Bell Work

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Bell Work

• Draw your ph versus amount of base used graph on
  the white-boards.

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Redox Reactions

• Chapter 4

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Redox reaction

• A redox reaction (oxidation-reduction reaction)
  is a common type of reaction in aqueous solution
  which involves a transfer of electrons between
  two species
• One species loses (donates) electrons and is said
  to be oxidized.
  
• The other species gains (receives) electrons and
  is reduced.

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Redox reactions

• The ion or molecule that accepts electrons is
  called the oxidizing agent.
  
• The species that donates electrons is called the
  reducing agent.
  
• Therefore the substance being oxidized is a good
  reducing agent and the substance being reduced is
  a good oxidizing agent.

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LEO the lion says GER

• LEO Loss of Electrons is Oxidized
• Zn (s) ? Zn2 (aq) 2e-
• GER Gain of Electrons is Reduced
• 2H(aq) 2e- ?H2 (g)

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Oxidation Half-Reaction

• There is a lost of electrons
• Electrons appear has a product
• Na ? Na e-
• Fe2 ? Fe3 e-
• The charges are equal on both sides of reaction.

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Reduction Half-Reactions

• There is a gain of electrons
• Electrons appear as a reactant
• F2 2 e- ? 2 F-
• Cr3 e- ? Cr2
• Fe2 2e- ? Fe
• Again same charge on both sides

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Oxidation and reduction

• Occur together in the same reaction
• You cant have one without the other
• There is not net change in the number of
  electrons in a redox reaction
  
• Those given off in the oxidation half-reaction
  are taken on by another species in the reduction
  half-reaction
  
• Oxidation is the lost of electrons from an atom -
  more positive
  
• Often gains oxygen atoms
• Reduction is the gain of electrons - more
  negative
  
• Often loses oxygen atoms
• Oxidizing agent
• Ion or molecule that accepts the electrons
• Reducing agent
• The species that donates the electrons

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Redox Reactions

• Oxidation is raising of oxidation number
• Reduction is lowering of oxidation number
• The reducing agent is being oxidized
• The oxidizing agent is being reduced
• 2 Cl2 2 H2O ? 4 HCl O2
• 0 1 2- 1 1- 0
• Chlorine is reduced so oxidizing agent
• Oxygen is oxidized so reducing agent

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Identify the Agents

• MnO4- NO2- H ? Mn2 NO3- H2O
• Mn is reduced so oxidizing agent
• N is oxidized so reducing agent
• Se NO3- ? SeO2 NO
• Se is oxidized so reducing agent
• N is reduced so oxidizing agent

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Oxidation Numbers

• The apparent charge of an atom
• Elements 0
• Hydrogen 1 Hydride 1-
• Oxide 2- Peroxide 1-
• Group 1 2 1 and 2 respectively
• Sum in compound Zero
• Sum in polyatomic ion charge of ion

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Examples

• Sulfite - SO32-
• Oxygen is 2- Total is 6-
• Charge is 2- therefore S is 4
• Nitrite - NO21-
• Oxygen is 2- Total is 4-
• Charge is 1- therefore N is 3
• Vinegar - HC2H3O2
• H is 1 total is 4
• O is 2- total is 4-
• Compound is Zero so C 0

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Examples

• 1. What is the oxidation number for Cl in HClO4?
  
  
• 2. What is the oxidation number for Cr in
  Cr2O72-?
  
• 3. Ca (s) 2HNO3 (aq) ? Ca(NO3)2 (aq) H2 (g)
• a. Which atom is oxidized (include oxidation )?
  
  
• b. Which atom is reduced (include oxidation )?
• c. The atom that is oxidized goes from oxidation
  number ___ to __.
  
• d. The atom that is reduced goes from oxidation
  number __ to ___.
  
• e. Identify the oxidizing agent
• f. Identify the reducing agent

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The Art of Balancing Redox Equations

• Rules for balancing redox reactions using half
  reaction method
  
• Acidic solution
• 1) Separate the given reaction into 2 half
  reactions.
  
• 2) Balance all atoms except O, H.
• 3) Balance O atoms by adding H2O.s.
• 4) Balance H atoms by adding H ions.
• 5) Balance the charge by adding electrons (e-) to
  the more positive side.
  
• 6) Multiply each reaction by an integer so that
  the electrons gained electrons lost.
  
• 7) Combine the half reactions and simplify by
  canceling common species.
  
• Basic solution
• 1) Do steps 1-7 above.
• 2) Note the of H ions. Add this number of OH-
  ions to both sides.
  
