Oxidation- Reduction Reaction

1
Oxidation- Reduction Reactionredox reaction

• Part 1

2
Objectives for the Unit

• Identify elements that change oxidation state
  from one side of a chemical equation to the
  other.
• Identify reducing agents and oxidizing agents in
  a chemical equation.
• Use the half-reaction method to balance redox
  reactions.

3
Oxidation

• Old definition
• combination of oxygen with other substances.
• Oxidation as defined today
• The loss of electrons from an atom or ion.
• e- goes bye-bye

4
Reduction

• Old definition
• reducing" a substance into its components.
• Reduction as defined today
• The gaining of electrons by an atom or ion.

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Reduction-Oxidation ReactionRedox reaction

• A reduction and oxidation reaction is commonly
  called redox reaction for short
• Oxidation always accompanies reduction
• One atom must lose electrons and another must
  pick up the electrons.
• The number of electrons lost (oxidation) must be
  equal to the number of electrons gained
  (reduction)

6
Concept check(answer these questions on you
paper using Cornell note format. Answer the
question first before advancing the slide)

• Oxidation means gaining/ losing electrons
• Answer losing
• Reduction means gaining/losing electrons
• Answer gaining
• Can there be a oxidation without reduction?
  Explain.
• Answer no, one atom must give up electrons
  (oxidation) and another must be there to receive
  or gain (reduction) electrons
• Compare a reducing agent with an oxidation agent.
• Answer reducing agents are atoms that provide
  the electrons (started out with electrons) and
  when they lose the electrons, then they are
  oxidized. Oxidation agents are atoms that will
  accept the electrons, (started out with less
  electrons) and when they gain electrons, they
  will be reduced.

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Tip

• LEO The Lion Goes GER
• Loose Electrons, Oxidize Gain
  Electrons, Reduce

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How do you know thisis a "Redox" equation?
0
0
1 -2

• 2 H2 O2 ? 2 H2O

What is the charge of H2 here?
What is the charge of O2 here?
What is the charge of H2 and O here?
Think back, how does hydrogen bond with oxygen
and why 2 hydrogen for every oxygen?
This is a redox reaction because atoms are being
reduced (oxygen gained two electrons to go from a
charge of 0 to -2) and oxidized (hydrogen loses
one electron to go from a charge of 0 to 1)
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How to identify a redox reaction

• Oxidation and Reduction must both occur in a
  Redox reaction. If one particle gains electrons
  in a reaction, some other particle must lose
  them.
• So?
• Look for the change in oxidation number (charges).

10
How to assign oxidation numbers.

• You have learned to read the oxidation numbers of
  many elements from the periodic table, ie group
  1 elements is 1, group 2 elements are 2, group
  13 elements are 3, group 15 elements are -3,
  group 16 elements are -2 and group 17 elements
  are -1. While that information is important, the
  following rules are to be your guide when working
  with Redox equations.

• Rules for assigning oxidation numbers
• The oxidation number of a free element 0.
• The oxidation number of a monatomic ion the
  charge on the ion.
• The oxidation number of hydrogen 1 and rarely
  - 1.
• The oxidation number of oxygen - 2 and in
  peroxides (H2O2) - 1.
• The sum of the oxidation numbers in a polyatomic
  ion (ions made up of two or more elements)
  charge on the ion.
• Elements in group 1, 2, and aluminum are always
  as indicated on the periodic table.

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The oxidation number of elements not covered by
the rules must be "calculated" using the known
oxidation numbers in a compound.

• Example 1 K2CO3
• To calculate C
• By rule K is 1 and O is -2
• The sum of all the oxidation numbers in this
  formula equal 0 because there is no charge on
  this polyatomic ion. Multiply the subscript by
  the oxidation number for each element.
• K    (2) ( 1 )    2
• O    (3) ( - 2 )    - 6
• therefore, C    (1) ( 4 )    4
• Example 2 HSO4-
• To calculate S
• by rule, H is 1 by rule O is - 2
• The sum of all the oxidation numbers in this
  formula equal -1 because this polyatomic ion has
  a charge of -1. Multiply the subscript by the
  oxidation number for each element.
• H    (1) ( 1 )    1
• O    (4) ( - 2 )    - 8
• therefore, S    (1) ( 6 )    6

