1. REDOX AND ELECTROCHEMISTRY Oxidation Number A. Convenient way for keeping track of the number of electrons transferred in a chemical reaction A. Convenient way for keeping track of the number of electrons transferred in a chemical reaction B. Oxidation numbers are either: +, -, or 0 B. Oxidation numbers are either: +, -, or 0

2. Rules for assigning oxidation numbers 1. Each atom of a free element has an oxidation number of zero Ex: Na  Na o Na in NaCl is not a free element and does not have a charge of zero H 2  each H-atom has a charge of zero 2. For any element in which only one oxidation number is present use that number.

3. 3. Oxygen has an oxidation number of –2 EXCEPT in peroxides (H 2 O 2 ) where it is –1 and in compounds with fluorine where it may be +1 or +2 –1 and in compounds with fluorine where it may be +1 or +2 Example in H 2 SO 4 oxygen has an oxidation state of –2 A peroxide may occur when hydrogen or a group one or two reacts with oxygen, in this case there is more oxygen than in a normal metal oxide and the oxidation state is -1

4. 4. Hydrogen has a +1 oxidation state in all compounds EXCEPT metal hydrides (LiH, CaH 2 ) where its oxidation state is -1 Example in H 2 SO 4 H has a charge of +1 5. If the compound is ionic use the first negative oxidation state for the nonmetal. example: Fe x Cl 3 -1 example: Fe x Cl 3 -1 x+3(-1)=0 x+3(-1)=0 x=+3 oxidation state of Fe is +3 x=+3 oxidation state of Fe is +3

5. HOW TO ASSIGN OXIDATION NUMBERS IN COMPOUNDS IN COMPOUNDS 1. Use the “rules” to identify the oxidation numbers of all atoms for which a rule exists 2. Multiply the oxidation number of each atom by its subscript 3. The sum of all oxidation number times the subscripts must be zero

6. 4. Example: Na 2 S Na is group 1 and has an oxidation number of +1 and there is no rule for S Na 2 +1 S x Na 2 +1 S x 2(+1) + x =0 2(+1) + x =0 x=-2 so the oxidation state of S is -2 x=-2 so the oxidation state of S is -2 5. For polyatomic ions follow the same rules and procedure used for a compound except the sum of the (oxidation numbers X subscript) must equal the charge on the ion.

7. 6. Example (SO 4 ) -2 Oxygen has an oxidation number of –2 so S x O 4 -2 S x O 4 -2 x-8=-2 x-8=-2 x=+6 so the oxidation state of S in (SO 4 ) -2 is +6 x=+6 so the oxidation state of S in (SO 4 ) -2 is +6

8. If a compound consists of 2 polyatomic ions break it into the two ions and solve separately. Example (NH 4 ) 2 SO 4 (NH 4 ) +1 (SO 4 ) -2 N x H 4 +1 S x O 4 -2 X+4=+1 x-8=-2 X=-3 x=+6 So the oxidation state of N is -3 and of S is +6

9. REDOX Oxidation and Reduction result from the competition for electrons between atoms Oxidation and Reduction result from the competition for electrons between atoms A. Oxidation 1. Represents a loss of electrons 2. Refers to any chemical change in which there is an increase in oxidation number 3. The particle that increases in oxidation number is said to be oxidized.

10. 4. Reducing Agent -the substance that is oxidized 5. Examples Ca + Cl 2  CaCl 2 Assign oxidation numbers Ca 0 +Cl 2 0  Ca +2 Cl 2 -1 Calcium, Ca 0, loses electrons to the chlorine atom, its oxidation state increases from 0 to +2, therefore the calcium, Ca 0 is oxidized or it is said to be the reducing agent Calcium, Ca 0, loses electrons to the chlorine atom, its oxidation state increases from 0 to +2, therefore the calcium, Ca 0 is oxidized or it is said to be the reducing agent

11. B. Reduction 1. Represents a gain of electrons 2. Refers to any chemical change in which there is a decrease in the oxidation number 3. The particle that decreases in oxidation number is said to be reduced. 4. Oxidizing Agent –The substance that is reduced

12. 5. Examples Ca + Cl 2  CaCl 2 Assign oxidation numbers Ca 0 +Cl 2 0  Ca +2 Cl 2 -1 The Cl 0 gains electrons from Ca 0, its oxidation state decreases from 0 to –1, therefore Cl 0 is reduced or it is said to be the oxidizing agent

13. Redox Reactions- Reactions that involve both oxidation and reduction All types of reactions EXCEPT double replacement reactions can be a redox reaction. Double replacement reactions are NEVER redox reactions.

