Electrochemistry Chemical reactions and Electricity

1. ElectrochemistryChemical reactions and Electricity

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) Fe2O3 (s) • Why is rust Fe2O3 , 2Fe to 3O?

3. Oxidation of Iron • Electron transfer of iron- Electron transfer to oxygen • Fe  Fe3+ + 3e- 1/2 O2 + 2e- 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- I2 + 2 e- • Bal. chemical O and H : 8e- + 8H+ + S2O8-2  S2O42- + 4H2O • Mult 1st rxn by 4: 8I- 4 I2 + 8e- • Add rxn 1 & 2: 8I-4I2 + 8e- • 8e- + 8H+ + S2O8-2  S2O42- + 4H2O • 8I-+ 8H+ + S2O8-2  4I2 + S2O42- + 4H2O

10. Redox Reactions - Ion electron method.Under Basic conditions 1, 2. Procedure identical to that under acidic conditions Balance the half reaction 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- + 14H+ + Cr2O7-2 (aq)  2Cr3++ 7 H2O • Half Rxn (red): ( H2O2 (aq) O2 + 2H+ + 2e- ) x 3 • 8 H+ + 3H2O2 + Cr2O72- 2Cr+3 + 3O2+ 7H2O • add: 8H2O  8 H+ + 8 OH- • 8 H+ + 3H2O2 + Cr2O72- 2Cr+3 + 3O2+ 7H2O • 8H2O  8 H+ + 8 OH- • Net Rxn: 3H2O2 + Cr2O72 - + H2O  2Cr+3 + 3O2 + 8 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 • Balance redox chem eqn: Solve problem using stoichiometric strategy. • Q: 1.225 g Fe ore requires 45.30 ml of 0.0180 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+2 + MnO4- Fe+3 + Mn2+ • Bal. rxn: 5 Fe2+ + MnO4- + 8 H+ 5 Fe3+ + Mn2+ + 4 H2O • Note Fe2+ 5 Fe3+ : Oxidized Lose e- : Reducing Agent • Mol of MnO4- = 45.30 ml • 0.180(mol/L) = 0.8154 mmol MnO4- • Amt of Fe: = 0.8154 mmol • 5 mol Fe+2 • 55.8 g = 0.2275 g • 1 mol MnO4- 1 mol Fe2+ • % Fe = (0.2275 g / 1.225 g) • 100 = 18.6 %

14. Redox Titration: Example • 1. A piece of iron wire weighting 0.1568 g is converted to Fe2+ (aq) and requires 26.24 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 0.2482 g of Na2C2O4 is dissolved in water and titrated with 23.68 mL KMnO4, what is the molarity of the KMnO4(aq) ?