Presentation is loading. Please wait.

Presentation is loading. Please wait.

Reduction- Oxidation Reactions (1) 213 PHC 10th lecture (1) Gary D. Christian, Analytical Chemistry, 6 th edition. 1.

Similar presentations


Presentation on theme: "Reduction- Oxidation Reactions (1) 213 PHC 10th lecture (1) Gary D. Christian, Analytical Chemistry, 6 th edition. 1."— Presentation transcript:

1 Reduction- Oxidation Reactions (1) 213 PHC 10th lecture (1) Gary D. Christian, Analytical Chemistry, 6 th edition. 1

2 By the end of the lecture the student should be able to: Define oxidation, reduction, oxidizing agent, and reducing agent. Understand the principals of electrochemical cell and electrode potential Calculate the electrode potential by Nernest equation. 2

3 WHAT IS A REDOX REACTION? It is a reaction occurs between an oxidizing agent and a reducing agent. 3

4 OXIDATIONREDUCTION A loss of electrons to give a higher oxidation state Fe 2+  Fe 3+ + e - A gain of electrons to give a lower oxidation state Ce 4+ + e -  Ce 3+ 4

5 Oxidizing AgentReducing Agent Take on electrons Gets reduced Ce 4+ + e -  Ce 3+ Give up electrons Gets oxidized Fe 2+  Fe 3+ + e - 5

6 Fe 2+ + Ce 4+  Fe 3+ + Ce 3+    reducing oxidizing oxidized reduced agentagent form form (Reduced form) (oxidized form) 6

7 The reducing or oxidizing tendency of a substances will depend on its reduction potential. 7

8 Electrode Potential (E o ) Each half-reaction will generate a potential that adopted by an inert electrode dipped in the solution. Individual electrode potential can’t be measured. The difference between 2 electrode potentials can be measured. The standard hydrogen electrode is used to measure the potential of any half reaction because it’s potential is zero. 8

9 The more +ve E o = (oxidation). The more -ve E o = (reduction). 9

10 Fe 3+ + e -  Fe 2+ E o = 0.771 V Sn 4+ + 2e -  Sn 2+ E o = 0.154 V 2Fe 3+ + Sn 2+  2Fe 2+ + Sn 4+ 10

11 Questions? 11

12 Example Fe 3+ + e - = Fe 2+ E o = 0.771 V I 3 - + 2e - = 3I - E o = 0.535 V 2Fe 3+ + 3I - = 2Fe 2+ + I 3 - E o cell = 0.771 – 0.535 = 0.235 V 12

13 Quiz 1. What is the overall cell reaction and the cell potential for the two half-reactions? A) Cu 2+ + 2e = Cu E o = 0.34 V Zn 2+ + 2e = Zn E o = -0.76 V B) Fe 3+ + e = Fe 2+ E o = 0.77 V Ti 4+ + e = Ti 3+ E o = 0.15 V 13

14 The Nernst equation 14

15 The Nernst equation Describes the dependence of potential on concentration. 15

16 E = E o – 2.3026 RT / n F log [Red] / [Ox] E = reduction potential at specific conc. E o = standard potential. n = no. of electrons R = gas const. (8.3143) T = absolute temp. F = Faraday const. (96.487) 16

17 At 25 o C E = E o – (0.05916 / n) log [Red] / [Ox] 17

18 Questions? 18

19 Summary: Definition of oxidation-reduction reactions. Electrochemical cells. Electrode potential. Nernast equation 19

20 Thank you 20


Download ppt "Reduction- Oxidation Reactions (1) 213 PHC 10th lecture (1) Gary D. Christian, Analytical Chemistry, 6 th edition. 1."

Similar presentations


Ads by Google