1. Electro Chemistry Chemical reactions and Electricity

2. Introduction Electron transfer
The basis of electrochemical processes is the transfer of electrons between substances.
A  e A+
B e -  B-
Oxidation; the reaction with oxygen.
4 Fe(s) + 3O 2 (g) Fe2O3 (s)
Why is rust Fe2O3 , 2Fe to 3O?

3. Oxidation of Iron
Electron transfer of iron Electron transfer to oxygen
Fe  Fe e- 1/2 O e-  O2-
Net reaction:
4 Fe(s) + 3O2(g) Fe2O3(s)
Fe(+3) O(-2) 
Fe2O3 : Electrical neutrality

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 O2 + 2e-  O2- (oxidizing agent)
Redox Process is the combination of an oxidation and reduction process.

5. Symbiotic Process
Redox process always occurs together. In redox process, one can’t occur without the other.
Example: 2 Ca (s) + O2  2CaO
Which is undergoing oxidation ? Reduction?
Oxidation: Ca  Ca+2
Reduction: O2  O-2
Oxidizing agent; That which is responsible to oxidize another.
O2 ; 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. 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. Detailed: Assigning Oxidation Number
Rules for Assigning an Oxidation Number (Ox#)
General rules
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. 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 H2O
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. Example: Acidic Conditions
I- + S2O8-2  I2 + S2O42-
Half Rxn (oxid): I-  I2
Half Rxn (red): S2O8-2  I2 + S2O42-
Bal. chemical and e- : 2 I-  I e-
Bal. chemical O and H : 8e H+ + S2O8-2  S2O H2O
Mult 1st rxn by 4: 8I-  4 I2 + 8e-
Add rxn 1 & 2: 8I-  4 I2 + 8e-
8e H+ + S2O8-2  S2O H2O
8I- + 8H+ + S2O  4 I S2O H2O

10. Redox Reactions - Ion electron method. Under Basic conditions
1, 2. Procedure identical to that under acidic conditions
Balance the half rxn separately except H & O’s.
Balance Oxygen by H2O
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- H2O

11. Example: Basic Conditions
H2O2 (aq) + Cr2O7-2(aq )  Cr 2+ (aq) + O2 (g)
Half Rxn (oxid): 6e H+ + Cr2O7-2 (aq)  2Cr H2O
Half Rxn (red): ( H2O2 (aq)  O H e- ) x 3
8 H H2O2 + Cr2O72-  2Cr O H2O
add: 8H2O  8 H OH-
8H2O  8 H OH-
Net Rxn: 3H2O2 + Cr2O H2O  2Cr O OH-

12. Exercise Try these examples:
1. BrO4- (aq) + CrO2- (aq)  BrO3- (aq) + CrO42- (aq) (basic)
2. MnO4- (aq) + CrO42- (aq)  Mn2+ (aq) + Co2 (aq) (acidic)
3. Fe2+ (aq) + MnO4- (aq)  Fe3+(aq) + Mn2+ (aq) (acidic)

13. Redox Titration KMnO4 (purple)  Mn2+ (pink) Mn (+7) Mn(+2)
Balance redox chem eqn: Solve problem using stoichiometric strategy.
Q: g Fe ore requires ml of M KMnO4. How pure is the ore sample?
When iron ore is titrated with KMnO4 . The equivalent point results when:
KMnO4 (purple)  Mn2+ (pink)
Mn (+7) Mn(+2)
Rxn: Fe MnO4-  Fe Mn2+
Bal. rxn: 5 Fe MnO H+  5 Fe Mn H2O
Note Fe2+  5 Fe3+ : Oxidized Lose e- : Reducing Agent
Mol of MnO4- = ml • 0.180(mol/L) = mmol MnO4-
Amt of Fe: = mmol • 5 mol Fe+2 • g = g
1 mol MnO4- 1 mol Fe2+
% Fe = ( g / g) • 100 = %

14. Redox Titration: Example
Petrucci 7th Ed. p 152
1. A piece of iron wire weighting g is converted to Fe2+ (aq) and requires mL of a KMnO4 (aq) solution for its titration. What is the molarity of the KMNO4 (aq) ?
2. Another substance that may be used to standardized KMNO4 (aq) is sodium oxalate, Na2C2O4. If g of Na2C2O4 is dissolved in water and titrated with mL KMnO4, what is the molarity of the KMnO4 (aq) ?