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Oxidation-Reduction
Topic 9.1
1+
2+
3+
...etc.
4+/- 3-
2-
1-
0
Memorize these!
NO31-
nitrate
NO21-
nitrite
OH1-
hydroxide
ClO21-
chlorite
ClO31-
chlorate
HCO31SO42-
hydrogencarbonate
(bicarbonate)
sulfate
SO32-
sulfite
CO3 2-
carbonate
PO43-
phosphate
NH41+
ammonium
Introduction
• oxidation and reduction can be considered
in terms of…
1. oxidation- substance gains oxygen
reduction- substance loses oxygen
2. oxidation- the loss of electrons
reduction- the gain of electrons
3. oxidation- the oxidation state/# increases
reduction- the oxidation state/# decreases
O
I
L
- oxidation
- is
- loss of electrons
R
I
G
- reduction
- is
- gain of electrons
Oxidation states
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 + 2 e-  O2- (g)
Rules for assigning oxidation states
1
The oxidation number of an 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-Obonding
4
Group 1 metals are always +1
Group 2 metals are always +2
Aluminum is always +3
Li, Na…
Mg, Ba..
Al
5
Fluorine is always -1
Other group 17 (halogens) are often
-1
HF, OF2
HI, NaCl,
KBr
6
Oxidation numbers of monatomic ions follow S2-, Zn2+
the charge of the ion
7
The SUM of oxidation numbers is
zero for a neutral compound.
LiMnO4…
8
For polyatomic ions, the SUM is their
charge.
SO42-,
NO31-
Practice Assigning Oxidation Numbers
NO2
O is -2 x 2 = -4
N must equal +4
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;
N must equal +10/2 = +5
KMnO4
O is -2 x 4 = -8; K is +1;
Mn must equal +7
Fe(OH)3
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-
N is -3; Charge = -1 means
that C must be +2
K3Fe(CN)6
N is -3 x 6 = -18;
C is +2 x 6 = +12;
K is +1 x 3 =+3; Fe must be +3
CH4
H is +1 x 4 = +4;
C must be -4
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
0
+2 +6-2
+2+6-2
0
Zn: 0  + 2  Oxidized
Cu: +2  0  Reduced
S and O are unchanged
Exercise
For each of the following reactions (not
balanced to simplify) find the element
oxidized and the element reduced
Cl2
Cu
+
KBr

KCl +
Br2
+ HNO3  Cu(NO3)2 + NO2 + H2O
HNO3 +
I2
 HIO3 +
NO2
Exercise
For each of the following reactions find
the element oxidized and the element
reduced
Cl2
0
+
KBr
+1-1

KCl +
+1-1
Br loses an electron -- oxidized
Cl gains an electron -- reduced
K remains unchanged at +1
Br2
0
Exercise
For each of the following reactions find
the element oxidized and the element
reduced
Cu + HNO3  Cu(NO3)2 + NO2 + H2O
0
+1+5-2
+2 +5 -2
+4 –2 +1 -2
• Cu increases from 0 to +2. 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
Exercise
For each of the following reactions find
the element oxidized and the element
reduced
HNO3 +
+1+5-2
•
•
•
I2
0
 HIO3 +
+1+5-2
NO2
+4-2
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.
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
Review of Stock nomenclature
(naming) of transitional metals
• 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+
• 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
Examples
• lead (II) hydroxide
– Pb(OH)2
• cadmium nitrate
– cadmium is always +2
– Cd(NO3)2
• MnO2
– manganese (IV) oxide
Balancing Redox Reactions
(only in Neutral or Acid 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
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
– Fe3+ + 3 e-  Fe ….this is the reduction
–
Al  Al3+ + 3 e- ….this is the oxidation
• a wide variety of half equations can be found
in the data booklet
More…
Br2 +
0
• 2e- + Br2 
•
2I- 
2I-  2Br- + I2
-1
-1
0
2BrI2 + 2e-
• 2e- + Br2 + 2I-  2Br- + I2 + 2e-
Balancing Redox Reactions in 8 “easy” steps.
