1. Oxidation-Reduction
Dr. Ron Rusay
Spring 2008

2. Oxidation-Reduction Oxidation is the loss of electrons.
Reduction is the gain of electrons.
The reactions occur together. One does not occur without the other.
The terms are used relative to the change in the oxidation state or oxidation number of the reactant(s).

4. Oxidation Reduction Reactions

7. Oxidation State (Oxidation Number)

8. QUESTION

9. ANSWER 2) NO Section 4.9 Oxidation Reduction Reactions (p. 154)
Oxygen almost always has an oxidation state of
2 when part of a compound. The exception is
when it is part of a peroxide. For example,
hydrogen peroxide H
. Then it has an
oxidation state of 2 -
STUDENT CD: Understanding Concepts: Oxidation-Reduction Reactions – O 2 2 – 1 .

11. Zinc

13. QUESTION

14. ANSWER 3) 2 Section 4.9 Oxidation Reduction Reactio ns (p. 154)
If an element is found on the reactant’s side, this
is almost always a redox reaction, since an
element usually becomes part of a compound
during a chemical reaction. -

15. - 2 e- +2 e- Cu 2+  Cu (s) H2 (g)  2 H +
Number of electrons gained must equal the number of electrons lost.
- 2 e-
+2 e-
Use oxidation numbers to determine what is oxidized and what is reduced.
+2 e-
Cu 2+  Cu (s)
- 2 e-
H2 (g)  2 H +
Refer to Balancing
Oxidation-Reduction Reactions

16. QUESTION

17. ANSWER 2) the oxidizing agent. Section 4.9 Oxidation
Reduction Reactions
(p. 154)
Metals lose electrons, so they are oxidized,
making the other reactant an oxidizing agent. -
HMCLASS PREP: Figure 4.19

18. Balancing Redox Equations in acidic solutions
1) Determine the oxidation numbers of atoms in both reactants and products.
2) Identify and select out those which change oxidation number (“redox” atoms) into separate “half reactions”.
3) Balance the “redox” atoms and charges (electron gain and loss must equal!).
4) In acidic reactions balance oxygen with water then hydrogen from water with acid proton(s).

19. Balancing Redox Equations
Fe+2(aq)+ Cr2O72-(aq) +H+(aq)---->
Fe3+(aq) + Cr3+(aq) + H2O(l) ?
Fe 2+(aq)+ Cr2O72-(aq) +H+(aq)---->
Fe 3+(aq) + Cr 3+(aq) + H2O(l)
x = ? Cr ; 2x+7(-2) = -2; x = +6

20. Fe 2+(aq) ---> Fe 3+(aq)
Balancing Redox Equations
Fe 2+(aq) > Fe 3+(aq)
Cr2O72-(aq) > Cr 3+(aq)
Cr = (6+)
-e - 2
6 e -
6 (Fe 2+(aq) -e - ---> Fe3+(aq))
6 Fe 2+(aq) ---> 6 Fe3+(aq) + 6 e -
Cr2O72-(aq) + 6 e - --> 2 Cr3+(aq)

21. Balancing Redox Equations
6 Fe2+(aq) ---> 6 Fe3+(aq) + 6 e -
Cr2O72-(aq) + 6 e - --> 2 Cr3+(aq)
6 Fe2+(aq)+ Cr2O72-(aq) + ? 2nd H+(aq) ---->
6 Fe3+(aq) + 2 Cr3+(aq)+ ? 1st Oxygen H2O(l)
Oxygen = 7
2nd (Hydrogen) = 14

22. Balancing Redox Equations
Completely Balanced Equation:
6 Fe2+(aq)+ Cr2O72-(aq) H+(aq) ---->
6 Fe3+(aq) + 2 Cr3+(aq)+ 7 H2O(l)

23. Balancing Redox Equations in basic solutions
1) Determine oxidation numbers of atoms in Reactants and Products
2) Identify and select out those which change oxidation number into separate “half reactions”
3) Balance redox atoms and charges (electron gain and loss must equal!)
4) In basic reactions balance the Oxygen with hydroxide then Hydrogen from hydroxide with water

24. Balancing Redox Equations in basic solutions
MnO2 (aq)+ ClO31-(aq) + OH 1-aq) ->
MnO41- (aq)+ Cl 1-(aq) + H2O(l)
Mn4+ (MnO2) ---> Mn7+ (MnO4 ) 1-
Cl+5 (ClO3 ) e- ---> Cl 1-

25. Balancing Redox Equations in basic solutions
Electronically Balanced Equation:
2 MnO2 (aq)+ ClO31-(aq) + 6 e >
2 MnO Cl e-

26. Balancing Redox Equations in basic solutions
Completely Balanced Equation:
2 MnO2 (aq)+ ClO31-(aq) + 2 OH 1- (aq)---->
2 MnO4 (aq)1- + Cl 1- (aq)+ 1 H2O (l)
9 O in product

27. MnO4– + C2O42–  MnO2 + CO32– (basic solution)
QUESTION
Oxalate ion can be found in rhubarb and spinach (among other green leafy plants). The following unbalanced equation carried out in a basic solution, shows how MnO4– could be used to analyze samples for oxalate.
MnO4– + C2O42–  MnO2 + CO32– (basic solution)
When properly balanced, how many OH– are present?
1. 1
2. 2
3. 3
4. 4

28. ANSWER
Choice 4 is correct. The redox equation could be balanced according to the steps for an acid solution, but then OH– must be added to neutralize the H+ ions. When done properly 4 OH– ions will be present in the basic equation.
Section 4.10: Balancing Oxidation–Reduction Equations