1. Oxidation-Reduction Topic 9.1

2. 1+ 2+ 3+
4+/- 3- 2- 1-
  ...etc.

3. hydrogencarbonate (bicarbonate)
Memorize these!
NO31-
nitrate
NO21-
nitrite
OH1-
hydroxide
ClO21-
chlorite
ClO31-
chlorate
HCO31-
hydrogencarbonate (bicarbonate)
SO42-
sulfate
SO32-
sulfite
CO3 2-
carbonate
PO43-
phosphate
NH41+
ammonium

5. Oxidation state/# increases Oxidation state/# decreases
oxidation and reduction can be considered in terms of...
Oxidation
Reduction
Adding oxygen
Removing oxygen
Removing hydrogen
Adding hydrogen
Loss of electrons
Gain of electrons
Oxidation state/# increases
Oxidation state/# decreases

6. O - oxidation I - is L - loss of electrons R - reduction I - is G - gain of electrons

7. Note: oxidation states must be written with the sign in FRONT: +2 not 2+
the oxidation state (number) is the apparent or theoretical charge of an free element, molecule, or ion
oxidation
a process where the number increases (more positive because loses neg. electrons)
Ex: Mg  Mg2+ (aq) + 2e-
reduction
a process where the number decreases (less positive/more negative because gains neg. electrons)
Ex: O e-  O2- (g)

8. Rules for assigning oxidation states1
The oxidation number of a free element is always 0
O2, H2, Ne, Zn 2
The oxidation number of Hydrogen is usually +1 Metal hydrides are an exception. They are -1
HCl, H2SO4
NaH 3
The oxidation number of Oxygen is usually -2. Peroxides are an exception They are –1.
H2O, NO2, etc.
-O-O- bonding 4
Group 1 metals are always +1
Group 2 metals are always +2
Aluminum is always +3
Li, Na…
Mg, Ba.. Al

9. Other group 17 (halogens) are often -1 HF, OF2 HI, NaCl, KBr 65
Fluorine is always -1
Other group 17 (halogens) are often -1
HF, OF2
HI, NaCl, KBr 6
Oxidation numbers of monatomic ions follow the charge of the ion
Cu2+ , Zn2+ 7
The SUM of oxidation numbers is zero for a neutral compound.
LiMnO4… 8
For polyatomic ions, the SUM is their charge.
SO42-, NO31-

10. Practice Assigning Oxidation Numbers
N2O5
O is -2 x 5 = -10
N must equal +10/2 = +5
HClO3
(+1) + (x) + 3(-2) = 0;
x = (+5)
HNO3
O is -2 x 3 = -6; H is +1;
N must equal +5
Ca(NO3)2
O is -2 x 3 = -6 x 2 = -12;
Ca is +2;
KMnO4
O is -2 x 4 = -8; K is +1;
Mn must equal +7

11. Fe(OH)3 K2Cr2O7 CO32- CN-1 K3Fe(CN)6 CH4 O is -2 x 3 = -6; Fe is +3;
H is +1 x 3 = +3
K2Cr2O7
O is -2 x 7 = -14;
K is +1 x 2 = +2;
Cr must equal +12/2 = +6
CO32-
x + 3(-2) = -2
x = +4
CN-1
N is -3; Charge = -1 means that C must be +2
K3Fe(CN)6
CN is -1 x 6 = -6;
K is +1 x 3 =+3; Fe must be +3
CH4
H is +1 x 4 = +4;
C must be -4

12. Using Oxidation Numbers
an increase in the oxidation number indicates that an atom has lost electrons and therefore is oxidized
a decrease in the oxidation number indicates that an atom has gained electrons and therefore reduced
Example
Zn CuSO4  ZnSO4 + Cu
Zn: 0   Oxidized
Cu: +2   Reduced
S and O are unchanged
Recognize these!
NO31-
nitrate
NO21-
nitrite
OH1-
hydroxide
ClO21-
chlorite
ClO31-
chlorate
HCO31-
hydrogencarbonate (bicarbonate)
SO42-
sulfate
SO32-
sulfite
CO3 2-
carbonate
PO43-
phosphate
NH41+
ammonium

13. Exercise
For each of the following reactions (not balanced to simplify) find the element oxidized and the element reduced
Cl KBr  KCl Br2
Cu HNO3  Cu(NO3) NO2 + H2O
HNO I  HIO NO2

14. Exercise
For each of the following reactions find the element oxidized and the element reduced
Cl KBr  KCl Br2  
Br loses an electron -- oxidized
Cl gains an electron -- reduced
K remains unchanged at +1

