1. Oxidation- Reduction Reaction “redox 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: Oxidation as defined today:
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: Reduction as defined today:
“reducing" a substance into its components.
Reduction as defined today:
The gaining of electrons by an atom or ion.

5. Reduction-Oxidation Reaction “Redox 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.

7. Loose Electrons, Oxidize Gain Electrons, Reduce
Tip
LEO The Lion Goes GER
Loose Electrons, Oxidize Gain Electrons, Reduce

8. How do you know this is a "Redox" equation?
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)

9. 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.

11. by rule, H is + 1 by rule O is - 2
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
answers

13. Oxidation- Reduction Reaction “redox reaction”
Part 2
Reducing Agents and Oxidizing Agents

14. 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.

18. Fe2O3 (s)+ 3CO (g)  2Fe(s) + 3CO2 (g)
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 -.

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.

20. Charting Reducing Agents and Oxidizing Agents Practice Problems
Mg + O2  MgO
Cl2 + I-  Cl-+ I2
MnO4 - + C2O  Mn+2 + CO2
Cr + NO2 -  CrO2 - + N2O2 -2
BrO3 - + MnO2  Br - + MnO4 -
Fe+2 + MnO4 -  Mn+2 + Fe+3
Cr + Sn+4  Cr+3 + Sn+2
NO3 - + S  NO2 + H2SO4
IO4- + I-  I2
NO2 + ClO - NO3 - + Cl -

21. Oxidation- Reduction Reaction “redox reaction”
Part 3
Balancing Redox Equations by the Half-reaction Method

22. 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.
Remember that the arrow in step #1 indicates the number of electrons lost by one atom.

23. 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.

24. MnO4- + H2SO3 + H +  Mn+2 + HSO4- + H2O
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 +     Mn+2 + 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.

25. MnO4- + H2SO3 + H+  Mn+2 + HSO4- + H2O
+7 to +3 gain 5 e-
Oxidizing agent
Reducing agent
+4 to +7 lose 3 e-
Reducing half reaction: Step 2
MnO e-  Mn+2
Reducing half reaction: Step 5
3MnO4- + 5H2SO e-  3Mn HSO e-
Oxidation half reaction: Step 3
H2SO3  HSO e-
Reducing half reaction: Step 5a
3MnO4- + 5H2SO e-  3Mn HSO e-
Reducing half reaction: Step 4
3 (MnO e-  Mn+2) = 3MnO e-  3Mn+2
Oxidation half reaction: Step 4
5 (H2SO3  HSO e-) = 5H2SO3  5HSO e-
Reducing half reaction: Step 5b
3MnO4- + 5H2SO3-  3Mn+2 + 5HSO4-

26. 3MnO4- + 5H2SO3 + H+  3Mn+2 + 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+     3Mn+2 + 5HSO4- + 7H2O

27. 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 -      Cr+3 + I2 + H2O
As2O3 + H + + NO3- + H2O    H3AsO4 + NO
CuS + NO3-    Cu+2 + NO2 + S
H2SeO3 + Br -      Se + Br2
Fe+2 + Cr2O7-2  Fe+3 + Cr+3
HS - + IO3-     I - + S
CrO4-2 + I -     Cr+3 + I2
IO4- + I -      I2
MnO4- + H2O2  Mn+2 + O2
H3AsO4 + Zn      AsH3 + Zn+2
Click below to see answer

28. oxidation number answers:
Sb + 5
N + 5
P + 5
S + 6
Cr + 6
Cl + 7
B + 5
Si + 4
I + 5
N – 3
Mn + 7
Br + 5
Cl + 1
Se + 4
Back

29. Balanced Redox Equations Answers:
2HNO3 + 3H3PO3  2NO + 3H3PO4 + H2O
14H+ + Cr2O I -  2Cr+3 + 3I2 + 7H2O
3As2O3 + 4H+ + 4NO3- + 7H2O  6H3AsO4 + 4NO
CuS + 2NO3- + 4H+  Cu+2 + 2NO2 + S + 2H2O
H2SeO3 + 4Br - + 4H+  Se + 2Br2 + 3H2O
6Fe+2 + Cr2O H+  6Fe+3 + 2Cr+3 + 7H2O
3HS - + IO3- + 3H+  I - + 3S + 3H2O
16H+ + 2CrO I -  2Cr+3 + 3I2 + 8H2O
8H+ + IO4- + 7I -  4I2 + 4H2O
6H+ + 2MnO4- + 5H2O2  2Mn+2 + 5O2 + 8H2O
8H+ + H3AsO4 + 4Zn  AsH3 + 4Zn+2 + 4H2O
Click below to go back to problems.

30. Charting Reducing Agents and Oxidizing Agents Practice Problems
0 to -2, lose 2 e-
Mg + O2  MgO
Cl2 + I-  Cl-+ I2
MnO4 - + C2O  Mn+2 + CO2
0 to -2, gain 2 e-
-1 to 0, lose 1 e-
0 to -1, gain 1 e-
+7 to +2, gain 5 e-
+3 to +4, lose 1 e-

31. Charting Reducing Agents and Oxidizing Agents Practice Problems
Mg + O2  MgO
Cl2 + I-  Cl-+ I2
MnO4 - + C2O  Mn+2 + CO2
Cr + NO2 -  CrO2 - + N2O2 -2
BrO3 - + MnO2  Br - + MnO4 -
Fe+2 + MnO4 -  Mn+2 + Fe+3
Cr + Sn+4  Cr+3 + Sn+2
NO3 - + S  NO2 + H2SO4
IO4- + I-  I2
NO2 + ClO - NO3 - + Cl -

32. Charting Reducing Agents and Oxidizing Agents Practice Problems
Mg + O2  MgO
Cl2 + I-  Cl-+ I2
MnO4 - + C2O  Mn+2 + CO2
Cr + NO2 -  CrO2 - + N2O2 -2
BrO3 - + MnO2  Br - + MnO4 -
Fe+2 + MnO4 -  Mn+2 + Fe+3
Cr + Sn+4  Cr+3 + Sn+2
NO3 - + S  NO2 + H2SO4
IO4- + I-  I2
NO2 + ClO - NO3 - + Cl -

33. Charting Reducing Agents and Oxidizing Agents Practice Problems
Mg + O2  MgO
Cl2 + I-  Cl-+ I2
MnO4 - + C2O  Mn+2 + CO2
Cr + NO2 -  CrO2 - + N2O2 -2
BrO3 - + MnO2  Br - + MnO4 -
Fe+2 + MnO4 -  Mn+2 + Fe+3
Cr + Sn+4  Cr+3 + Sn+2
NO3 - + S  NO2 + H2SO4
IO4- + I-  I2
NO2 + ClO - NO3 - + Cl -

34. Charting Reducing Agents and Oxidizing Agents Practice Problems
Mg + O2  MgO
Cl2 + I-  Cl-+ I2
MnO4 - + C2O  Mn+2 + CO2
Cr + NO2 -  CrO2 - + N2O2 -2
BrO3 - + MnO2  Br - + MnO4 -
Fe+2 + MnO4 -  Mn+2 + Fe+3
Cr + Sn+4  Cr+3 + Sn+2
NO3 - + S  NO2 + H2SO4
IO4- + I-  I2
NO2 + ClO - NO3 - + Cl -

35. Part 3 Redox Agents Answers
ox: H2SO4 red: HI
ox: Au2S3 red: H2
ox: HCl red: Zn