1. 1 Electrochemistry Chemical reactions and Electricity

2. 2 Introduction Electron transfer The basis of electrochemical processes is the transfer of electrons between substances. A  e - + B Oxidation; the reaction with oxygen. 4 Fe (s) + 3O 2 (g)  Fe 2 O 3 (s) Why is rust Fe 2 O 3, 2Fe to 3O?

3. 3 Oxidation of Iron Electron transfer of iron- Electron transfer to oxygen Fe  Fe 3+ + 3e - 1/2 O 2 + 2e -  O 2- Net reaction: 4 Fe (s) + 3O 2(g)  Fe 2 O 3 (s) Fe(+3) O(-2)  Fe 2 O 3 : Electrical neutrality

4. 4 Oxidation States Definition - Oxidation Process- (charge increase) Lose electron (oxidation) i.e., Fe  Fe +3 + 3e - (reducing agent) Reduction Process- (charge decrease) Gain electrons (reduction) i.e., 1/2 O 2 + 2e -  O 2- (oxidizing agent) Redox Process is the combination of an oxidation and reduction process.

5. 5 Symbiotic Process Redox process always occurs together. In redox process, one can’t occur without the other. Example:2 Ca (s) + O 2  2CaO Which is undergoing oxidation ? Reduction? Oxidation: Ca  Ca +2 Reduction: O 2  O -2 Oxidizing agent; That which is responsible to oxidize another. O 2 ; Oxidizing agent; The agent itself undergoes reduction Reducing agent; That which is responsible to reduce another. Ca; Reducing agent; The agent itself undergoes oxidation

6. 6 Rules of Oxidation State Assignment 1. Ox # = 0: Element in its free state (not combine with different element) 2. Ox # = Charge of ion: Grp1 = +1, Grp2 = +2, Grp7 = -1,... 3. F = -1: For other halogens (-1) except when bonded to F or O. 4. O = -2: Except with fluorine or other oxygen. 5. H = +1: Except with electropositive element (i.e., Na, K) H = -1.  Ox. # = charge of molecule or ion. Highest and lowest oxidation numbers of reactive main-group elements. The A group number shows the highest possible oxidation number (Ox.#) for a main-group element. (Two important exception are O, which never has an Ox# of +6 and F, which never has an Ox# of +7.) For nonmetals, (brown) and metalloids (green) the A group number minus 8 gives the lowest possible oxidation number

7. 7 Detailed: Assigning Oxidation Number Rules for Assigning an Oxidation Number (Ox#) General rules 1. For 1. For an atom in its elemental form (Na, O2, Cl2 …) Ox# = 0 2. For a monatomic ion: Ox# = ion charge 3. The sum of Ox# values for the atoms in a compound equals zero. The sum of Ox# values for the atoms in a polyatomic ion equals the ion charge. Rules for specific atoms or periodic table groups. 1. For fluorine:Ox# = -1 in all compounds 2. For oxygen:Ox# = -1 in peroxides Ox# = -2 in all other compounds (except with F) 3. For Group 7A(17):Ox# = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group. 4. For Group 1A(1):Ox# = +1 in all compounds 5. For Group 2A(2):Ox# = +2 in all compounds 6. For hydrogen:Ox# = +1 in combination with nonmetals Ox# = -1 in combinations with metals and boron General rules 1. For 1. For an atom in its elemental form (Na, O2, Cl2 …) Ox# = 0 2. For a monatomic ion: Ox# = ion charge 3. The sum of Ox# values for the atoms in a compound equals zero. The sum of Ox# values for the atoms in a polyatomic ion equals the ion charge. Rules for specific atoms or periodic table groups. 1. For fluorine:Ox# = -1 in all compounds 2. For oxygen:Ox# = -1 in peroxides Ox# = -2 in all other compounds (except with F) 3. For Group 7A(17):Ox# = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group. 4. For Group 1A(1):Ox# = +1 in all compounds 5. For Group 2A(2):Ox# = +2 in all compounds 6. For hydrogen:Ox# = +1 in combination with nonmetals Ox# = -1 in combinations with metals and boron

8. 8 Redox Reactions - Ion electron method. Under Acidic conditions 1. Identify oxidized and reduced species Write the half reaction for each. 2. Balance the half rxn separately except H & O’s. Balance: Oxygen by H 2 O Balance: Hydrogen by H + Balance: Charge by e - 3. Multiply each half reaction by a coefficient. There should be the same # of e - in both half-rxn. 4. Add the half-rxn together, the e - should cancel.

