Chapter 20 OxidationReduction Reactions Redox Reactions

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Chapter 20 Oxidation-Reduction Reactions
(Redox Reactions)
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Launch Lab

• Complete the Penny Chemistry Lab
• ¼ cup 57 mL
• 1 tsp 5 mL

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The chemical changes that occur when electrons
are transferred between reactants are called
oxidation reduction reactions
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oxidation reactions
-    -principal source of energy on earth -   
-combustion of gasoline -    -burning of
wood -burning food in your body
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Oxidation reactions are always accompanied by a
reduction reaction Oxidation -    originally
meant combining with oxygen -    iron rusting
(iron oxygen)   Reduction -    originally
meant the loss of oxygen from a compound removing
iron from iron ore ( iron II oxide)
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20.2 Electron Transfer in Redox Reactions  
Today OXIDATION means -    a complete or
partial LOSS of ELECTRONS   REDUCTION means -   
a complete or partial GAIN of ELECTRONS   Memory
Device LEO the lion says GER or OIL RIG
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The substance that donates electrons in a redox
reaction is the REDUCING AGENT  
The substance that takes electrons in a redox
reaction is the OXIDIZING AGENT
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• Reduction is
• the gain of electrons
• a decrease in oxidation state
• the loss of oxygen
• the addition of hydrogen
• MgO H2 Mg H2O
• notice the Mg2 in MgO is gaining electrons

• Oxidation is
• the loss of electrons
• an increase in oxidation state
• the addition of oxygen
• the loss of hydrogen
• 2 Mg O2 2 MgO
• notice the magnesium is losing electrons

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20.3 Assigning Oxidation Numbers (ON)
Oxidation States
Oxidation states are numbers assigned to atoms
that reflect the net charge an atom would have if
the electrons in the chemical bonds involving
that atom were assigned to the more
electronegative atoms. Oxidation states can be
thought of as imaginary charges. They are
assigned according to the following set of rules
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1 The ON of a simple ion is equal to its
ionic charge 1 2
-3 Na Cu 2 N3-
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2 The ON of hydrogen is always 1, except in
metal hydrides like NaH where it is 1
1 -1 HCl
NaH
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3 The ON of oxygen is always 2 except in
peroxides like X2O2 where it is 1
-2 -1 H2O H2O2  
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4 The ON of an uncombined element is always
zero 0 0 0 Na
Cu N2
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5 For any neutral(zero charge) compound, the
sum of the ONs is always zero
4-2 CO2
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6 For a complex ion, the sum of the ONs
equals the charge of the complex ion
7 -2 MnO41-
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Examples - assigning oxidation numbers
Assign oxidation states to all elements
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Assignment 1-OBWS 1-5 2-Text
1-3,(pp481-86) 8-15(pp498) Chp 20 3-Worksheet
2 Oxidation Numbers
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20.4 Oxidation Changes
an increase in oxidation number of an atom
signifies oxidation
2 to 4
a decrease in oxidation number of an atom
signifies reduction
0 to -1
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Identifying Redox Reactions
Oxidation and reduction always occur together in
a chemical reaction. For this reason, these
reactions are called redox reactions.
Although there are different ways of identifying
a redox reaction, the best is to look for a
change in oxidation state
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2 LEO
OA
-1
2
-1
4
-1
4
-1
2
SnCl2 PbCl4 SnCl4
PbCl2
RA
-2 GER
-3 GER
RA
-2
2
-2
2
1
-2
5
2
0
-2
1
CuS H NO3- Cu2
S NO H2O
OA
2 LEO
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Examples - labeling redox reactions

• In each reaction, look for changes in oxidation
  state.
• If changes occur, identify the substance being
  reduced, and the substance being oxidized.
• Identify the oxidizing agent and the reducing
  agent.

1 (H is oxidized)
(reducing agent)
0
-2
2
-2
1
0
H2 CuO Cu H2O
-2 (Cu is reduced)
(oxidizing agent)
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Try These!!
1 Fe 2 is oxidized (reducing agent)
5 Fe2 MnO4- 8 H 5 Fe3 Mn2
4 H2O Zn 2 HCl ZnCl2 H2
- 5 Mn 7 is reduced (oxidizing agent)
2 Zn 0 is oxidized (reducing agent)
- 1 H 1 is reduced (oxidizing agent)
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How to write net ionic equations

• 1) write a balanced equation
• Cu(s) 2NaCl(aq) ? 2Na(s)
  CuCl2 (aq)
• 2) Ionize any aqueous substances
• Cu(s) 2Na1(aq) 2Cl1-(aq) ? 2Na(s)
  Cu2 (aq) 2Cl 1- (aq)
• 3) Remove any like substances (spectators)
• Cu(s) 2Na1(aq) 2Cl1-(aq) ? 2Na(s)
  Cu2 (aq) 2Cl 1- (aq)
• 4) Sum up whats left
• Cu(s) 2Na1(aq) ? 2Na(s)
  Cu2 (aq)

The Net Ionic Equation (the reaction that is
really occurring)
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Table 12.1 Strength of oxidizing and reducing
agents Inquiry into Chemistry Chapter 12
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Oxidation Reduction Table 12.2 Inquiry into
Chemistry
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Spontaneous Reaction
Compare Reducing Agents
Loses 2 e -
Pt (s)
Sn (s)
Sn 2 (aq)
Pt 2 (aq)


?
Gains 2 e-
Compare Oxidizing Agents
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Non Spontaneous Reaction
Compare Reducing Agents
Loses 2 e -
Mg (s)
Fe2 (aq)
Mg 2 (aq)
Fe (s)

?

