1. 1 HRW Ch 19 Oxidation-Reduction Reactions

2. 2 Oxidation States - Memorize! l Elements & Ions  The oxidation state of elements in their standard states is zero. Na (s) = 0; Cl 2(g) = 0, but O 2(s) ≠ 0!  Oxidation states for monoatomic ions are the same as their charge. Na 1+ = +1 Cl 1- = -1 N 3- = -3

3. 3 Oxidation states - memorize! Oxygen & Hydrogen  Oxygen is assigned an oxidation state of -2 in its covalent compounds (H 2 O) except as peroxide (H 2 O 2 ) where = -1 or with fluorine (OF 2 ) where = +2  Hydrogen in compounds with: nonmetals is assigned +1 (H 2 O) with metals it is -1 (CaH 2 )

4. 4 Oxidation states - memorize!  fluorine is always -1 in compounds (HF).  The sum of the oxidation states must be: zero in compounds (Na 2 SO 4 ) or equal the charge of the ion (SO 4 2- )

5. 5 Half-Reactions l Write the half reactions for the following. Na + Cl 2  Na + + Cl - Steps... Na  Na 1+ + 1e- 2e- + Cl 2  2Cl 1-

6. 6 Half-Reactions (more steps) l Write the half reactions for: SO 3 -2 + H + + MnO 4 -  SO 4 -2 + H 2 O + Mn +2 l 1st, identify the elements that are changing their oxidation numbers, they are... l S from +4 to +6 so it is losing 2 e- (it is being oxidized so it is the reducing agent) l Mn from +7 to +2 so it is gaining 5 e- (being reduced so it is the oxidizing agent). SO 3 -2 + H + + MnO 4 -  SO 4 -2 + H 2 O + Mn +2 +4 +7 +6 +2 l So, the half reactions are the stuff with sulfur and manganese.

7. 7 Half-Reactions (more steps) SO 3 -2 + H + + MnO 4 -  SO 4 -2 + H 2 O + Mn +2 +4 +7 +6 +2 S from +4 to +6 (losing) and... Mn from +7 to +2 (gaining) l The half-reactions are... SO 3 -2  SO 4 -2 + 2 e- +4 +6 5 e- + MnO 4 -  Mn +2 +7 +2

8. 8 Balancing Redox Equations l In aqueous solutions the key is the number of electrons produced must be the same as those required. l For reactions in acidic solution: 8-step procedure.  Write separate half-reactions (assign oxidation numbers, temporarily delete substances that don’t change)  For each half reaction balance all reactants except H and O  Balance O using H 2 O

9. 9 Acidic Solution  Balance H using H +  Balance charge using e -  Multiply equations to make electrons equal  Add equations and cancel identical species (cancel coefficients as needed)  Check that charges and elements are balanced.

10. 10 Example l A breathalyzer uses potassium dichromate to test for ethanol because the orange potassium dichromate changes to the green chromium 3 + ion in the presence of alcohol. l Write and balance the Breathalyzer equation given the following: l The reactants are K 2 Cr 2 O 7, HCl and C 2 H 5 OH. l The products are CrCl 3, CO 2, KCl & H 2 O.

11. 11 Example l Write and balance the Breathalyzer equation given the following: Reactants: K 2 Cr 2 O 7, HCl and C 2 H 5 OH. Products: CrCl 3, CO 2, KCl & H 2 O. l Write the formula equation: K 2 Cr 2 O 7 + HCl + C 2 H 5 OH  CrCl 3 + CO 2 + KCl + H 2 O l Write the ionic equation (solubility poem): 2K 1+ Cr 2 O 7 2- + H 1+ + Cl 1- + C 2 H 5 OH  Cr 3+ 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O l Assign oxidation numbers 2K 1+ Cr 2 O 7 2- + H 1+ + Cl 1- + C 2 H 5 OH  Cr 3+ 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O +1 +6 -2 +1 -1 - 2+1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2

12. 12 Example Write the formula equation: K 2 Cr 2 O 7 + HCl + C 2 H 5 OH  CrCl 3 + CO 2 + KCl + H 2 O Write the ionic equation (solubility poem): 2K 1+ Cr 2 O 7 2- + H 1+ + Cl 1- + C 2 H 5 OH  Cr 3+ 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O Assign oxidation numbers 2K 1+ Cr 2 O 7 2- + H 1+ + Cl 1- + C 2 H 5 OH  Cr 3+ + 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O +1 +6 -2 +1 -1 -2 +1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2 Write the formula equation: K 2 Cr 2 O 7 + HCl + C 2 H 5 OH  CrCl 3 + CO 2 + KCl + H 2 O Write the ionic equation (solubility poem): 2K 1+ Cr 2 O 7 2- + H 1+ + Cl 1- + C 2 H 5 OH  Cr 3+ 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O Assign oxidation numbers 2K 1+ Cr 2 O 7 2- + H 1+ + Cl 1- + C 2 H 5 OH  Cr 3+ + 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O +1 +6 -2 +1 -1 -2 +1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2 l Delete substances where no element changes its oxidation number: 2K 1+ Cr 2 O 7 2- + H 1+ Cl 1- + C 2 H 5 OH  Cr 3+ + 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O +1 +6 -2 +1 -1 -2 +1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2 2K 1+ Cr 2 O 7 2- + H 1+ Cl 1- + C 2 H 5 OH  Cr 3+ + 3Cl 1- + CO 2 + K 1+ + Cl 1- + H 2 O +1 +6 -2 +1 -1 -2 +1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2 Only those where oxidation number changes : Cr 2 O 7 2- + C 2 H 5 OH  Cr 3+ + CO 2 +6 -2 +3 +4 Only those where oxidation number changes : Cr 2 O 7 2- + C 2 H 5 OH  Cr 3+ + CO 2 +6 -2 +3 +4

