1. 8 - 1 Assigning Oxidation Numbers We will hold off for the time being the formal definition of an oxidation number. The oxidation number (state) of a free or uncombined element is 0.  Fe, Na, O 2 The oxidation number of a monatomic ion is equal to its charge.  K +, Al 3+, Ca 2+, Cu +

2. 8 - 2 The oxidation number of hydrogen is +1 when it is covalently bonded.  HCl, H 2 O 2, H 2 O, H 2 SO 4 The oxidation number of hydrogen is -1 when it is ionically bonded with a metal.  NaH, CaH 2, KH  These are examples of hydrides and are named sodium hydride, calcium hydride, and potassium hydride.

3. 8 - 3 The oxidation number of oxygen is -2 in most of its covalent compounds.  One exception is when oxygen is bonded to fluorine as in OF 2. In OF 2, the oxidation number of oxygen is +2. Fluorine being the most electronegative element is always assigned an oxidation number of -1.

4. 8 - 4  Another exception is when the oxidation number for oxygen is -1 when it is a peroxide. O 2 2- can be found covalently bonded as it is in H 2 O 2 (hydrogen peroxide). O 2 2- can be found ionically bonded as it is in K 2 O 2 (potassium peroxide). Something to remember is that you do not reduce the subscripts in a metallic peroxide (K 2 O 2 ).

5. 8 - 5  A second exception is when oxygen has an oxidation number of -½. This occurs when oxygen is the anion, O 2 -, and is called a superoxide. LiO 2 is such a compound and is called lithium superoxide. For binary compounds, the element with the greater electronegativity is assigned a negative oxidation number equal to its charge.

6. 8 - 6 The algebraic sum of the oxidation numbers in a neutral compound must equal zero.  AlCl 3 H 2 O 2 NH 3 The algebraic sum of the oxidation numbers in a polyatomic ion must equal the charge on the ion.  H 3 O + NH 4 + PO 4 3- +3 +3 -3 +1 +2 -2 -3 +1 -3 +3 +1 -2 +3 -2 -3 +1 -3 +4 +5 -2 +5 -8

7. 8 - 7 Oxidation State Method The Oxidation States Method is a means of balancing the more simpler redox reactions. The steps for a variation on this method are as follows:  Assign oxidation numbers to all of the atoms in the equation.  Identify the substance reduced (oxidizing agent) and determine the number of electrons gained.

8. 8 - 8  Identify the substance oxidized (reducing agent) and determine the number of electrons given off.  Use coefficients to balance the atoms and electrons in the species reduced.  Use coefficients to balance the atoms and electrons in the species oxidized.  Use coefficients to balance the remaining atoms by inspection.

9. 8 - 9 Oxidizing And Reducing Agents The terminology for oxidation-reduction (redox) can be somewhat confusing. The oxidizing agent is the species (atom or ion) being reduced. In the following example, elemental chlorine is said to be reduced.  Cl 2 → Cl -

10. 8 - 10  For a species to be reduced, its oxidation is reduced to a smaller number.  Another way of saying the same thing is that the oxidation number of chlorine becomes less positive or more negative.  Oxidation-reduction always occur together.  Because chlorine is reduced, it is called an oxidizing agent.

11. 8 - 11  When chlorine is reduced, it is bringing about the oxidizing of another species.  Hence, chlorine is called an oxidizing agent.  The balanced equation for chlorine is given by: Cl 2 + 2e - → 2Cl - 0 0-2

12. 8 - 12  The previous equation shows the conservation of mass and charge. In the following example, sodium is said to be oxidized.  Na → Na + + e -  For a species to be oxidized, its oxidation number is increased to a larger number.  Another way of saying the same thing is that the oxidation number of sodium becomes more positive or less negative.

13. 8 - 13  Because sodium is oxidized, it is called a reducing agent.  When sodium is oxidized, it is bringing about the reduction of another species.  Hence, sodium is called an reducing agent.  The balanced equation for sodium is given by: Na → Na + + e - 01

14. 8 - 14  The previous equation shows the conservation of mass and charge.

15. 8 - 15 Summary Of Double Agents In summary,  An oxidizing agent is reduced.  A reducing agent is oxidized. Atoms in Group I have a weak attraction for their valence and are easily oxidized. Atoms in Group VII have a strong attraction for their valence electrons and are easily reduced.

16. 8 - 16 Single Replacement Reactions A single replacement reaction is one in which a free element becomes an ion and an ion in solution becomes a neutral atom. For example, Zn(s) + H 2 SO 4 (aq) → ZnSO 4 (aq) + H 2 (g) Zn becomes Zn 2+ and H + becomes H 2.