Page 217 in “the IB text book”. I do NOT like this way any more
1. assign oxidation states for each atom
2. deduce which species is oxidized and which is reduced
3. 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
4. balance elements other than O and H
5. balance so # of electrons lost equals the # gained by
adding e6. add the two half equations together to write the overall
redox reaction and simplify
7. total up the total charges on the reactant and product
sides to see if they are the same (not opposite)
8. balance the charges by adding H+ and balance
oxygens by adding H2O to the appropriate sides
Another way (p. 847 in AP book).
I like this way the best but do not yet have examples on this PowerPoint
done this way yet, only did in class
1. Divide into half equations
2. Balance
1.
2.
3.
4.
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
3. Make sure electrons lost in one half-reaction are
equally gained in the other half-reaction
1. if not, multiply the half reactions by integers as necessary
to balance electrons gained and lost
4. Add the two half-reactions back together and
simplify/canceling species appearing on both side of
the combined reaction as needed
Exercise
• Deduce and balanced redox equation and identify
the oxidizing and reducing agents.
Fe2+ + MnO4-  Fe3+ + Mn2+
• Step 1.
+2 +7 -2
+3
+2
• Step 2. Fe is oxidized (Mn was the reducing agent
that caused this) & Mn is reduced (Fe was the
oxidizing agent that caused this)
Fe2+  Fe3+ + eMnO4- + 5e-  Mn2+
• Step 4.
5Fe2+  5Fe3+ + 5eMnO4- + 5e-  Mn2+
• Step 3.
Exercise
• Step 5. 5Fe2+ + MnO4-  5Fe3+ + Mn2+
• Step 6. Total charge on reactant side = 9+
Total charge on product side = 17+
• Step 7. To balance the equation charges, 8H+
must be added to the reactant side.
5Fe2+ + MnO4- + 8H+  5Fe3+ + Mn2+
Now need to balance the hydrogens by
adding water
5Fe2+ + MnO4- + 8H+  5Fe3+ + Mn2+ + 4H2O
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.
Exercise
• Deduce and balanced redox equation and identify the
oxidizing and reducing agents.
Ag(s) + NO3–(aq) → Ag+(aq) + NO(g)
Step 1.
0
+5 -2
+1
+2-2
Step 2. Ag is oxidized & N is reduced
Step 3.
Ag  Ag+ + eNO3- + 3e-  NO
Step 4.
3Ag  3Ag+ + 3eNO3- + 3e-  NO
Exercise
3Ag + NO3– → 3Ag+ + NO
Total charge on reactant side = 1Total charge on product side = 3+
Step 7.
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
Step 5.
Step 6.
Practice #1
Using the “new way”
Balance
• Cr2O72- + Cl-  Cr3+ + Cl2
• 6e- + 14H+ + Cr2O72-  Cr3+ + 7H2O
•
3[2Cl-  Cl2 + 2e-]
Answer
• 14H+ + Cr2O72- + 6Cl-  2Cr3+ + 7H2O + 3Cl2
#2
Balance and state the oxidizing agent and reducing agent
Cu (s) + NO31- (aq)  Cu 2+ (aq) + NO2 (aq)
4H+ + Cu (s) + 2NO31- (aq)  Cu 2+ (aq) + 2NO2 (aq) + 2H2O
#3
Mn 2+ + NaBiO3  Bi 3+ + MnO4 1- + Na 1+
2Mn 2+ + 5NaBiO3 + 14H+  5Bi 3+ + 2MnO4 1- + 5Na 1+ + 7H2O
#4
NO2- + Cr2O72-  Cr 3+ + NO3 13NO2- + Cr2O72- + 8H+  2Cr 3+ + 3NO3 1- + 4H2O
#5
S + HNO3  H2SO3 + N2O
2S + 2HNO3 + H2O  2H2SO3 + N2O
#6
1.
2.
3.
4.
Balance
other than H&O
O by H2O
H by H+
charges by e-
Cr2O72- + CH3OH  HCO2H + Cr3+
16H+ + 2Cr2O72- + 3CH3OH  3HCO2H + 4Cr3+ + 11H2O