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

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

By the end of the lecture the student should be able to: Calculate the potential (E) by Nernest equation. Understand the regions of Redox reaction titration curve Calculate the potential (E) change throughout the titration. 2

The Nernst equation It describes the dependence of potential on concentration 3

E = E o – ( 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) 4

At 25 o C E = E o – ( / n) log [Red] / [Ox] 5

Questions? 6

EXAMPLE 12.3 Page no

Redox Titration Curve 8

Titration of 100 ml 0.1M Fe 2+ versus 0.1M Ce 4+ Example: Fe 2+ + Ce 4+  Fe 3+ + Ce 3+ Fe 2+  Fe 3+ + e Ce 4+ + e  Ce 3+ Oxidation Reduction (Reducing agent) (Oxidizing agent) 9

Fe 2+ is the analyte Ce 4+ is the titrant Ce 4+ is added with a burette to a solution of Fe 2+ The potential changes as a function of added titrant At the end of titration a color change is observed → End Point 10

11

Titration Curves Regions: Titration Curve has 3 Regions  Before the Equivalence Point  At the Equivalence Point  After the Equivalence Point 12

Before the Equivalence Point At beginning of titration, only Fe 2+ (E = 0) Ce 4+ is added to Fe 2+ solution Fe 2+ is oxidized to Fe 3+ Ce 4+ is reduced to Ce 3+ Unreacted Fe 2+ remains in solution Amounts of Fe 2+ and Fe 3+ are known, and used to determine the potential. E = − /1(log [Fe 2+ ] / [Fe 3+ ]) At midpoint of titration, E = E o (why?) 13

At the Equivalence Point Enough Ce 4+ has been added to react with all Fe 2+ Primarily only Ce 3+ and Fe 3+ present Both half-reactions are in equilibrium, and both contribute to the potential at the equivalence point Equivalence-point potential is independent of the concentrations and volumes of the reactants E = n 1 E o 1 + n 2 E o 2 / n 1 + n 2 14

After the Equivalence Point Excess unreacted Ce 4+ remains in solution Amount of Ce 3+ and Ce 4+ are known, and used to calculate the potential E =1.70 − /1 (log[Ce 3+ ] / [Ce 4+ ]) At the point when V = 2 the volume of the equivalence point, [Ce3+] = [Ce4+] and E = E° 15

Questions? 16

Summary: Potential calculation by Nernest equation. Redox titration curve. Potential change during redox titration. Thanks 17