12
Practice Problems Use the rules above to
determine the oxidation number of the element
indicated in each formula.

• Sb in Sb2O5
• N in Al(NO3)3
• P in Mg3(PO4)2
• S in (NH4)2SO4
• Cr in CrO4-2
• Cl in ClO4-
• B in NaBO3
• Si in MgSiF6
• I in IO3-
• N in (NH4)2S
• Mn in MnO4 -
• Br in BrO3 -
• Cl in ClO
• Cr in Cr2O7 -2
• Se in H2SeO3

13
Oxidation- Reduction Reactionredox reaction

• Part 2
• Reducing Agents and Oxidizing Agents

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Reducing Agents

• the reactant that gives up electrons.
• The reducing agent contains the element that is
  oxidized (looses electrons).
• If a substance gives up electrons easily, it is
  said to be a strong reducing agent.

15
Oxidizing and Reducing Agents

• Reducing agent atoms that provide electrons to
  reduce the other atom

16
Oxidizing agents

• the reactant that gains electrons.
• The oxidizing agent contains the element that is
  reduced (gains electrons).
• If a substance gains electrons easily, it is said
  to be a strong oxidizing agent.

17
Oxidizing and Reducing Agents

• Oxidizing agent atoms that will gain electrons
  thus oxidizing the other atom.

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• Example
• Fe2O3 (s) 3CO (g) ? 2Fe(s) 3CO2 (g)
• Notice that the oxidation number of C goes from
  2 on the left to 4 on the right.
• The reducing agent is CO, because it contains C,
  which loses e -.
• Notice that the oxidation number of Fe goes from
  3 on the left to 0 on the right.
• The oxidizing agent is Fe2O3, because it contains
  the Fe, which gains e -.

3 -2 2 -2
0 4 -2
19
Charting Reducing Agents and Oxidizing Agents

• Practice Problems
• In any Redox equation, at least one particle
  will gain electrons and at least one particle
  will lose electrons. This is indicated by a
  change in the particle's oxidation number from
  one side of the equation to the other. For each
  reaction below, draw arrows and show electron
  numbers as in the example here. The top arrow
  indicates the element that gains electrons,
  reduction, and the bottom arrow indicates the
  element that looses electrons, oxidation. An
  arrow shows what one atom of each of these
  elements gains or looses.

This is the first thing that must be done in
balancing a Redox reaction. Learn to do it well.
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Charting Reducing Agents and Oxidizing Agents
Practice Problems

1. Mg O2 ? MgO
2. Cl2 I- ? Cl- I2
3. MnO4 - C2O4 -2 ? Mn2 CO2
4. Cr NO2 - ? CrO2 - N2O2 -2
5. BrO3 - MnO2 ? Br - MnO4 -
6. Fe2 MnO4 - ? Mn2 Fe3
7. Cr Sn4 ? Cr3 Sn2
8. NO3 - S ? NO2 H2SO4
9. IO4- I- ? I2
10. NO2 ClO - NO3 - Cl -

21
Oxidation- Reduction Reactionredox reaction

• Part 3
• Balancing Redox Equations by the Half-reaction
  Method

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Balancing Redox Equations by the Half-reaction
Method

• Decide what is reduced (oxidizing agent) and what
  is oxidized (reducing agent). Do this by drawing
  arrows as in the practice problems.
• 2. Write the reduction half-reaction.
• The top arrow in step 1 indicates the reduction
  half-reaction. Show the electrons gained on the
  reactant side.
• Balance with respect to atoms / ions.
• To balance oxygen, add H2O to the side with the
  least amount of oxygen.
• THEN add H to the other side to balance
  hydrogen.
• Remember that the arrow in step 1 indicates the
  number of electrons gained by one atom.
• 3. Write the oxidation half-reaction.
• The bottom arrow in step 1 indicates the
  oxidation half-reaction.
• Show the electrons lost on the product side.
• Balance with respect to atoms / ions.
• To balance oxygen, add H2O to the side with the
  least amount of oxygen.
• THEN add H to the other side to balance
  hydrogen.
• Remember that the arrow in step 1 indicates the
  number of electrons lost by one atom.

