18.1 Electro Chemistryfaculty.sdmiramar.edu/fgarces/zCourse/All_Year/Ch... · 0.2482 g Na2CO 4∗...

1 January 13 ElectroChem

18.1 Electro Chemistry

Chemical reactions and Electricity

Dr. Fred Omega Garces Chemistry 201

Miramar College


Introduction Electron transfer The basis of electrochemical processes is the

transfer of electrons between substances.

A g e - + A+ B + e - g B-

Oxidation; the reaction with oxygen.

4 Fe(s) + 3O2 (g) g Fe2O3 (s) Why is rust Fe2O3 , 2Fe to 3O?

F-F

9 p10 n

Na+Na

11 p12 n

11 p12 n

9 p10 n

11 p12 n

9 p10 n

NaF


Oxidation of Iron Electron transfer of iron- Electron transfer to oxygen Fe g Fe3+ + 3e- 1/2 O2 + 2e- → O2-

Net reaction: 4 Fe(s) + 3O2(g) g Fe2O3(s) Fe (+3) O (-2) ⇓

Fe2O3 : Electrical neutrality


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.


Symbiotic Process Redox process always occurs together. In redox process, one can’t occur without the other.

Example: 2 Ca (s) + O2 → 2CaO(s) 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


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, (peroxide and superoxide).

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

6. Σ 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


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


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, the non H and non O elements first

followed by H & O’s. Balance the non-hydrogen and non-oxygen elements first 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 electrons should cancel.


Example: Acidic Conditions

I- + S2O8-2 D I2 + S2O4

2-

Half Rxn (oxid): I- D I2 Half Rxn (red): S2O8

-2 D S2O42-

Bal. chemical and e- : 2 I- D I2 + 2 e- Bal. chemical O and H : 8e- + 8H+ + S2O8

-2 D S2O42- + 4H2O

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

Add rxn 1 & 2: 8I- D 4 I2 + 8e- 8e- + 8H+ + S2O8

-2 D S2O42- + 4H2O

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

2- + 4H2O


1, 2. Procedure identical to that under acidic conditions Balance the half rxn separately except H & O’s. Balance the non-hydrogen and non-oxygen elements first 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 electrons should cancel.

5. Eliminate H+ by adding: H+ + OH- D H2O

Redox Reactions - Ion electron method. Under Basic conditions


Example: Basic Conditions

H2O2 (aq) + Cr2O7-2

(aq ) D Cr 3+ (aq) + O2 (g)

Half Rxn (oxid): 6e- + 14H+ + Cr2O7-2

(aq) D 2Cr3+ + 7 H2O Half Rxn (red): ( H2O2 (aq) D O2 + 2H+ + 2e- ) x 3

8 H+ + 3H2O2 + Cr2O72- D 2Cr+3 + 3O2

+ 7H2O

add: 8H2O D 8 H+ + 8 OH- 8 H+ + 3H2O2 + Cr2O7

2- D 2Cr+3 + 3O2 + 7H2O

8H2O D 8 H+ + 8 OH-

Net Rxn: 3H2O2 + Cr2O72 - + H2O D 2Cr+3 + 3O2

+ 8 OH-


Text Sample Problem


Exercise Try these examples:

1. In the reaction: Cu + HNO3 g Cu2+ + NO3 + NO2 + H2O, which pair acts as the reducing agent?

a) Cu/ Cu2+ b) HNO3/ NO3- c) HNO3/ NO2 d) H+ / H2O

2. Which one atom has the largest change in its oxidation state in the following reaction? Fe2S3(s) + 12 HNO3(aq) D 2 Fe(NO3)3(aq) + 3S(s) + 6NO2(g) + 6 H2O(l) a)  Fe b) S c) N d) All right atom

3. KMnO4 (aq) + HCl (aq) D Cl2 (aq) + KCl (aq) + MnCl2 (aq) (acidic)

4. I2 (aq) D IO3-(aq) + I-

(aq) (basic)

5. CrO42-

(aq) D Cr(OH)3 (aq) + O22-

(aq) (basic)

6. Fe2+ (aq) + MnO4

- (aq) D Fe3+

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


Redox Titration Balance redox chem eqn: Solve problem using stoichiometric strategy. Q: The oxidation to Fe+3 of 1.225 g Fe ore (Fe+2) 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) g Mn2+ (pink) Mn (+7) Mn(+2)

Rxn: Fe+2 + MnO4

- g Fe+3 + Mn2+


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) g Mn2+ (pink) Mn (+7) Mn(+2)

Rxn: Fe+2 + MnO4

- g Fe+3 + Mn2+

Bal. rxn: 5 Fe2+ + MnO4- + 8 H+ g 5 Fe3+ + Mn2+ + 4 H2O

Note Fe2+ g 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 %


Redox Titration: Example 1. A piece of iron wire weighting 0.1568 g is converted to Fe2+ (aq). The Fe3+ is titrated against KMnO4 and requires 26.24 mL KMnO4 (aq) to reach the equivalence point. 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, Na2C2O4 is dissolved in water and titrated with 23.68 mL KMnO4 , what is the molarity of the KMnO4 (aq)? C2O4

-2 is oxidize to CO2.

€

5 C2O4

-2 → 10 CO2 + 10 e−

______________________________________________

2 MnO4

- + 16H+ + 10e− → 2Mn+2 + 8 H2O n = 10

0.2482 g Na2C2O4 ∗ mol Na2C2O4

134.00 g∗

2 mol KMnO4 5 mol Na2C2O4

∗1

0.02368 L KMnO4 = 0.0312 M

€

5 Fe+2 + MnO4

- + 8H+ → Mn+2 + 5Fe+3 + 4H2O

MKMnO4 = 0.1568 g ∗ 1 mol Fe

55.8 g ∗ 1 mol MnO4

-

5 mol Fe+2 ∗ 1

0.02624 L =

MKMnO4 = 0.0212 M

18.1 Electro Chemistryfaculty.sdmiramar.edu/fgarces/zCourse/All_Year/Ch... · 0.2482 g Na2CO 4∗...

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