17. 8 - 17 Balancing the net ionic equation could be done by inspection but we will use the ion-electron method. Zn + H 2 SO 4 → ZnSO 4 + H 2 Zn + 2H + + SO 4 2- → Zn 2+ + SO 4 2- + H 2 Zn → Zn 2+ + 2e - 2H + + 2e - → H 2 Zn(s) + 2H + (aq) → Zn 2+ (aq) + H 2 (g)

18. 8 - 18 Alkali metals (Group I) and alkaline earth metals (Group II) are very active metals.  These metals are easily oxidized and are good reducing agents.  Such active metals react with water (H 2 O(l)) or steam (H 2 O(g)) to replace the hydrogen from the water forming a metallic hydroxide and hydrogen gas.  Sometimes water will be written as HOH to show that the first hydrogen is the one that gets replaced.

19. 8 - 19 Use the ion-electron method to write the balanced net ionic equation for the reaction between sodium and water. Na + HOH → NaOH + H 2 Na + HOH → Na + + OH - + H 2 Na → Na + + e - H + + e - → H 2

20. 8 - 20 2Na → 2Na + + 2e - 2H + + 2e - → H 2 Very unreactive metals such as Ag, Pt, and Au, are not found in nature chemically combined. 2Na(s) + 2H + (aq) → 2Na + (aq) + H 2 (g)

21. 8 - 21 When Is There A Reaction? One example of a single replacement reaction is when one metal tries replacing another. For example, what are the products of the following reaction?  Cu + AgNO 3 → To determine the products, one can use the values from Standard Reduction Potential Table.

22. 8 - 22  From the title, one expects to find the half-reactions to be written as reductions. From the table, we find that: Ag + + e - → AgE° red = 0.80 V Cu 2+ + 2e - → CuE° red = 0.34 V These E° values are measured with respect to the SHE (Standard Hydrogen Electrode).

23. 8 - 23 Hydrogen is arbitrarily assigned an E° red = 0 V (where V is volts). This is covered in much more detail in Electrochemistry. For now, TP (Think Positive)! Because E° red (Ag) > E° red (Cu) or E° red (Ag) is more positive than E° red (Cu), the single replacement takes place.

24. 8 - 24 The more formal approach gives the same result. If the reaction was to take place, it would be: Cu + AgNO 3 → Ag + Cu(NO 3 ) 2 (unbalanced) The two half-reactions would be: Cu → Cu 2+ + 2e - E° ox = -0.34 V 2Ag + + 2e - → 2Ag E° red = 0.80 V Cu(s) + 2Ag + (aq) → Ag(s) + Cu 2+ (aq) E° = 0.46 V

25. 8 - 25 Important fall out from this problem:  Notice that the half-reaction for copper is reversed and the sign changes to negative. This is necessary because Cu is required to be oxidized in the reaction, not reduced. You must multiply the Ag + half-reaction by 2 so the electrons cancel out in the net ionic equation.

26. 8 - 26 A very important note is that you do not multiply the E° by 2 because voltage is an intensive property. For the single replacement to occur, E° must be positive!

27. 8 - 27 When Is There A No Reaction? Not all single replacements occur! For example, what are the products of the following reaction?  Ag + Cu(NO 3 ) 2 → From the table, we find that: Ag + + e - → AgE° red = 0.80 V Cu 2+ + 2e - → CuE° red = 0.34 V

28. 8 - 28 Because E° red (Ag) > E° red (Cu) or E° red (Ag) is more positive than E° red (Cu), the single replacement does not take place! This reaction is significantly different. Ag → Ag + + e - E ox = -0.80 V Cu → Cu 2+ + 2e - E ox = -0.34 V TP (Think Positive) still works! The oxidation potential of copper is more positive than the silver.

29. 8 - 29 But the problem is that copper needs to be reduced in this problem! The more formal approach gives the same result. If the reaction was to take place, it would be: Ag + Cu(NO 3 ) 2 → Cu + AgNO 3 The two half-reactions would be:

30. 8 - 30 2Ag → 2Ag + + 2e - Because E° < 0, the reaction does not occur! E° red = 0.34 V E° ox = -0.80 V E° = -0.46 V Cu 2+ + 2e - → Cu

31. 8 - 31 When Is There A Reaction? Another example of a single replacement reaction is when one halogen tries replacing another. For example, what are the products of the following reaction?  Cl 2 + NaBr → To determine the products, one can use the values from Standard Reduction Potential Table.

32. 8 - 32  From the table, we find that: Cl 2 + 2e - → 2Cl - E° red = 1.36 V Br 2 + 2e - → 2BrE° red = 1.07 V Because E° red (Cl 2 ) > E° red (Br 2 ) or E° red (Cl 2 ) is more positive than E° red (Br 2 ), the single replacement takes place.