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Balancing Redox Equations by the Half-reaction
Method Continue

• 4. The number of electrons gained must equal the
  number of electrons lost.
• Find the least common multiple of the electrons
  gained and lost.
• In each half-reaction, multiply the electron
  coefficient by a number to reach the common
  multiple.
• Multiply all of the coefficients in the
  half-reaction by this same number.
• 5. Add the two half-reactions.
• Write one equation with all the reactants from
  the half-reactions on the left and all the
  products on the right.
• The order in which you write the particles in the
  combined equation does not matter.
• 6. Simplify the equation.
• A) Cancel things that are found on both sides of
  the equation as you did in net ionic equations.
• B) Rewrite the final balanced equation.
• Check to see that electrons, elements, and total
  charge are balanced.
• There should be no electrons in the equation at
  this time.
• The number of each element should be the same on
  both sides.
• It doesn't matter what the charge is as long as
  it is the same on both sides.
• If any of these are not balanced, the equation is
  incorrect. The only thing to do is go back to
  step 1 and begin looking for your mistake.

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Practice Problems

• Identify the oxidizing agent and reducing agent
  in each equation Answers
• H2SO4 8HI ? H2S 4I2 4H2O
• Au2S3 3H2 ? 2Au 3H2S
• Zn 2HCl ? H2 ZnCl2
• To make working with redox equations easier, we
  will omit all physical state symbols. However,
  remember that they should be there. An
  unbalanced redox equation looks like this
• MnO4- H2SO3 H   ?  Mn2 HSO4- H2O
• Study how this equation is balanced using the
  half-reaction method on the next slide. It is
  important that you understand what happens in
  each step.

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• MnO4- H2SO3 H   ?  Mn2 HSO4- H2O

7 to 3 gain 5 e-

7 -2 1 4 -2 1
2 1 7 -2
1 -2
Oxidizing agent
Reducing agent
4 to 7 lose 3 e-
Reducing half reaction Step 2 MnO4- 5e- ?
Mn2
Reducing half reaction Step 5 3MnO4- 5H2SO3-
15 e- ? 3Mn2 5HSO4- 15e-
Oxidation half reaction Step 3 H2SO3 ? HSO4-
3e-
Reducing half reaction Step 5a 3MnO4- 5H2SO3-
15 e- ? 3Mn2 5HSO4- 15e-
Reducing half reaction Step 5b 3MnO4- 5H2SO3-
? 3Mn2 5HSO4-
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• 3MnO4- 5H2SO3 H   ?  3Mn2 5HSO4- H2O

Notice, some of the above is balance but the H
and O are not. Balance the equation the way that
you know, by counting the number and kind of
atoms.

• 3MnO4- 5H2SO3 9H   ?  3Mn2 5HSO4- 7H2O

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Practice Problems Balance these Redox equations
using the half-reaction method.

• The balanced equation and the detailed
  half-reaction solution are provided for each
  equation. Use these helps only after doing your
  best to balance the equation without help.
• It is to your advantage to work all these
  problems to gain experience with different
  situations that arise when working with redox
  equations.
• HNO3 H3PO3   ?   NO H3PO4 H2O
• Cr2O7-2 H I -   ?   Cr3 I2 H2O
• As2O3 H NO3- H2O ?   H3AsO4 NO
• CuS NO3- ?   Cu2 NO2 S
• H2SeO3 Br -   ?   Se Br2
• Fe2 Cr2O7-2 ? Fe3 Cr3
• HS - IO3-   ?  I - S
• CrO4-2 I -    ? Cr3 I2
• IO4- I -   ?   I2
• MnO4- H2O2 ? Mn2 O2
• H3AsO4 Zn   ?   AsH3 Zn2

Click below to see answer
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oxidation number answers

1. Sb 5
2. N 5
3. P 5
4. S 6
5. Cr 6
6. Cl 7
7. B 5
8. Si 4
9. I 5
10. N 3
11. Mn 7
12. Br 5
13. Cl 1
14. Cr 6
15. Se 4

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Balanced Redox Equations Answers

