1. Daniel L. Reger Scott R. Goode David W. Ball www.cengage.com/chemistry/reger Lecture 03B (Chapter 18, sections 18.1, 18.2) Balancing Redox Reactions

2. Oxidation is the loss of electrons by a chemical process. When sodium forms a compound, Na + is formed. Sodium is oxidized. Reduction is the gain of electrons by a chemical process. When Cl - ions are formed from elemental chlorine, chlorine is reduced. Oxidation and Reduction

3. An oxidation-reduction reaction, or redox reaction, is one in which electrons are transferred from one species to another. In every redox reaction, at least one species is oxidized and at least one species is reduced. 2Na(s) + Cl 2 (g) → 2NaCl(s) is a redox reaction because Na is oxidized and Cl is reduced. Oxidation-Reduction (“Redox”)

4. An oxidizing agent is the reactant that accepts electrons, causing an oxidation to occur. The oxidizing agent is reduced. A reducing agent is the reactant that supplies electrons, causing a reduction to occur. The reducing agent is oxidized. In the reaction of sodium with chlorine, Na is the reducing agent and Cl 2 is the oxidizing agent. Oxidizing and Reducing Agents

5. In a half-reaction, either the oxidation or reduction part of a redox reaction is given, showing the electrons explicitly. Half-reactions emphasize the transfer of electrons in a redox reaction. For 2Na(s) + Cl 2 (g) → 2NaCl(s) : Na → Na + + 1e - oxidation half-reaction Cl 2 + 2e - → 2Cl - reduction half-reaction Half-reactions

6. The oxidation state is the charge on the monatomic ion, or the charge on an atom when the shared electrons are assigned to the more electronegative atom. Electron pairs shared by atoms of the same element are divided equally. In CaCl 2, an ionic compound: calcium has an oxidation state of +2. chlorine has an oxidation state of -1. Oxidation States

7. Rule 1: The O.N. of an atom in its (uncombined) elemental state is zero. Exs. Al(0), O2(0), H2(0), Br2(0), Na(0) Rule 2: The O.N. of a monatomic ion is equal to the charge on the ion. Exs. Na + (+1), Mg 2+ (+2), Al 3+ (+3), O 2- (-2), Br - (-1) Rule 3: Atoms in polyatomic ions or molecular compounds usually have O.N.s identical to the charges they would have as ions. Exceptions: -Hydrogen (except as H2) is usually +1, but sometimes -1 (ex. H-Ca-H) -Oxygen (except as O2) is usually -2, but can be -1 when –O-O- bonds exist (peroxides) -Halogens usually -1, unless combine with more EN element (dichlorine monoxide, Cl- O-Cl, Cl is +1; F is always -1) Rule 4: The algebraic sum of all O.N.s of all atoms in a neutral compound or polyatomic ion is equal to the net charge. - For neutral compounds, the sum of the O.N.s is 0 - For a charged polyatomic ion, the sum of the O.N.s is equal to the charge. Rules for Determining Oxidation Numbers

8. Assign oxidation numbers for each atom in K 2 CrO 4. K = +1 by rule 2. O = -2 by rule 5. Cr = +6 by rule 6. 2(+1) + 6 + 4(-2) = 0, the overall charge on the substance. Assigning Oxidation Numbers

9. K2CO32(O.N of K) + (O.N. of C) + 3(O.N. of O) = 0 2(+1) + ? + 3(-2) = -4 2(+1) + (+4) + 3(-2) = 0 CH2O(O.N of C) + 2(O.N. of H) + (O.N. of O) = 0 2(+1) + (-2) = 0 (0) + 2(+1) + (-2) = 0 NO3 - (O.N of N) + 3(O.N. of O) = -1 + 3(-2) = -6 (+5) + 3(-2) = -1

10. Assign oxidation numbers to each atom in the following substances. (a) PF 3 (b) CO(c) NH 4 Cl Test Your Skill

11. Assign oxidation numbers to each atom in the following substances. (a) PF 3 (b) CO(c) NH 4 Cl Answers:(a) P = +3, F = -1 (b) C = +2, O = -2 (c) N = -3, H = +1, Cl = -1 Test Your Skill

12. Redox Rxns Determined by Change in O.N. (O.N. of S) + 2(O.N. of O) = 0 S(s) + O 2 (g)  SO 2 (g) (O.N of uncombined element = 0) (+4) + 2(-2) = 0 Sulfur is oxidized, Oxygen is reduced 2(O.N. of Al) + 3(O.N. of O) = 0 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s) (O.N of uncombined element = 0) + 3(-2) = -6  Al is oxidized (O.N. increased by 3), Al is reducing agent  Oxygen is reduced (O.N. decreases by 2), O is oxidizing agent 2(+3) + 3(-2) = 0

13. Redox e - Transfer: Examples SO 2 (g) + H 2 O( l )  H 2 SO 3 (aq) S+4 O-2 H+1 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s)

14. Balancing Net Ionic Equations Using Redox Half-reactions Cr 2 O 7 2- (aq) + Cl - (aq)  Cr 3+ (aq) + Cl 2 (aq) With more complex net ionic equations, it is not always easy to see how the equation should be balanced. We have another set of steps to address this!!!!

