KTT 1113 Inorganic Chemistry I

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Chapter 6 Oxidation-Reduction
Reactions
KTT 111/3 Inorganic Chemistry I
Dr. Farook Adam
August 2005
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Chapter 6 Oxidation-Reduction Reactions

• Reactions that involve the transfer of electrons
  are called oxidation-reduction or redox reactions
  
• Oxidation is the loss of electrons by a reactant
• Reduction is the gain of electrons by a reactant

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• Oxidation and reduction always occur together
• The total number of electrons lost by one
  substance is the same as the total number of
  electrons gained by the other
• For a redox reaction to occur, something must
  accept the electrons that are lost by another
  substance
• The substance that accepts the electrons is
  called the oxidizing agent

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• The substance that lost the electrons is called
  the reducing agent
• Note that the oxidizing agent is reduced and the
  reducing agent is oxidized
• For example
• 2 Na Cl2 ? 2 NaCl
• Na is the reducing agent because it lost
  electrons and was oxidized
• Cl2 is the oxidizing agent because it gained
  electrons and was reduced

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• Oxidation numbers provide a way to keep track of
  electron transfers
• The oxidation number of any free element is zero.
• The oxidation number of any simple, monoatomic
  ion is equal to the charge on the ion.
• The sum of all oxidation numbers of the atoms in
  a molecule or polyatomic ion must equal the
  charge on the particle.
• In its compounds, fluorine has an oxidation
  number of 1.
• In its compounds, hydrogen has an oxidation
  number of 1.
• In its compounds, oxygen has an oxidation number
  of 2.

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• If there is a conflict between two rules apply
  the rule with the lower number and ignore the
  conflicting rule
• In binary ionic compounds with metals, the
  nonmetals have oxidation numbers equal to the
  charges on their anions
• Example What is the oxidation number of Fe in
  Fe2O3?
• ANALYSIS This binary compound is ionic. Apply
  rule 3 and 6

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• Fe 2x
• O 3(-2) -6
• 0 2x (-6) or x 3 ox. number of Fe
• Note that fractional values of oxidation numbers
  are allowed
• In terms of oxidation numbers
• Oxidation is an increase in oxidation number
• Reduction is a decrease in oxidation number
• This provides a simple way to follow redox
  reactions

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• Many redox reactions take place in aqueous
  solution
• A procedure called the ion-electron method
  provides a way to balance these equations
• The oxidation and reduction are divided into
  equations called half-reactions
• The half-reactions are balanced separately, then
  combined into the fully balanced net ionic
  equation

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• Both mass and charge must be balanced
• Charge is balanced by adding electrons to the
  side of the equation that is more positive or
  less negative
• Example Balance the following skeleton equation

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• Balancing redox reactions Acidic solution.
• The Ion-Electron Method in Acidic Solution
• Divide the equation into two half-reactions.
• Balance atoms other than H and O.
• Balance O by adding water.
• Balance H by adding hydrogen ion.
• Balance net charge by adding electrons.
• Make electron gain and loss equal add
  half-reactions.
• Cancel anything thats the same on both sides of
  the equation.

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• Example Balance the following equation-
• I- OCI- ? I3- C1-
  (asid)
• Oxidation I- ? I3-
• Step (2) 3I- ? I3-
• Steps (3) and (4) not required!!!
• Step (5) 3I- ? I3- 2e-
• Reduction OC1- ? C1-
• Step (3) OC1- ? C1- H2O

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• OC1- ? C1- H2O
• Step (5)
• 2H OC1- ? C1- H2O
• 1
  -1
• Step (6) 2H OC1- 2e- ? C1- H2O
  (2)
• 3I- ? I3- 2e-
  (1)
• (1) (2)
• 2H OC1- 3I- ? C1- I3- H2O
• The simple addition here eliminates the electrons.

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• For balancing equations in basic medium
• All the steps are the same except step 4
• To balance H in step (4) follow this procedure
• For each H atom that is lacking, add one
  molecule of H2O to the side that requires it. At
  the same time add one unit of OH- in the opposite
  side of the half-reaction.

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• Example Balance the following in basic
• MnO4- C2O42- ? MnO2 CO32-
• solution

First half reaction MnO4- ? MnO2

• MnO4- ? MnO2 2H2O
• 4 H2O MnO4- ? MnO2 2H2O 4 OH-
• 4 H2O MnO4- 3 e- ? MnO2 2H2O 4
  OH-
• (d) 2 H2O MnO4- 3 e- ? MnO2 4 OH-
  (1)

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Second half reaction
C2O42- ? CO32-

• C2O42- ? 2 CO32-
• 2 H2O C2O42- ? 2 CO32-
• 4 OH- 2 H2O C2O42- ? 2 CO32-
  4 H2O
• 4 OH- 2 H2O C2O42- ? 2 CO32-
  4 H2O 2 e-
• 4 OH- C2O42- ? 2 CO32- 2 H2O 2
  e- (2)

2 H2O MnO4- 3 e- ? MnO2 4 OH-
(1)
We eliminate electrons by multiplying (1) by 2
and (2) by 3.
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(2) X 3 12 OH- 3 C2O42- ? 6 CO32-
6 H2O 6 e- (3)
(1) X 2
4 H2O 2 MnO4- 6 e- ? 2 MnO2 8 OH-
(4)
(3) (4) to eliminate the electrons and common
species.
4 OH- 3 C2O42- 2 MnO4- ? 6 CO32-
2 H2O 2 MnO2
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• Metals more active than hydrogen (H2) dissolve in
  oxidizing acids

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• Some examples
• More active metals will displace a less active
  metal from its compound
• This often occurs in solution and is called a
  single replacement reaction

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Zinc is a more active metal than copper. Copper
ions (blue) collide with zinc metal (gray)
picking up electrons. Copper ions become copper
atoms (red-brown) and stick to the zinc surface.
Zinc ions (yellow) replace the copper ions in
solution.

• An activity series arranges metals according to
  their ease of oxidation
• They can be used to predict reactions

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• Activity Series for Some Metals and Hydrogen
• A given element will be displaced from its
  compounds by any element below it in the table

(See Table 6.2 for a more extensive list.)
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• Oxygen reacts with many substances
• The products depends, in part, on how much oxygen
  is available
• Combustion of hydrocarbons
• Organic compounds containing O also produce
  carbon dioxide and water

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• Organic compounds containing S produce sulfur
  dioxide
• Many metals corrode or tarnish when exposed to
  oxygen
• Most nonmetals react with oxygen directly
• Redox reactions are more complicated than most
  metathesis reactions

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• In general, it is not possible to balance a redox
  reaction by inspection
• This is especially true when acid or bases are
  involved in the reaction
• Once balanced, they can be used for
  stoichiometric calculations
• Redox titrations are common because they often
  involve dramatic color changes
• Mole-to-mole ratios are usually involved

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• Example A 0.3000 g sample of tin ore was
  dissolved in acid solution converting all the tin
  to tin(II). In a titration, 8.08 mL of 0.0500 M
  KMnO4 was required to oxidize the tin(II) to
  tin(IV). What was the percentage tin in the
  original sample?
• ANALYSIS This is a redox titration in acidic
  solution.
• SOLUTION
• Form skeleton equation and use the ion-electron
  method to produce a balanced equation

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Therefor, the percentage of Sn in the ore is
Percent Sn
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Assignment

• Page 246
• Level 1/2 problems
• No. 4, 8, 10 and 12
• And
• 6.7 6.60 6.101
• 6.21 6.82
• 6.44 6.83
• 6.46 6.88
• 6.50 6.96

Hand in this assignment on or before 12th August
2005
The End