15. hydrogencarbonate (bicarbonate)
Exercise
For each of the following reactions find the element oxidized and the element reduced
Cu HNO3  Cu(NO3) NO2 + H2O –
Cu increases from 0 to It is oxidized
Only part of the N in nitric acid changes from +5 to +4. It is reduced
The nitrogen that ends up in copper nitrate remains unchanged
Recognize these!
NO31-
nitrate
NO21-
nitrite
OH1-
hydroxide
ClO21-
chlorite
ClO31-
chlorate
HCO31-
hydrogencarbonate (bicarbonate)
SO42-
sulfate
SO32-
sulfite
CO3 2-
carbonate
PO43-
phosphate
NH41+
ammonium

16. Exercise
For each of the following reactions find the element oxidized and the element reduced
HNO I  HIO NO2
N is reduced from +5 to +4. It is reduced.
I is increased from 0 to +5 It is oxidized
The hydrogen and oxygen remain unchanged.

17. Agents
all redox reactions have one element oxidized and one element reduced
the compound that supplies the electrons (is oxidized) is the reducing agent
the compound that accepts the electrons (is reduced) is the oxidizing agent
occasionally, the same element may undergo both oxidation and reduction. This is known as an auto-oxidation reduction

18. Exercise
For each of the following reactions find the element oxidized, element reduced, and the agents.
HNO I  HIO NO2
N is reduced from +5 to +4.
I2 is the reducing agent that caused this
I is oxidized from 0 to +5.
HNO3 is the oxidizing agent that caused this
The hydrogen and oxygen remain unchanged.

19. use roman numerals to indicate the charge
Review of Stock nomenclature (naming) of transitional metals. This was already covered in Topic 4
transition metals can form more than one type of ion (i.e. lose different amounts of electrons)
Cu1+, Cu 2+
use roman numerals to indicate the charge
Cu1+ = “Copper I” ; Cu 2+ = “Copper II”
Exceptions: Ag1+, Zn2+, Cd2+, Al3+, Sc3+

20. ex. copper (II) oxide (“copper two oxide”)
CuO
oxygen has a -2 charge, so it would only take one Cu2+ to bond with Oxygen.
ex. copper (I) oxide
copper 1+ -we would need two of these to react with oxygen so the formula would be:
Cu2O

21. Examples lead (II) hydroxide cadmium nitrate MnO2 Pb(OH)2
cadmium is always +2
Cd(NO3)2
MnO2
manganese (IV) oxide

22. Balancing Redox Reactions (only in Neutral or Acidic Solutions)
many chemical reactions involving oxidations and reductions are complex and very difficult to balance by the “guess and check” methods we learned earlier
for complicated reactions, a more systematic approach is required

23. Half-equations
half equations show the changes to individual species in a redox reaction
can use charges or oxidation #’s to do this
Fe2O3 + 2 Al  2 Fe + Al2O3
Fe e-  Fe ….this is the reduction
Al  Al e- ….this is the oxidation
a wide variety of half equations can be found in the data booklet

24. More… Br2 + 2I-  2Br- + I2 0 -1 -1 0 2e- + Br2  2Br- 2I-  I2 + 2e-
2e- + Br2  2Br-
2I-  I e-
2e- + Br I-  2Br- + I e-

25. Balancing Oxidation – Reduction Equations
often given an incomplete equation for redox reaction in a acidic solution
when balancing re-dox reactions, the gains and loses of electrons must be balanced

26. Balancing Redox Reactions in 8 “easy” steps
Balancing Redox Reactions in 8 “easy” steps. Page 217 in “the IB text book”. I do NOT like this way any more
assign oxidation states for each atom
deduce which species is oxidized and which is reduced
write half reactions for the oxidation and reduction
take compounds where “action” took place, split them and write them as individual reactions; there will be 2 half reactions
balance elements other than O and H
balance so # of electrons lost equals the # gained by adding e-
add the two half equations together to write the overall redox reaction and simplify
total up the total charges on the reactant and product sides to see if they are the same (not opposite)
balance the charges by adding H+ and balance oxygens by adding H2O to the appropriate sides

27. The following example problems are done using “the old way” that I no longer like

28. Exercise
Deduce and balanced redox equation and identify the oxidizing and reducing agents.
Fe2+ + MnO41-  Fe3+ + Mn2+
Step
Step 2. Fe is oxidized (Mn was the reducing agent that caused this) & Mn is reduced (Fe was the oxidizing agent that caused this)
Step Fe2+  Fe3+ + e-
MnO e-  Mn2+
Step Fe2+  5Fe3+ + 5e-

