1. Electrochemical Reactions Redox reaction: electrons transferred from one species to another Oxidation ≡ loss of electrons Reduction ≡ gain of electrons

2. What is reduced is the oxidizing agent H + oxidizes Zn by taking electrons from it What is oxidized is the reducing agent Zn reduces H + by giving it electrons Oxidation and Reduction

3. Zn (s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu (s) Zn is oxidized; Zn is reducing agentZn Zn 2+ + 2e - Cu 2+ is reduced; Cu 2+ is oxidizing agent Cu 2+ + 2e - Cu Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu (s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag (s) Cu Cu 2+ + 2e - Ag + + 1e - AgAg + is reduced Ag + is oxidizing agent

4. Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

5. Assigning Oxidation Numbers 1.Elements in their elemental form have an oxidation number of 0. 2.The oxidation number of a monatomic ion is the same as its charge. 3.Nonmetals tend to have negative oxidation numbers. 4.The sum of the oxidation numbers in a neutral compound is 0. 5.The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

6. Balancing Redox Equations Neutral solution: Cr (s) + Ag + (aq) ⇌ Cr 3+ (aq) + Ag (s)

7. Balancing Redox Equations Acidic solution: MnO 4 − (aq) + C 2 O 4 2− (aq)  Mn 2+ (aq) + CO 2 (aq)

8. Electrochemistry The Half-Reaction Method Consider the reaction between MnO 4 − and C 2 O 4 2− : MnO 4 − (aq) + C 2 O 4 2− (aq)  Mn 2+ (aq) + CO 2 (aq) Balancing Redox Equations Fig 20.2

9. Electrochemistry The Half-Reaction Method First, we assign oxidation numbers: MnO 4 − + C 2 O 4 2-  Mn 2+ + CO 2 +7+3+4+2 Manganese goes from +7 to +2, it is reduced Carbon goes from +3 to +4, it is oxidized

10. Electrochemistry Oxidation Half-Reaction C 2 O 4 2−  CO 2 To balance the carbon, we add a coefficient of 2: C 2 O 4 2−  2 CO 2

11. Electrochemistry Oxidation Half-Reaction C 2 O 4 2−  2 CO 2 The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side. C 2 O 4 2−  2 CO 2 + 2 e −

12. Electrochemistry Reduction Half-Reaction MnO 4 −  Mn 2+ The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side. MnO 4 −  Mn 2+ + 4 H 2 O

13. Electrochemistry © 2009, Prentice- Hall, Inc. Reduction Half-Reaction MnO 4 −  Mn 2+ + 4 H 2 O To balance the hydrogen, we add 8 H + to the left side. 8 H + + MnO 4 −  Mn 2+ + 4 H 2 O

14. Electrochemistry Reduction Half-Reaction 8 H + + MnO 4 −  Mn 2+ + 4 H 2 O To balance the charge, we add 5 e − to the left side. 5 e − + 8 H + + MnO 4 −  Mn 2+ + 4 H 2 O

15. Electrochemistry Combining the Half-Reactions Now we evaluate the two half-reactions together: C 2 O 4 2−  2 CO 2 + 2 e − 5 e − + 8 H + + MnO 4 −  Mn 2+ + 4 H 2 O To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.

16. Electrochemistry Combining the Half-Reactions 5 C 2 O 4 2−  10 CO 2 + 10 e − 10 e − + 16 H + + 2 MnO 4 −  2 Mn 2+ + 8 H 2 O When we add these together, we get: 10 e − + 16 H + + 2 MnO 4 − + 5 C 2 O 4 2−  2 Mn 2+ + 8 H 2 O + 10 CO 2 +10 e −

17. Electrochemistry Combining the Half-Reactions 10 e − + 16 H + + 2 MnO 4 − + 5 C 2 O 4 2−  2 Mn 2+ + 8 H 2 O + 10 CO 2 +10 e − The only thing that appears on both sides are the electrons. Subtracting them, we are left with: 16 H + + 2 MnO 4 − + 5 C 2 O 4 2−  2 Mn 2+ + 8 H 2 O + 10 CO 2

18. Balancing Redox Equations Basic solution: V 2+ (aq) + V(OH) 4 + (aq) ⇌ VO 2+ (aq)