9. 9 Example: Acidic Conditions I - + S 2 O 8 -2  I 2 + S 2 O 4 2- Half Rxn (oxid): I -  I 2 Half Rxn (red): S 2 O 8 -2  I 2 + S 2 O 4 2- Bal. chemical and e- : 2 I -  I 2 + 2 e - Bal. chemical O and H : 8e - + 8H + + S 2 O 8 -2  S 2 O 4 2 - + 4H 2 O Mult 1st rxn by 4: 8 I -  4 I 2 + 8e - Add rxn 1 & 2: 8I -  4 I 2 + 8e - 8e - + 8H + + S 2 O 8 -2  S 2 O 4 2 - + 4H 2 O 8I - + 8H + + S 2 O 8 -2  4 I 2 + S 2 O 4 2 - + 4H 2 O

10. 10 1, 2. Procedure identical to that under acidic conditions Balance the half reaction separately except H & O’s. Balance Oxygen by H 2 O Balance Hydrogen by H + Balance charge by e - 3. Mult each half rxn such that both half- rxn have same number of electrons 4. Add the half-rxn together, the e - should cancel. 5. Eliminate H+ by adding: H + + OH -  H 2 O Redox Reactions - Ion electron method. Under Basic conditions

11. 11 Example: Basic Conditions H 2 O 2 (aq) + Cr 2 O 7 -2 (aq )  Cr 2+ (aq) + O 2 (g) Half Rxn (oxid): 6e - + 14H + + Cr 2 O 7 -2 (aq)  2Cr 3+ + 7 H 2 O Half Rxn (red): ( H 2 O 2 (aq)  O 2 + 2H + + 2e - ) x 3 8 H + + 3H 2 O 2 + Cr 2 O 7 2-  2Cr +3 + 3O 2 + 7H 2 O add: 8H 2 O  8 H + + 8 OH - 8 H + + 3H 2 O 2 + Cr 2 O 7 2-  2Cr +3 + 3O 2 + 7H 2 O 8H 2 O  8 H + + 8 OH - Net Rxn: 3H 2 O 2 + Cr 2 O 7 2 - + H 2 O  2Cr +3 + 3O 2 + 8 OH -

12. 12 Exercise Try these examples: 1. BrO 4 - (aq) + CrO 2 - (aq)  BrO 3 - (aq) + CrO 4 2- (aq) (basic) 2. MnO 4 - (aq) + CrO 4 2- (aq)  Mn 2+ (aq) + CO 2 (aq)(acidic) 3. Fe 2+ (aq) + MnO 4 - (aq)  Fe 3+ (aq) + Mn 2+ (aq) (acidic )

13. 13 Redox Titration Balance redox chem eqn: Solve problem using stoichiometric strategy. Q: 1.225 g Fe ore requires 45.30 ml of 0.0180 M KMnO 4. How pure is the ore sample? When iron ore is titrated with KMnO 4. The equivalent point results when : KMnO 4 (purple)  Mn 2+ (pink) Mn (+7) Mn(+2) Rxn:Fe +2 + MnO 4 -  Fe +3 + Mn 2+ Bal. rxn: 5 Fe 2+ + MnO 4 - + 8 H +  5 Fe 3+ + Mn 2+ + 4 H 2 O Note Fe 2+  5 Fe 3+ : Oxidized Lose e- : Reducing Agent Mol of MnO 4 - = 45.30 ml 0.180(mol/L) = 0.8154 mmol MnO 4 - Amt of Fe:= 0.8154 mmol 5 mol Fe +2 55.8 g = 0.2275 g 1 mol MnO 4 - 1 mol Fe 2+ % % Fe= (0.2275 g / 1.225 g) 100 = 18.6 %

14. 14 Redox Titration: Example 1. A piece of iron wire weighting 0.1568 g is converted to Fe 2+ (aq) and requires 26.24 mL of a KMnO 4 (aq) solution for its titration. What is the molarity of the KMNO 4 (aq) ? 2. Another substance that may be used to standardized KMNO 4 (aq) is sodium oxalate, Na 2 C 2 O 4. If 0.2482 g of Na 2 C 2 O 4 is dissolved in water and titrated with 23.68 mL KMnO 4, what is the molarity of the KMnO 4 (aq) ?