Electrochemistry Chemical reactions and Electricity

1
Electrochemistry Chemical 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?

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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

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Oxidation States

• Definition -
• Oxidation Process- (charge increase)
• Lose electron (oxidation)
• i.e., Fe ?? Fe3 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 cant occur without the other.
• Example 2 Ca (s) O2 ? 2CaO
• Which is undergoing oxidation ? Reduction?
• Oxidation Ca ? Ca2
• 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) 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
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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
  Os.
• 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 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- ?? 4 I2 8e-
• 8e- 8H S2O8-2 ? S2O42- 4H2O
• 8I- 8H S2O8-2 ?? 4 I2
  S2O42- 4H2O

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Redox Reactions - Ion electron method.Under
Basic conditions

• 1, 2. Procedure identical to that under acidic
  conditions
• Balance the half reaction separately except H
  Os.
• 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- ? 2Cr3 3O2
  7H2O
• add 8H2O ? 8 H 8 OH-
• 8 H 3H2O2 Cr2O72- ? 2Cr3 3O2
  7H2O
• 8H2O ? 8 H 8 OH-
• Net Rxn 3H2O2 Cr2O72 - H2O ? 2Cr3
  3O2 8 OH-

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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)

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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 Fe2 MnO4- ? Fe3 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 Fe2 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 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) ?