• 2HNO3 3H3PO3 ? 2NO 3H3PO4 H2O
• 14H Cr2O7-2 6I - ? 2Cr3 3I2 7H2O
• 3As2O3 4H 4NO3- 7H2O ? 6H3AsO4 4NO
• CuS 2NO3- 4H ? Cu2 2NO2 S 2H2O
• H2SeO3 4Br - 4H ? Se 2Br2 3H2O
• 6Fe2 Cr2O7-2 14H ? 6Fe3 2Cr3 7H2O
• 3HS - IO3- 3H ? I - 3S 3H2O
• 16H 2CrO4-2 6I - ? 2Cr3 3I2 8H2O
• 8H IO4- 7I - ? 4I2 4H2O
• 6H 2MnO4- 5H2O2 ? 2Mn2 5O2 8H2O
• 8H H3AsO4 4Zn ? AsH3 4Zn2 4H2O

Click below to go back to problems.
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Charting Reducing Agents and Oxidizing Agents
Practice Problems
0 to -2, lose 2 e-
0 0 2
-2

1. Mg O2 ? MgO
2. Cl2 I- ? Cl- I2
3. MnO4 - C2O4 -2 ? Mn2 CO2

0 to -2, gain 2 e-
-1 to 0, lose 1 e-
0 -1 -1 0
0 to -1, gain 1 e-
7 to 2, gain 5 e-
7 -2 3 -2
2 4 -2
3 to 4, lose 1 e-
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Charting Reducing Agents and Oxidizing Agents
Practice Problems

1. Mg O2 ? MgO
2. Cl2 I- ? Cl- I2
3. MnO4 - C2O4 -2 ? Mn2 CO2
4. Cr NO2 - ? CrO2 - N2O2 -2
5. BrO3 - MnO2 ? Br - MnO4 -
6. Fe2 MnO4 - ? Mn2 Fe3
7. Cr Sn4 ? Cr3 Sn2
8. NO3 - S ? NO2 H2SO4
9. IO4- I- ? I2
10. NO2 ClO - NO3 - Cl -

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Charting Reducing Agents and Oxidizing Agents
Practice Problems

1. Mg O2 ? MgO
2. Cl2 I- ? Cl- I2
3. MnO4 - C2O4 -2 ? Mn2 CO2
4. Cr NO2 - ? CrO2 - N2O2 -2
5. BrO3 - MnO2 ? Br - MnO4 -
6. Fe2 MnO4 - ? Mn2 Fe3
7. Cr Sn4 ? Cr3 Sn2
8. NO3 - S ? NO2 H2SO4
9. IO4- I- ? I2
10. NO2 ClO - NO3 - Cl -

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Charting Reducing Agents and Oxidizing Agents
Practice Problems

1. Mg O2 ? MgO
2. Cl2 I- ? Cl- I2
3. MnO4 - C2O4 -2 ? Mn2 CO2
4. Cr NO2 - ? CrO2 - N2O2 -2
5. BrO3 - MnO2 ? Br - MnO4 -
6. Fe2 MnO4 - ? Mn2 Fe3
7. Cr Sn4 ? Cr3 Sn2
8. NO3 - S ? NO2 H2SO4
9. IO4- I- ? I2
10. NO2 ClO - NO3 - Cl -

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Charting Reducing Agents and Oxidizing Agents
Practice Problems

1. Mg O2 ? MgO
2. Cl2 I- ? Cl- I2
3. MnO4 - C2O4 -2 ? Mn2 CO2
4. Cr NO2 - ? CrO2 - N2O2 -2
5. BrO3 - MnO2 ? Br - MnO4 -
6. Fe2 MnO4 - ? Mn2 Fe3
7. Cr Sn4 ? Cr3 Sn2
8. NO3 - S ? NO2 H2SO4
9. IO4- I- ? I2
10. NO2 ClO - NO3 - Cl -

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Part 3 Redox Agents Answers

1. ox H2SO4 red HI
2. ox Au2S3 red H2
3. ox HCl red Zn