29. Exercise Step 5. 5Fe2+ + MnO4-  5Fe3+ + Mn2+
Step Total charge on reactant side = 9+
Total charge on product side = 17+
Step To balance the equation charges, 8H+
must be added to the reactant side.
5Fe2+ + MnO H+  5Fe3+ + Mn2+
Now need to balance the hydrogens by
adding water
5Fe2+ + MnO H+  5Fe3+ + Mn2+ + 4H2O

30. Ag(s) + NO3–(aq) → Ag+(aq) + NO(g)
More…
Nitric acid reacts with silver in a redox reaction.
Ag(s) + NO3–(aq) → Ag+(aq) + NO(g)
Using oxidation numbers, deduce the complete balanced equation for the reaction showing all the reactants and products.

31. Ag(s) + NO3–(aq) → Ag+(aq) + NO(g)
Exercise
Deduce and balanced redox equation and identify the oxidizing and reducing agents.
Ag(s) + NO3–(aq) → Ag+(aq) + NO(g)
Step
Step 2. Ag is oxidized & N is reduced
Step Ag  Ag+ + e-
NO e-  NO
Step Ag  3Ag+ + 3e-

32. Exercise Step 5. 3Ag + NO3– → 3Ag+ + NO
Step Total charge on reactant side = 1-
Total charge on product side = 3+
Step To balance the equation charges, 4H+
must be added to the reactant side.
3Ag + NO3– + 4H+ → 3Ag+ + NO
Now need to balance the hydrogens by
adding water
3Ag + NO3– + 4H+ → 3Ag+ + NO + 2H2O

33. Another way (p. 847 in AP book). This the preferred method by most
Divide into half equations
Balance
elements other than O and H
balance O by adding water as needed
balance H by adding H+ as needed
balance charges by adding e- as needed
Make sure electrons lost in one half-reaction are equally gained in the other half-reaction
if not, multiply the half reactions by integers as necessary to balance electrons gained and lost
Add the two half-reactions back together and simplify/canceling species appearing on both side of the combined reaction as needed

34. 14H+ + Cr2O72- + 6Cl-  2Cr3+ + 7H2O + 3Cl2
Practice #1
Balance
Cr2O Cl-  2Cr Cl2
6e H+ + Cr2O72-  2Cr3+ + 7H2O
3 [2Cl-  Cl2 + 2e-]
Answer
14H+ + Cr2O Cl-  2Cr H2O + 3Cl2
divide into half reactions
balance:
other than H&O
O by H2O
H by H+
charges by e-
make e- in both half-reactions the same
add half-reactions and simplify

35. 14H+ + Cr2O72- + 6Cl-  2Cr3+ + 7H2O + 3Cl2
Practice #2
Balance
MnO C2O  Mn CO2
2 [5e- + 8H+ + MnO4-  Mn H2O ]
5 [C2O42-  2 CO2 + 2e-]
10e H+ + 2MnO4-  2Mn H2O
5 C2O42-  10 CO e-
16H+ + 2MnO C2O42-  2Mn H2O + 10 CO2
Answer
14H+ + Cr2O Cl-  2Cr H2O + 3Cl2
divide into half reactions
balance:
other than H&O
O by H2O
H by H+
charges by e-
make e- in both half-reactions the same
add half-reactions and simplify
divide into half reactions
balance:
other than H&O
O by H2O
H by H+
charges by e-
make e- in both half-reactions the same
add half-reactions and simplify

36. Balance and state the oxidizing agent and reducing agent #3
Cu (s) + NO31- (aq)  Cu 2+ (aq) + NO2 (aq)
4H+ + Cu (s) + 2NO31- (aq)  Cu 2+ (aq) + 2NO2 (aq) + 2H2O #4
Mn NaBiO3  Bi MnO Na 1+
2Mn NaBiO3 + 14H+  5Bi MnO Na H2O #5
NO2- + Cr2O72-  Cr NO3 1-
3NO2- + Cr2O H+  2Cr NO H2O #6
S + HNO3  H2SO N2O
2S HNO3 + H2O  2H2SO N2O #7
Cr2O CH3OH  HCO2H + Cr3+
16H+ + 2Cr2O CH3OH  3HCO2H Cr H2O
Balance
other than H&O
O by H2O
H by H+
charges by e-
make e- in both reactions the same