Electrochemistry
Chemical reactions and Electricity
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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?
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Oxidation of Iron
Electron transfer of ironFe  Fe3+ + 3e-
Electron transfer to oxygen
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 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.
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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
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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)
when bonded to F or O.
4. O = -2:
except
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
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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
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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.
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Example: Acidic Conditions
I- + S2O8-2  I2 + S2O42Half Rxn (red):
I-  I 2
S2O8-2  I2 + S2O42-
Bal. chemical and e- :
2 I-  I2 + 2 e-
Half Rxn (oxid):
Bal. chemical O and H :
Mult 1st rxn by 4:
Add rxn 1 & 2:
8e- + 8H+ + S2O8-2  S2O42- + 4H2O
8I-  4 I2 + 8e8I-  4 I2 + 8e-
8e- + 8H+ + S2O8-2  S2O42- + 4H2O
8I- + 8H+ + S2O8-2  4 I2 +
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S2O42- + 4H2O
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
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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 OH8 H+ + 3H2O2 + Cr2O72-  2Cr+3 + 3O2 + 7H2O
8H2O  8 H+ + 8 OHNet Rxn:
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3H2O2 + Cr2O72 - + H2O  2Cr+3 + 3O2 + 8 OH-
Exercise
Try these examples:
1. BrO4- (aq) + CrO2- (aq)  BrO3- (aq) + CrO42- (aq)
2. MnO4- (aq) + CrO42- (aq)  Mn2+ (aq) + CO2 (aq)
3. Fe2+ (aq) + MnO4- (aq)  Fe3+(aq) + Mn2+ (aq)
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(basic)
(acidic)
(acidic)
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) 
Mn (+7)
Mn2+ (pink)
Mn(+2)
Rxn:
Fe+2 + MnO4-  Fe+3 + Mn2+
Bal. rxn:
Note
5 Fe2+ + MnO4- + 8 H+  5 Fe3+ + Mn2+ + 4 H2O
Fe2+  5 Fe3+ : Oxidized Lose e- : Reducing Agent
Mol of MnO4- = 45.30 ml • 0.180(mol/L) = 0.8154 mmol MnO4Amt of Fe:
= 0.8154 mmol • 5 mol Fe+2 • 55.8 g
=
1 mol MnO4- 1 mol Fe2+
% Fe = (0.2275 g / 1.225 g) • 100 = 18.6 %
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0.2275 g
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) ?
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