +2  2Fe +3 + 3Cu

21. Balancing Redox Equations AA toms and electrons have to balance II f the electrons balance, the charge will also balance (but be sure to check it!) CC u + HNO 3  Cu(NO 3 ) 2 + NO 2 + H 2 O OO xidation: Cu  Cu +2 + 2e - RR eduction: NO 3 - + 1e -  NO 2

22.  Reduction half reaction must be balanced – in acid solution use 2H + and H 2 O for each missing oxygen  2H + + NO 3 - + 1e -  NO 2 + H 2 O  Number of electrons in oxidation and reduction must be equal  Add half reactions to get balanced equation

23. Balancing Redox Equations 2(2H + + NO 3 - + 1e -  NO 2 + H 2 O) Cu  Cu +2 + 2e - 4H + +2NO 3 - +2e - +Cu  Cu +2 +2e - +2NO 2 +2H 2 O  Electrons cancel; addition of nitrates to each side (spectators) gives overall equation 4HNO 3 +Cu  Cu(NO 3 ) 2 +2NO 2 +2H 2 O

24. Balancing Redox Example #2 Zn + VO 3 -  Zn +2 + VO +2 (in acid solution)  Half reactions:  Oxidation: Zn  Zn +2 + 2e -  VO 3 - V is +5, VO +2 V is +4  Reduction: VO 3 - + 1e -  VO +2

25. Balancing Redox Example #2  balance with H + and H 2 O  2(4H + + VO 3 - + 1e -  VO +2 + 2H 2 O)  Balanced equation is sum of half reactions  8H + +2VO 3 - +Zn  2VO +2 +4H 2 O+Zn +2

26. Balancing in Base Solution  Use 2OH - and H 2 O for each missing oxygen  Cr(OH) 3 + ClO 3 -  CrO 4 2- + Cl -  Oxidation  Cr(OH) 3  CrO 4 -2 + 3e - + 3OH -  Hydroxides are added to balance hydrogens.  Balance oxygen (four missing on left) with 2OH - /H 2 O.

27. Balancing in Base Solution 88 OH - + Cr(OH) 3  CrO 4 -2 + 3e - + 3OH - + 4H 2 O CC ancel hydroxides on both sides. 55 OH - + Cr(OH) 3  CrO 4 -2 + 3e - + 4H 2 O RR eduction: CC lO 3 - + 6e -  Cl - BB alance oxygen (three missing on right) with 2OH - /H 2 O.

28. Balancing Redox in Base Solution 3H 2 O + ClO 3 - + 6e -  Cl - + 6OH -  Add equations and eliminate spectators 2[5OH - + Cr(OH) 3  CrO 4 -2 + 3e - + 4H 2 O] 3H 2 O + ClO 3 - + 6e -  Cl - + 6OH - 10OH - + 2Cr(OH) 3 + 3H 2 O + ClO 3 -  2CrO 4 -2 + 8H 2 O + Cl - + 6OH - 4 4OH - + 2Cr(OH) 3 + ClO 3 -  2CrO 4 -2 + 5H 2 O + Cl - 5