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ElectrochemistryElectrochemistry

Chemical reactions and ElectricityChemical reactions and Electricity

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IntroductionIntroduction

Electron transferThe 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?

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Oxidation of IronOxidation of Iron

Electron transfer of iron- Electron transfer to oxygenFe Fe3+ + 3e- 1/2 O2 + 2e- O2-

Net reaction:4 Fe(s) + 3O2(g) Fe2O3(s)

Fe(+3) O(-2)

Fe2O3 : Electrical neutrality

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Oxidation StatesOxidation States

Definition - Definition -

Oxidation Process- (charge increase)Oxidation Process- (charge increase)Lose electron (oxidation)Lose electron (oxidation)

i.e., Fe i.e., Fe Fe Fe+3 +3 + 3e+ 3e-- (reducing agent) (reducing agent)

Reduction Process-Reduction Process-(charge decrease)(charge decrease)

Gain electrons (reduction)Gain electrons (reduction)

i.e., 1/2 Oi.e., 1/2 O22 + 2e + 2e-- O O2-2- (oxidizing agent(oxidizing agent))

Redox Process is the combination of an Redox Process is the combination of an oxidation and reduction process.oxidation and reduction process.

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Symbiotic ProcessSymbiotic 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

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Rules of Oxidation State AssignmentRules of Oxidation State Assignment1. Ox # = 0: 1. Ox # = 0: Element in its free Element in its free

state state (not combine with different (not combine with different element)element)

2. Ox # = Charge of ion: 2. Ox # = Charge of ion: Grp1 = +1, Grp2 = +2, Grp1 = +1, Grp2 = +2, Grp7 = -1, ...Grp7 = -1, ...

3. F = -1: 3. F = -1: For other halogens (-1) For other halogens (-1) except when bonded to F or O.except when bonded to F or O.

4. O = -2: 4. O = -2: Except with fluorine or Except with fluorine or other oxygen.other oxygen.

5. H = +1: 5. H = +1: Except with Except with electropositive element (i.e., Na, K) electropositive element (i.e., Na, K) H = -1.H = -1.

Ox. #Ox. # = charge of molecule = charge of molecule or ion.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

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Detailed: Assigning Oxidation Detailed: Assigning Oxidation NumberNumber

Rules for Assigning an Oxidation Number (Ox#)General rulesGeneral rules1. For1. For an atom in its elemental form (Na, O2, Cl2 …) Ox# = 02. For a monatomic ion: Ox# = ion charge3. 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.Rules for specific atoms or periodic table groups.1. For fluorine: Ox# = -1 in all compounds2. 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 compounds5. For Group 2A(2): Ox# = +2 in all compounds6. For hydrogen: Ox# = +1 in combination with nonmetals

Ox# = -1 in combinations with metals and boron

General rulesGeneral rules1. For1. For an atom in its elemental form (Na, O2, Cl2 …) Ox# = 02. For a monatomic ion: Ox# = ion charge3. 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.Rules for specific atoms or periodic table groups.1. For fluorine: Ox# = -1 in all compounds2. 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 compounds5. For Group 2A(2): Ox# = +2 in all compounds6. For hydrogen: Ox# = +1 in combination with nonmetals

Ox# = -1 in combinations with metals and boron

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Redox Reactions - Redox Reactions - Ion electron Ion electron method.method.

Under Acidic conditionsUnder Acidic conditions1. Identify oxidized and reduced species

Write the half reaction for each.

2. Balance the half rxn separately except H & O’s.

Balance: Oxygen by H2OBalance: 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.

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Example: Acidic ConditionsExample: Acidic Conditions

I- + S2O8-2 I2 + S2O4

2-

Half Rxn (oxid): I- I2

Half Rxn (red): S2O8-2 I2 + S2O4

2-

Bal. chemical and e- : 2 I- I2 + 2 e-

Bal. chemical O and H : 8e- + 8H+ + S2O8-2 S2O4

2- + 4H2O

Mult 1st rxn by 4: 8I- 4 I2 + 8e-

Add rxn 1 & 2: 8I- 4 I2 + 8e-

8e- + 8H+ + S2O8-2 S2O4

2- + 4H2O

8I- + 8H+ + S2O8-2 4 I2 + S2O4

2- + 4H2O

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1, 2. Procedure identical to that under acidic 1, 2. Procedure identical to that under acidic conditionsconditionsBalance 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 3. Mult each half rxn such that both half- rxn have same number of electronshave same number of electrons

4. Add the half-rxn together, the e4. Add the half-rxn together, the e-- should should cancel.cancel.

5. Eliminate H+ by adding: 5. Eliminate H+ by adding: HH++ + OH + OH-- HH22OO

Redox Reactions - Redox Reactions - Ion electron Ion electron method.method.

Under Basic conditionsUnder Basic conditions

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Example: Basic ConditionsExample: 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-

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ExerciseExercise

Try these examples:1. BrO4

- (aq) + CrO2

- (aq) BrO3

- (aq) + CrO4

2- (aq) (basic)

2. MnO4- (aq) + CrO4

2- (aq) Mn2+ (aq) + CO2 (aq) (acidic)

3. Fe2+ (aq) + MnO4

- (aq) Fe3+

(aq) + Mn2+ (aq) (acidic)

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Redox TitrationRedox TitrationBalance redox chem eqn: Solve problem using stoichiometric Balance redox chem eqn: Solve problem using stoichiometric strategy.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:KMnOKMnO44 (purple) (purple) MnMn2+2+ (pink) (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 %%

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Redox Titration: ExampleRedox 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) ?