Announcements Exam #3 is THURSDAY! Chapter 13 – EDTA Titrations Chapter 23 – What is Chromatography Chapter 24 – Gas Chromatography* Chapter 25 – HPLC* Final is Wednesday Afternoon on May 9 th Read sections 15-3 thru 15-4 for lab tomorrow. Homework 14-15, 14-26, 15-6, 16-3, 16-6,

Some Examples Assign oxidation numbers to each element in the compounds: 1. Cr 2 O Cl 2 O 3 3. K 2 SO 4

Some empirical observations on Oxidation numbers for Elements in Compounds A. Group I metals are always +1 B. Group II metals are always +2 C. Fluorine is always in -1 oxidation state D. Oxygen is usually in -2 oxidation state Except when superceded by rules A, B, or C.

Oxidation Numbers Cr 2 O 7 -2 Oxygen = -2Overall charge = -2 Cr = ? From our Rules 2(Cr) + 7(O) = -2 2(Cr) + 7(-2) = -2 2Cr - 14 = Cr = 12 Oxidation number on Cr in Cr 2 O 7 2- is 6

Oxidation numbers (cont’d) Cl 2 O 3 2(Cl) + 3 (O) = 0 2Cl + 3(-2) = 0 2Cl = +6 Oxidation on Cl is + 3

Oxidation numbers (cont’d) K 2 SO 4 2(K) + 1 (S) + 4(O) = 0 2(+1) + 1 (S) + 4(-2) = 0 Oxidation number of S = +6

Redox Reactions Every Redox reaction can be broken into two “half-reactions” One describing the oxidation process One describing the reduction process Together the two processes form an Reduction-Oxidation reaction. Sometimes called a Redox reaction.

Balance the following reaction occurring in Acidic Solution NO 2 - (aq) + Cr 2 O 7 2- (aq) -> Cr 3+ (aq) + NO 3 - (aq) Redox reaction in solution can be VERY complex. Therefore it is necessary to develop a systematic method for balancing REDOX equation in Solution (Either Basic or Acidic)

Balance the following reaction occurring in Acidic Solution NO 2 - (aq) + Cr 2 O 7 2- (aq) -> Cr 3+ (aq) + NO 3 - (aq) Redox reaction in solution can be VERY complex. Therefore it is necessary to develop a systematic method for balancing REDOX equation in Solution (Either Basic or Acidic)

Balancing REDOX Equations Divide the equation into two half reactions. (One for oxidation and one for reduction) Balance each half reaction Element Balance - Balance elements other than hydrogen (H) or oxygen (O) Oxygen Balance - Balance Oxygen by adding water to the side of the equation that needs Oxygen Hydrogen Balance For reaction occurring in acidic solutions ADD H+  For reactions occurring in basic solutions ADD OH- TO EACH SIDE Electron Balance – Balance charge on each side of each half-reaction by adding electrons (e-) Multiply each half reaction by an integer number in order to CANCEL ELECTRONS Add and Simplify Check – same number of each type of atom on both sides Same total charge on each side of the equation

Balance the following reaction occurring in Acidic Solution 1. Divide into two half reactions NO 2 - (aq) + Cr 2 O 7 2- (aq) -> Cr 3+ (aq) + NO 3 - (aq) NO 2 - -> NO 3 - Oxidation Cr 2 O > Cr 3+ Reduction

Balance the following reaction occurring in Acidic Solution 2. Balance Each Half Reaction NO 2 - -> NO 3 - Oxidation Cr 2 O > Cr 3+ Reduction 2

Balance the following reaction occurring in Acidic Solution 2. Balance Oxygens NO 2 - -> NO 3 - Oxidation Cr 2 O > Cr 3+ Reduction 2 H 2 O + + 7H 2 O

Balance the following reaction occurring in Acidic Solution 2. Balance Hydrogens NO 2 - -> NO 3 - Oxidation Cr 2 O > Cr 3+ Reduction 2 H 2 O + + 7H 2 O + 2H + 14H + +

Balance the following reaction occurring in Acidic Solution 2. Balance Electrons NO 2 - -> NO 3 - Oxidation Cr 2 O > Cr 3+ Reduction 2 H 2 O + + 7H 2 O + 2H + 14H e - 6e - +

Balance the following reaction occurring in Acidic Solution 3. Multiply by an integer number in order to cancel electrons NO 2 - -> NO 3 - Oxidation Cr 2 O > Cr 3+ Reduction 2 H 2 O + + 7H 2 O + 2H + 14H e - 6e - +() x 1 () x 3

Balance the following reaction occurring in Acidic Solution 3. Multiply by an integer number in order to cancel electrons NO 2 - -> NO 3 - Oxidation Cr 2 O > Cr 3+ Reduction 2 H 2 O + + 7H 2 O + 2H + 14H e - 6e - +() x 1 () x 3 4. Add and Simplify 3H 2 O + 3 NO H + + Cr 2 O > 3NO Cr H + + 7H 2 O 4 8

Balance the following reaction occurring in Acidic Solution NO 2 - -> NO 3 - Oxidation Cr 2 O > Cr 3+ Reduction 2 H 2 O + + 7H 2 O + 2H + 14H e - 6e - +() x 1 () x 3 5. Check 3 NO H + + Cr 2 O > 3NO Cr H 2 O Same number of each type of atom on both side Same total charge on each side of the equation ok

REDOX reactions There are two general ways to conduct an REDOX reaction Mix them together Separate them Half cells connected by salt bridge and wires

Standard Hydrogen Electrode ‘Standard by which all half- reactions are measured’! The Hydrogen Electrode is the ‘Standard by which all half- reactions are measured’! H 2 (g) -> 2H + (aq) + 2e - The Reaction: E o = 0V The Hydrogen reference electrode is arbitrarily assigned a value of V.

Measuring standard half cell reduction potentials! Cu Cu Cu e - -> Cu E o = Ag + + e - -> Ag E o = Fe Fe 2+ Fe e - -> Fe E o = ? E o = Reverse Reaction IS Favored (OXIDATION)! H 2 (g) -> 2H + (aq) + 2e - 2Ag + + 2e - -> 2Ag E o = 0.799

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Line Notation

What is the line notation? anode  solution at anode  solution at cathode  cathode Cd(s)  Cd +2 (CONC)  Ag + (CONC)  Ag (s)

What is the line notation? anode  solution at anode  solution at cathode  cathode Zn(s)  Zn 2+ (0.10M)  Cu 2+ (0.10M)  Cu (s)

Measuring half cell potentials! What is the line notation? anode  solution at anode  solution at cathode  cathode anode  H + (1 M)  Ag + (1 M)  Ag (s) Pt  H 2 (g, 1 bar)  H + (1 M)  Ag + (1 M)  Ag (s) electron

Spontaneity and Cell Potential The relationship between  G and E (electromotive force)

Spontaneity and Cell Potential Gibbs free energy is related to E (the electromotive force) by: Under standard conditions Where n = number of electrons F = Faraday’s constant = charge on one mole of electrons) = Coulombs/mol e - = J V -1 mole e -1