• 3) Combine OH- and H to form H2O and simplify
  the reaction.
  
• If the equation is correctly balanced, the
  number of each type of atom and the net charge
  must be equal for both sides.
  

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Example

• Fe 2 (aq) MnO4- (aq) ? Fe 3 (aq) Mn 2 (aq)
  (acidic solution)
  
• 1) Separate the given reaction into 2 half
  reactions
  
• Fe 2 (aq) ? Fe 3 (aq) losing electrons
  being oxidized (reducing agent)
  
• MnO4- (aq) ? Mn 2 (aq) gaining electrons being
  reduced (oxidizing agent)
  
• Write it according to what is happening and
• 2) Balance all atoms except O, H.
• Fe 2 (aq) ? Fe 3 (aq)
• MnO4- (aq) ? Mn 2

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• 3) Balance O atoms by adding H2O.s.
• Fe 2 (aq) ? Fe 3 (aq)
• MnO4- (aq) ? Mn 2 4 H2O
• 4) Balance H atoms by adding H ions.
• Fe 2 (aq) ? Fe 3 (aq)
• MnO4- (aq) 8 H? Mn 2 4 H2O
• 5) Balance the charge by adding electrons (e-) to
  the more positive side.
  
• Fe 2 (aq) ? Fe 3 (aq) e-
• MnO4- (aq) 5e- 8 H? Mn 2 4 H2O

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• 6) Multiply each reaction by an integer so that
  the electrons gained electrons lost.
  
• (Fe 2 (aq) ? Fe 3 (aq) e-)5
• (MnO4- (aq) 5e- 8 H? Mn 2 4 H2O)1
• 7) Combine the half reactions and simplify by
  canceling common species.
  
• 5Fe 2 (aq) ? 5Fe 3 (aq)
  5e-
  
• MnO4- (aq) 5e- 8 H? Mn 2 4 H2O
• --------------------------------------------------
  ----------------
  
• Final answer
• 5Fe 2 (aq) MnO4- (aq) 8 H (aq) ? 5Fe 3
  (aq) Mn 2 4 H2O
  

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Bell Work

• What is the oxidation numbers of the elements in
  the following molecules
  
• H2SO4
• SO3-2

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You try an example

• Cl - (aq) MnO4- (aq) ? Cl2 (g) Mn 2 (aq)
  (acidic solution)
  

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(4H2O Mn 2 ? MnO4 8H 5 e - )2(BiO3 -
2 e - 6 H ? Bi 3 3H2O )58H2O 2Mn 2 ?
2MnO4 16H 10 e - 5BiO3 - 10 e - 30 H
? 5 Bi 3 15H2O 2Mn 2 5BiO3 - 14 H ?
2MnO4 ? 5 Bi 3 7H2O
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Basic example

• H2O2 (aq) Cl2O7 (aq) ? ClO 2 - (aq) O2(g)
• Do step 1-7 as before
• H2O2 (aq) ? O2(g) - oxidized
• Cl2O7 (aq) ? ClO 2 - (aq) - reduced
• H2O2 (aq) ? O2(g) 2H
• Cl2O7 (aq) ? 2ClO2 - (aq) 3 H2O
• H2O2 (aq) ? O2(g) 2H
• Cl2O7 (aq) 6 H? 2ClO2 - (aq) 3 H2O

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• H2O2 (aq) ? O2(g) 2H 2 e -
• Cl2O7 (aq) 6 H 8e -? 2ClO2 - (aq) 3
  H2O
  
• (H2O2 (aq) ? O2(g) 2H 2 e - )4
• Cl2O7 (aq) 6 H 8e -? 2ClO2 - (aq) 3
  H2O
  
• 
  2
  
• 4H2O2 (aq) ? 4O2(g) 8H
  8 e -
  
• Cl2O7 (aq) 6 H 8e -? 2ClO2 - (aq) 3
  H2O
  
• 4H2O2 (aq) Cl2O7 (aq) ? 4O2(g) 2H 2ClO2
  - (aq) 3 H2O
  

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New step

• 2) Note the of H ions. Add this number of OH-
  ions to both sides.
  
• 2OH- 4H2O2 (aq) Cl2O7 (aq) ? 4O2(g)
  2H 2ClO2 - (aq) 3 H2O 2OH-
  
• 3) Combine OH- and H to form H2O and simplify
  the reaction.
  
• Simplify Cl2O7 4H2O2 2OH- ? 2ClO2 - 5 H2O
  4O2
  

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Practice
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Bell Work

• Briefly explain your procedures