13. 13 Balancing Redox Equations K 2 Cr 2 O 7 + HCl + C 2 H 5 OH  CrCl 3 + CO 2 + KCl + H 2 O K 2 Cr 2 O 7 + HCl + C 2 H 5 OH  CrCl 3 + CO 2 + KCl + H 2 O Only those where oxidation number changes : Cr 2 O 7 2- + C 2 H 5 OH  Cr 3+ - + CO 2 +6 -2 +3 +4 Only those where oxidation number changes : Cr 2 O 7 2- + C 2 H 5 OH  Cr 3+ - + CO 2 +6 -2 +3 +4 l For reactions in acidic solution: 8-step procedure.  Write separate half-reactions Cr 2 O 7 2-  Cr 3+ C 2 H 5 OH  CO 2 Cr 2 O 7 2-  Cr 3+ C 2 H 5 OH  CO 2

14. 14 Balancing Redox Equations Cr 2 O 7 2-  Cr 3+ C 2 H 5 OH  CO 2  For each half-reaction balance all reactants except H and O Cr 2 O 7 2-  Cr 3+ C 2 H 5 OH  CO 2

15. 15 Balancing Redox Equations Cr 2 O 7 2-  Cr 3+ - C 2 H 5 OH  CO 2  Balance O using H 2 O Cr 2 O 7 2-  Cr 3+ + 7H 2 O 3H 2 O + C 2 H 5 OH  CO 2  Balance H using H 1+ 14H 1+ + Cr 2 O 7 2-  Cr 3+ + 7H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ 14H 1+ + Cr 2 O 7 2-  Cr 3+ + 7H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+

16. 16 Example 14H 1+ + Cr 2 O 7 2-  Cr 3+ + 7H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+  Balance the charge using e - 6e - + 14H 1+ + Cr 2 O 7 2-  Cr 3+ + 7H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e - 6e - + 14H 1+ + Cr 2 O 7 2-  Cr 3+ + 7H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e -  Multiply equations to equalize the charges 2(6e - + 14H 1+ + Cr 2 O 7 2-  Cr 3+ + 7H 2 O) 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e - 2(6e - + 14H 1+ + Cr 2 O 7 2-  Cr 3+ + 7H 2 O) 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e - Get: 12e - + 28H 1+ + 2Cr 2 O 7 2-  Cr 3+ + 14H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e - Get: 12e - + 28H 1+ + 2Cr 2 O 7 2-  Cr 3+ + 14H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e -

17. 17 Example 12e - + 28H 1+ + 2Cr 2 O 7 2-  Cr 3+ + 14H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e -  Add equations and cancel identical species (cancel coefficients if needed) 12e - + 16 28H 1+ + 2Cr 2 O 7 2-  Cr 3+ + 11 14H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e - 12e - + 16 28H 1+ + 2Cr 2 O 7 2-  Cr 3+ + 11 14H 2 O 3H 2 O + C 2 H 5 OH  CO 2 + 12H 1+ + 12e - 16H 1+ + 2Cr 2 O 7 2- + C 2 H 5 OH  CO 2 +  Cr 3+ + 11H 2 O 16H 1+ + 2Cr 2 O 7 2- + C 2 H 5 OH  CO 2 +  Cr 3+ + 11H 2 O

18. 18 Acidic Solution 16H 1+ + 2Cr 2 O 7 2- + C 2 H 5 OH  CO 2 +  Cr 3+ + 11H 2 O  Combine the net ions to form the original compounds and check that everything is balanced... The original compounds were: K 2 Cr 2 O 7 + HCl + C 2 H 5 OH  CrCl 3 + CO 2 + KCl + H 2 O So the balanced equation is...(write it out) 2K 2 Cr 2 O 7 + 16HCl + C 2 H 5 OH  CrCl 3 + 2CO 2 + 4KCl + 11H 2 O

19. 19 Basic Solution l Do everything you would with acid, but we have to add a step because... l Because there is no H 1+ in basic solution. So, add OH 1- sufficient to convert the H 1+ to water (OH 1- + H 1+  H 2 O) l Be sure to add the OH 1- to both sides of the reaction (conservation of mass law) Cr(OH) 3 + OCl - + OH -  CrO 4 -2 + Cl - + H 2 O we’ll do this on the board to get... 2Cr(OH) 3 + 3OCl - + 4OH -  CrO 4 -2 + 3Cl - + 5H 2 O

20. 20 Agents l Oxidizing agent - has the potential to cause another substance to get oxidized. l Other substance loses electrons, but... l In causing that other substance to get oxidized the oxidizing agent gets reduced. l The oxidizing agent gains electrons (gets a more negative oxidation state). l MnO 4 1- will oxidize Fe 2+ to Fe 3+ so here it is an oxidizing agent. But the Mn in MnO 4 1- gets reduced from +7 to +2 (MnO 4 1- + 5e -  Mn 2+ ) +7 +2

21. 21 Agents l Reducing agent - has the potential to cause another substance to get reduced. l Other substance gains electrons, but... l In causing that other substance to get reduced the reducing agent gets oxidized. l The reducing agent loses electrons (gets a more positive oxidation state). l Na will reduce Zn 2+ to Zn so here it is a reducing agent. But the Na gets oxidized from 0 to +1 (Na  Na 1+ + 1e - )