14. How to analyze a redox reaction. 1. Assign oxidation numbers to all atoms Example 2H 2 + O 2  2H 2 O 0 0 +1 –2 0 0 +1 –2 2. Show the changes in oxidation number for the atoms involved. Use a line to connect the atoms undergoing oxidation and those undergoing reduction

15. 2H 2 + O 2  2H 2 O 2H 2 + O 2  2H 2 O 0 0 +1 –2 0 0 +1 –2 3. Indicate which element is oxidized and which is reduced H 2 - increases its oxidation number therefore it is oxidized so it is the reducing agent H 2 - increases its oxidation number therefore it is oxidized so it is the reducing agent O 2 - decreases its oxidation number therefore it is reduced so it is the oxidizing agent O 2 - decreases its oxidation number therefore it is reduced so it is the oxidizing agent ox RED

16. ONLY REACTANTS CAN BE THE SUBSTANCE OXIDIZED OR REDUCED. ONLY REACTANTS CAN BE THE SUBSTANCE OXIDIZED OR REDUCED. The best reducing agents are group I and 2 metals because they want to lose electrons the most. Using Table J the best reducing agents are found at the top of the table. The best reducing agents are group I and 2 metals because they want to lose electrons the most. Using Table J the best reducing agents are found at the top of the table.

17. The best oxidizing agents are the most electronegative elements such as fluorine and oxygen because they want to gain electrons the most. Using table J the best oxidizing agents are found at the bottom of table J The best oxidizing agents are the most electronegative elements such as fluorine and oxygen because they want to gain electrons the most. Using table J the best oxidizing agents are found at the bottom of table J

18. ELECTROCHEMISTRY A. Half –Reactions 1. Every redox reaction consists of 2 parts each called a half reaction 1. Every redox reaction consists of 2 parts each called a half reaction a. Oxidation half reaction shows an atom or ion losing one or more electrons a. Oxidation half reaction shows an atom or ion losing one or more electrons b. Reduction half reaction shows an atom or ion gaining one or more electrons b. Reduction half reaction shows an atom or ion gaining one or more electrons

19. 2. Separate electron equations can be written for each 3. Like other chemical reactions, half reactions must follow the law of conservation of mass. Atoms must be balanced. 4. In addition to conservation of mass, there must also be a conservation of charge. a. The net charge must be the same on both sides of the equation but it does not necessarily equal zero a. The net charge must be the same on both sides of the equation but it does not necessarily equal zero

20. b. Charge is balanced by adding electrons to one side of the half reaction b. Charge is balanced by adding electrons to one side of the half reaction 5. Example: 1. Assign oxidation states and identify the substance oxidized and reduces. Mg + Cl 2  MgCl 2 0 0 +2 -1 0 0 +2 -1 2. Write oxidation half reaction a. balance for mass by adding a. balance for mass by adding coefficients coefficients

21. b. balance for charge by adding electrons b. balance for charge by adding electrons Oxidation half reaction: Mg o  Mg +2 + 2e - 3. Write the reduction half reaction a. balance for mass by adding a. balance for mass by adding coefficients coefficients b. balance for charge by adding b. balance for charge by adding electrons electrons Reduction half reaction: Cl 2 o + 2e-  2Cl -

22. B. Balancing simple redox reactions 1. Write the oxidation and reduction half reactions (see previous notes) Example __Pb +__ Cr +3  __ Pb +2 +__ Cr Oxidation half reaction: Pb o  Pb +2 + 2e - Pb o  Pb +2 + 2e - Reduction half-reaction: Cr +3 + 3e -  Cr o

23. 2. Balance the half reactions by adding coefficients such that the number of electrons in both equations are equal 3(Pb o  Pb +2 +2e - ) = 3Pb o  3Pb +2 + 6e - 2(Cr +3 + 3e -  Cr o ) = 2Cr +3 + 6e -  2Cr o 3. Add the two half reactions together 3Pb o  3Pb +2 + 6e - 3Pb o  3Pb +2 + 6e - +2Cr +3 + 6e -  2Cr o 3Pb o + 2Cr +3  3Pb +2 + 2Cr o

24. D. Calculating Net Potential (voltage) of a Redox Reaction 1. Write both the oxidation and reduction half reactions 1. Write both the oxidation and reduction half reactions 2. Go to the table of Reduction potential 2. Go to the table of Reduction potential 3. For the reduction half reaction used the E 0 value as written 3. For the reduction half reaction used the E 0 value as written

25. 4. For the oxidation half reaction negate the E 0 value 5. Add the E 0 values 6. Meaning of E 0 a. A positive E 0 means the reaction is spontaneous in the direction written b. A negative E 0 means the reaction is not spontaneous in the direction it is written c. An E 0 equal to zero indicates the reaction is in equilibrium Examples Calculate E 0 for the following Examples Calculate E 0 for the following Zn + CuSO 4  ZnSO 4 + Cu Zn + CuSO 4  ZnSO 4 + Cu