15. Balancing Net Ionic Equations Using Redox Half-reactions (under acidic conditions) Step 1: Write the unbalanced net ionic equation Step 2: Determine which atoms are oxidized and reduced, and write the two unbalanced half-reactions Step 3: Balance both half-reactions for all atoms except H, O Step 4: Balance each half-reactions for O - Add H2O to side with less O - Add H+ to side with less H Step 5: Balance each half-reactions for charge - Add electrons to side with greater positive charge - Multiply half-reaction as needed so the electron count is the same on both sides Step 6: Add the two half-reactions - Cancel electrons and other species (spectator ions) that appear on both sides Step 7: Confirm that equation balanced for atoms and charge

16. Balancing Net Ionic Equations Using Redox Half-reactions (under acidic conditions) Cr 2 O 7 2- (aq) + Cl - (aq)  Cr 3+ (aq) + Cl 2 (aq) Step 1: Write the unbalanced net ionic equation Step 2: Determine which atoms are oxidized and reduced, and write the two unbalanced half-reactions Cl-(aq)  Cl 2 (aq) Cr 2 O 7 2- (aq)  Cr 3+ (aq)  Cl is oxidized (loses e-) from -1 to 0  Reducing agent  Cr is reduced (gains e-) from +6 to +3  Oxidizing agent Oxidation half-rxn Reduction half-rxn Step 3: Balance both half-reactions for all atoms except H, O 2 2

17. Balancing Net Ionic Equations Using Redox Half-reactions (under acidic conditions) Step 4: Balance each half-reactions for O, H - Add H 2 O to side with less O - Add H + to side with less H 2Cl - (aq)  Cl 2 (aq) Cr 2 O 7 2- (aq) + 14H + (aq)  2Cr 3+ (aq) + 7H 2 0 ( l ) Oxidation half-rxn Reduction half-rxn Cr 2 O 7 2- (aq)  2Cr 3+ (aq) +7H 2 0 ( l )

18. Balancing Net Ionic Equations Using Redox Half-reactions (under acidic conditions) Step 5: Balance each half-reactions for charge - Add electrons to side with greater positive charge - Multiply half-reaction as needed so the electron count is the same on both sides 2Cl - (aq)  Cl 2 (aq) Cr 2 O 7 2- (aq) + 14H + (aq)  2Cr 3+ (aq) + 7H 2 0 ( l ) Oxidation half-rxn Reduction half-rxn + 2 e- Cr 2 O 7 2- (aq) + 14H + (aq) + 6 e-  2Cr 3+ (aq) + 7H 2 0 ( l )

19. Balancing Net Ionic Equations Using Redox Half-reactions (under acidic conditions) Step 5: Balance each half-reactions for charge - Add electrons to side with greater positive charge - Multiply half-reaction as needed so the electron count is the same on both sides 3 X [2Cl - (aq)  Cl 2 (aq) + 2e-] = Oxidation half-rxn Reduction half-rxn Cr 2 O 7 2- (aq) + 14H + (aq) + 6e-  2Cr 3+ (aq) + 7H 2 0 ( l ) 6Cl - (aq)  3Cl 2 (aq) + 6e-

20. Balancing Net Ionic Equations Using Redox Half-reactions (under acidic conditions) Step 6: Add the two half-reactions - Cancel electrons and other species (spectator ions) that appear on both sides (Ox) (Red) Cr 2 O 7 2- (aq) + 14H + (aq) + 6e-  2Cr 3+ (aq) + 7H 2 0 ( l ) 6Cl - (aq)  3Cl 2 (aq) + 6e- Cr 2 O 7 2- (aq) + 14H + (aq) + 6Cl - (aq)  3Cl 2 (aq) + 2Cr 3+ (aq) + 7H 2 0 ( l ) Step 7: Confirm that equation balanced for atoms and charge

21. In basic solutions, follow all of the same steps. When finished, add an equivalent amount of OH - to each side of the reaction to eliminate any H + by combining them to make H 2 O. Balancing Redox Reactions under basic conditions

22. Test Your Skill Balance the following equation in acid solution: Cr 2 O 7 2- + C 2 H 5 OH → Cr 3+ + CO 2

23. Test Your Skill Balance the following equation in acidic solution: Cr 2 O 7 2- + C 2 H 5 OH → Cr 3+ + CO 2 Answer: 2Cr 2 O 7 2- + C 2 H 5 OH + 16H + → 4Cr 3+ + 2CO 2 + 11H 2 O

24. Test Your Skill Balance the following equation in basic solution: Zn + ClO - → Zn(OH) 4 2- + Cl -

25. Test Your Skill Balance the following equation in basic solution: Zn + ClO - → Zn(OH) 4 2- + Cl - Answer: Zn + ClO - + H 2 O + 2OH - → Zn(OH) 4 2- + Cl -