33. 8 - 33 The more formal approach gives the same result. If the reaction was to take place, it would be: Cl 2 + NaBr → Br 2 + NaCl (unbalanced) The two half-reactions would be: Cl 2 + 2e - → 2Cl - E° red = 1.36 V 2Br - → Br 2 + 2e - E° red = -1.07 V Cl 2 (g) + 2Br - (aq) → Br 2 (l) + 2Cl - (aq) E° = 0.46 V

34. 8 - 34 When Is There A No Reaction? For example, what are the products of the following reaction?  I 2 + LiF → From the table, we find that: F 2 + 2e - → 2F - E° red = 2.87 V I 2 + 2e - → 2I - E° red = 0.53 V

35. 8 - 35 Because E° red (F 2 ) > E° red (I 2 ) or E° red (F 2 ) is more positive than E° red (I 2 ), the single replacement does not take place! This reaction is significantly different. F 2 + 2e - → 2F - E° red = 2.87 V I 2 + 2e - → 2I - E° red = 0.53 V TP (Think Positive) still works! The oxidation potential of fluorine is more positive than the iodine.

36. 8 - 36 But the problem is that fluorine needs to be oxidized in this problem! The more formal approach gives the same result. If the reaction was to take place, it would be: I 2 (s) + 2LiF(aq) → F 2 (g) + 2LiI(aq) The two half-reactions would be:

37. 8 - 37 I 2 + 2e - → 2I - Because E° < 0, the reaction does not occur! E° ox = -2.87 V E° red = 0.53 V E° = -2.34 V 2F - → F 2 + 2e -

38. 8 - 38 Summarizing SRP Rxns For Halogens The Standard Reduction Potentials for nonmetals measures the ability of a nonmetal to accept valence electrons. The more reactive nonmetals have a greater tendency to gain an electron or be reduced.  Fluorine with its highest electronegativity (4.0) is the most reactive halogen and is the strongest oxidizing agent.

39. 8 - 39  Iodine has the smallest tendency to gain electrons because it has the largest radius and smaller electronegativity.

40. 8 - 40 Balance Rxns Using Oxidation Numbers When the equations become somewhat more difficult to balance by inspection, use the following rules.  Assign oxidation numbers to all elements.  Identify the element that is reduced and write its half-reaction.  Identify the element that is oxidized and write its half-reaction.

41. 8 - 41  Balance and add the half-reactions so the electrons cancel.  Use the resulting coefficients as coefficients in the original equation.  Balance the remaining atoms by inspection.

42. 8 - 42 More Interesting Equations To Balance Balance the following reaction using the oxidation-number method. HNO 3 + H 2 S → NO + S + H 2 O 1-2 5 1-6 5 1-2 1 2 01 2

43. 8 - 43 S → S + 2e - N + 3e- → N 3S → 3S + 6e - 2N + 6e - → 2N -20 52 0 5 2 3S + 2N → -250 2

44. 8 - 44 The sum of the reactant oxidation numbers is +4. The sum of the product oxidation numbers is +4. Using the coefficients of the net equation in the original equation gives: 2HNO 3 + 3H 2 S → 2NO + 3S + H 2 O 2HNO 3 (aq) + 3H 2 S(aq) → 2NO(g) + 3S(s) + 4H 2 O(l)

45. 8 - 45 Another Interesting Problem Potassium dichromate and hydrochloric acid react to produce potassium chloride, chromium(III) chloride, water and chlorine. Write the balanced equation for this reaction. K 2 Cr 2 O 7 + HCl → KCl + CrCl 3 + H 2 O + Cl 2 16-21 1 3 1-20 212-141 1 3-31-20

46. 8 - 46 Cl is being oxidized.Cr is being reduced. Cl - → Cl 2 Cr 2 → Cr 3+ 2Cl - → Cl 2 Cr 2 → 2Cr 3+ 2Cl - → Cl 2 + 2e - Cr 2 + 6e - → 2Cr 3+ 6Cl - → 3Cl 2 + 6e - Cr 2 + 6e - → 2Cr 3+ 6Cl - + Cr 2 → 3Cl 2 + 2Cr 3+

47. 8 - 47 Using the coefficients from the net equation gives, K 2 Cr 2 O 7 + 6HCl → KCl + 2CrCl 3 + H 2 O + 3Cl 2 and balancing by inspection gives: K 2 Cr 2 O 7 (aq) + 14HCl(aq) → 2KCl(aq) + 2CrCl 3 (aq) + 7H 2 O(l) + 3Cl 2 (g)