1. Topic: Redox – Half reactions Assign Oxidation number to H, Cl and O for the following compounds 1.HClO 2.HClO 2 3.HClO 3 4.HClO 4

2. Now we know how to assign oxidation number…we can look at redox rxns Haber Process – N 2(g) + 3H 2(g)  2NH 3(g) Start by assigning oxidation numbers N 2(g) + 3H 2(g)  2NH 3(g) What was oxidized? Reduced? 00+1-3

3. N 2(g) + 3H 2(g)  2NH 3(g) O I L R I G 00+1 -3 Oxidation Is Losing electrons Reduction Is Gaining electrons Began Ended N0 -3 H0 +1 Gained 3 electrons = Reduced Lost 1 electron = Oxidized 0 43214321 43214321 -2 -3 -4 -2 -3 -4

4. Why use the word “ reduced ” when electrons are gained? Electrons are Negative! Look how oxidation number changes Ex: Cl gains an electron → Cl -1 oxidation # ↓ from 0 to -1; the # was reduced oxidation # ↓ from 0 to -1; the # was reduced

5. Half Reactions Even though oxidation & reduction reactions occur together we write separate equations for each process and include # of e - gained/lost – known as Half-Reactions

6. Half-reactions must demonstrate: – conservation of mass & conservation of charge – # atoms on left must = # atoms on right (always balance mass first!!!!) Keep 7 diatomics together, everything else write as single ion) – total charge on left must = total charge on right Add electrons to more positive side Half Reactions

7. Oxidation = electrons lost (becomes more positive) – So electrons are on the product side H 2  H +1 + 1e - N 2 + 3H 2  2NH 3 00 +1-3 OIL 0 +1 1 and -1 equals 0 = But something is wrong!

8. Remember… H 2  H +1 + 1e - Total Charge on left = Total Charge on right # atoms on left = # atoms on right H 2  2H +1 + 1e - Something is still wrong! Charge is off now H 2  2H +1 + 2e - 2 x +1 = +2 and -2 equals 0

9. Reduction = electrons gained (becomes more negative) – So electrons are on the reactant side Each N gained 3 electrons, – N 2 + 3e -  N -3 N 2 + 3H 2  2NH 3 00 +1-3 RIG Balance mass 6 But not you have -6 2

10. You Try: Mg + S  MgS What is oxidized? What is reduced? Assign Oxidation Numbers Figure out change in oxidation numbers0 0 +2 -2 Mg: 0 to +2 = Oxidation S: 0 to -2 = Reduction

11. Now write the Half-Reactions Mg + S  MgS00 +2 -2 Mg is oxidized: Mg  Mg +2 + 2e - S is reduced: S + 2e -  S -2

12. Zn + 2HCl  H 2 + ZnCl 2 What is oxidized? Reduced? Zn goes from 0 to +2 = oxidation H goes from +1 to 0 = reduction Cl goes from -1 to -1; No change0 +1 +1 -1 -1 0 +2 +2 -1 -1

13. Write Half Reactions Zn + 2HCl  H 2 + ZnCl 20 +1 +1 -1 -1 0 +2 +2 -1 -1 Zn  Zn +2 + 2e - 2H +1 + 2e -  H 2

14. In Redox # of electrons gained has to equal number of electrons lost Zn  Zn +2 + 2e - 2H +1 + 2e -  H 2 H 2  2H +1 + 2e - N 2 + 6e -  2N -3

15. In Redox # of electrons gained has to equal number of electrons lost H 2  2H +1 + 2e - N 2 + 6e -  2N -3 3 3H 2  6H +1 + 6e - 3H 2 + N 2  6H +1 + 2N -3

16. N 2(g) + 3H 2(g)  2NH 3(g) 3H 2 + N 2  6H +1 + 2N -3

17. The steps… start with balanced rxn 1.Assign oxidation numbers to all atoms in equation 2.Determine elements changed oxidation number 3.Identify element oxidized & element reduced 4.Write half-reactions (diatomics must stay as is, everyone you can write with their oxidation number only – NH 3 = N -3 but Cl 2 stays Cl 2 not Cl) – Make sure # of atoms on each side is balanced – Make sure charge on each side is balanced 5.Number electrons lost & gained must be equal; multiply half-reactions if necessary 6.Add half-reactions

18. Cu + AgNO 3  Cu(NO 3 ) 2 + Ag 1.Assign oxidation numbers to all atoms in equation 2.Determine elements changed oxidation number – Cu 0  +2 Ag +1  0 – N +5  +5 unchanged O -2  -2 unchanged 3.Identify element oxidized & element reduced – Oxidation (OIL) =Cu – Reduction (RIG) = Ag0 +1 +5 -2 +2 +5-2 0 2 2

19. 4.Write half-reactions – Cu  Cu +2 + 2e - – Ag +1 + 1e -  Ag 5.Number electrons lost & gained must be equal; multiply half- reactions if necessary – 2(Ag +1 + 1e -  Ag) – 2Ag +1 + 2e -  2Ag Cu + 2AgNO 3  Cu(NO 3 ) 2 + 2Ag0 +1 +5 -2 +2 +5-2 0

20. 6.Add half-reactions; Transfer coefficients to skeleton equation Cu  Cu +2 + 2e - 2Ag +1 + 2e -  2Ag Cu + AgNO 3  Cu(NO 3 ) 2 + Ag Cu + 2AgNO 3  Cu(NO 3 ) 2 + 2Ag0 +1 +5 -2 +2 +5-2 0 + ______________________ Cu + 2Ag +1 + 2e -  2Ag + Cu +2 + 2e - 22 Cu + 2Ag +1  2Ag + Cu +2