Gains 2 e-
Compare Oxidizing Agents
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Assignment 1-OBWS 6,7 2-Text 4, 16, 17 Chp
20 3-Worksheet 3 Oxidizing Agents and Reducing
Agents 4- Investigation 12.A Testing Relative
Oxidizing and Reducing Strengths of Metal Atoms
and Ions (see table 12.1) 5-Question Sheet for
Table 12.2 6-Go back and answer part 4 of each Q
on Worksheet 3
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20.5 Balancing Redox Equations
There are two methods used to balance redox
reactions
1)the oxidation number change method
2)the half reaction method
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These methods are based on the fact that the
total number of electrons gained in reduction
must equal the total number of electrons lost in
oxidation
Redox reactions are often quite complicated and
difficult to balance. For this reason, youll
learn a step-by-step method for balancing these
types of reactions, when they occur in acidic or
in basic solutions.
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Oxidation Number Change Method
Balance the following Fe2O3 CO Fe
CO2
1)Assign ON to all atoms
0
3
-2
-2
2
-2
4
Fe2O3 CO Fe CO2
2)Identify which atoms are oxidized and which are
reduced
-3 (Fe reduced)
0
3
-2
-2
2
-2
4
Fe2O3 CO Fe CO2
2 (C oxidized)
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3) Make the total increase in oxidation number
equal the total decrease in oxidation number by
using appropriate coefficients on the reactant
side only.
-3
(x 2 atoms) 6 electrons gained
0
3
-2
-2
2
-2
4
Fe2O3 CO Fe CO2
3
2
(X 3 atoms) 6 electrons lost
4) Finally check to be sure that the equation is
balanced both for atoms and charge.
Fe2O3 CO Fe CO2
3
2
3
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Assignment 1-OBWS 8 2-Text 5,
18,19 3-Practice Sheet 20A 4-Investigation 12.B
Redox Reactions and Balanced Equations
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Balancing Equations with the Half-Reaction Method
1) First split the original equation into two
half-reactions, one reduction and the other
oxidation. In each half-reaction, follow these
steps 2) Balance all elements except H and
O. 3) Balance the Os by adding water,
H2O. 4) Balance the Hs by adding hydrogen
ions, H. If your rxn is taking place in an
acidic solution, skip to step 8 If your rxn is
taking place in a basic solution proceed to step
5 5) Adjust for basic conditions by adding to
both sides the same of OH- ions as the number
of H ions already present 6) Simplify the
equation by combining H and OH- that appear on
the same side of the equation into water
molecules. 7) Cancel any water molecules present
on both sides of the equation 8) Balance the
charges by adding electrons 9) Recombine the ½
reactions into a complete balanced equation.
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Example
Fe2 Cr2O72- Fe3 Cr3 acidic solution
Fe2 Fe3
6( )
1e-

Cr2O72- Cr3
2
7 H2O

14 H

1(
)
6 e-

Cr2O72- 6 Fe2 14 H 2 Cr3 6 Fe3 7 H2O
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What if the solution was basic?
Notice that the method has assumed the solution
was acidic - we added H to balance the equation.
The H in a basic solution is very small. The
OH- is much greater. For this reason, we will
add enough OH- ions to both sides of the equation
to neutralize the H added in the reaction. The
hydrogen and hydroxide ions will combine to make
water, and you may have to do some canceling
before youre done. Cr2O72-
Fe2 H2O Cr3 Fe3
Try this in a basic solution!!!
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Cr2O72- Fe2 H2O Cr3 Fe3 Basic Solution
(
)
6
1e-
14OH-
2
(

)
1

6 e-

14 H2O
14OH-
7 H2O
14H


Cr2O72- 6 Fe2 7 H2O 2 Cr3 6 Fe3 14
OH-
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Balancing Redox Equations Practice

• Balance in acidic solution
• H2C2O4 MnO4- Mn2 CO2

5 H2C2O4 2 MnO4- 6 H 2 Mn2 10 CO2 8
H2O

• Balance in basic solution
• CN- MnO4- CNO- MnO2

3 CN- 2 MnO4- H2O 3 CNO- 2 MnO2 2 OH-
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Assignment 1-OBWS 9, 10, 11 2-Worksheet 4
Half Reactions 3-Practice Sheet 20B
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Redox Reactions - Whats Happening?

• Zinc is added to a blue solution of copper(II)
  sulfate
• The blue colour disappearsthe zinc metal
  dissolves, and solid copper metal precipitates
  on the zinc strip
• The zinc is oxidized (loses electrons)
• The copper ions are reduced (gain electrons)

Zn (s) CuSO4 (aq) ? ZnSO4 (aq) Cu (s)
Zn (s) Cu2 (aq) ? Zn2 (aq) Cu (s)
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Copper ions (Cu2) collide with the zinc metal
surface
A zinc atom (Zn) gives up two of its electrons
to the copper ion
The result is a neutral atom of Cu deposited on
the zinc strip, and a Zn2 ion released into
the solution
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Voltaic Cell
SIMULATIONS
Electrolysis
Redox Titration
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Assignment 1-Web Quest Oxidation/Reduction
2-Blue Print Lab 3-Review Worksheet