Oxidation Numbers
and
Balancing Redox
Reactions
1
1
Oxidation Numbers
Keeping Track of Electrons in Redox Reactions
2
2
Oxidation
• Oxidation Numbers aka Oxidation States
• A concept devised to keep track of electrons in
a redox reaction.
• Increase in oxidation number = oxidation
✓
LEO (Lose Electrons = Oxidation)
• Decrease in oxidation number = reduction
✓
says GER (Gain Electrons = Reduction)
• The sum of oxidation numbers in neutral
compounds must = 0
• The sum of oxidation numbers in a polyatomic
ion = the charge of the ion
3
3
Determining Oxidation #
•
•
•
The charge on a monotomic ion is the #
Atoms in elemental form are zero
✓
H in H2, atoms in a lump of iron, Fe, P atoms in P4
Nonmetals usually have negative # - but
they can be positive.
✓
oxygen is -2 in both ionic and molecular compounds
except in peroxides in which the oxidation number is -1
✓ hydrogen is always +1 when bonded to nonmetals and
always -1 when bonded to metals
✓ fluorine is -1 in all compounds. The other halogens are
usually -1 in binary compounds, but when combined
with oxygen they are positive
4
4
Determine the oxidation number of
each element in the following
substances
-2
1. CaCl2
6. SO3
2. PbO2
7. H2O
3. Cl2
8. H2O2 hydrogen peroxide
4. S8
9. AlH3
5. SO4
-2
10.ClO3
-1
5
5
Determine the oxidation number of
phosphorus in Mg2P2O7
Submit a numeric value. If it is negative, put on
the - sign. If positive, just leave it.
6
6
Determine the oxidation number of
iron in K4Fe(CN)6
Submit a numeric value. If it is negative, put on
the - sign. If positive, just leave it.
7
7
Determine the oxidation number of
chromium in Na2Cr2O7
Submit a numeric value. If it is negative, put on
the - sign. If positive, just leave it.
8
8
What is the oxidation number of
sulfur in SF2?
1. 0
2. +2
3. -2
4. +4
5. -4
9
9
What is the oxidation number of
sulfur in SF2?
1. 0
2. +2
3. -2
4. +4
5. -4
• Since fluorine is -1 *2 = -2 the sulfur must
be +2 to balance.
10
10
What is the oxidation state of
chlorine in KClO4 ?
1.
2.
3.
4.
5.
-2
-4
+4
+7
+8
11
11
What is the oxidation state of
chlorine in KClO4 ?
1.
2.
3.
4.
5.
•
-2
-4
+4
+7
+8
O is -2 * 4 = -8, the K is +1 thus the Cl
must be the remaining +7.
12
12
What is the oxidation state of nitrogen
in N2O5 ?
1.
2.
3.
4.
5.
+5
+2
-10
+10
-5
13
13
What is the oxidation state of nitrogen
in N2O5 ?
1.
2.
3.
4.
5.
•
+5
+2
-10
+10
-5
Since oxygen is -2 *5 = -10, the nitrogens
must total +10, but 2 nitrogens cause the
+10, so each one individually is +5.
slide show
14
14
Nitrogen is the master of many
oxidation states. Determine the
oxidation number of nitrogen in
each of the following compounds.
1.
2.
3.
4.
5.
NO
N 2O
NO2
N 2H 4
NH3
6.
7.
8.
9.
Li3N
N2
-1
NO2
-1
NO3
15
15
Nitrogen is the master of many oxidation
states. Determine the oxidation number of
nitrogen in each of the following compounds.
+2 -2
1. NO
-2
+1
2. N2O
+2 -2
3. NO2
-2
+1
4. N2H4
+1
-3
6. Li3N
0
7. N2
+3 -2
8. NO2
+5 -2
9. NO3
-1
-1
-3 +1
5. NH3
16
16
Oxidation - Reduction
• Oxidation (LEO) [OIL]
✓
When an element loses electrons
• Reduction (GER) [RIG]
✓
When an element gains electrons
• The name oxidation is used because this type of
reaction was first studied by reacting metals with
oxygen.
✓
Rusting: Fe + O2 → Fe2O3
• It was called reduction because when elements
are isolated from minerals, we say they are
“reduced” to their elemental form.
✓
AlCl3 → Al(s) + Cl2(g)
17
17
What is the total number of electrons
transferred when magnesium is burned
in oxygen?
•
type in the number transfered
18
18
What is the total number of electrons
transferred when magnesium is burned
in oxygen?
•
•
•
•
For which balanced equation?
-1
2Mg + O2 → 2MgO (4 e )
-1
OR Mg + ½O2 → MgO (2 e )
Thus you might say:
•2
-1
e per
mole of MgO formed.
19
19
Balancing Aqueous
Redox Equations
Chapter 4 & 20
20
20
-1
AuCl4 + Cu → Au +
-1
Cl
+
+2
Cu
1. Identify which elements are oxidized and reduced
2. balance those atoms based on total electrons
transferred
3. balance any other elements (other than H and O)
4. balance O by adding water as necessary
5. balance H by adding H+1 as necessary
6. recheck by confirming that charge is balanced
In an acidic solution
21
21
-1
AuCl4 + Cu → Au +
-1
Cl
+
+2
Cu
1. Identify which elements are oxidized and reduced
→
0
+ -1
+ +2
• +3, -1 + 0
• Au reduced 3e- gained, Cu oxidized 2e- lost
2. balance those atoms based on total electrons
transferred
• 2AuCl4-1 + 3Cu → 2Au + Cl-1 + 3Cu+2
3. balance any other elements (other than H and O) - none
4. balance O by adding water as necessary - not
5. balance H by adding H+1 as necessary - not
6. recheck by confirming that charge is balanced
• 2- = 222
22
-1
-1
8Cl
+2
3Cu
2AuCl4 + 3Cu → 2Au +
+
From time to time, AP will ask you to
write out the half reactions.
1. oxidation
+2 + 2eCu
→
Cu
•
2. reduction
-1 + 3e- → Au + 4Cl-1
AuCl
4
•
3. this corresponds with what we had determined by the
previous method
• Au reduced 3e- gained, Cu oxidized 2e- lost
23
23
In an acidic sol’n
-1
Zn + NO3 →
+2
Zn
+ N2
1. Identify which elements are oxidized and reduced
2. balance those atoms based on total electrons
transferred
3. balance any other elements (other than H and O)
4. balance O by adding water as necessary
5. balance H by adding H+1 as necessary
6. recheck by confirming that charge is balanced
24
24
(in acidic)
-1
+2
Zn + NO3 → Zn + N2
1. Identify which elements are oxidized and reduced
0 + +5,-2
→ +2
+ 0
•
• N reduced 5e-/atom, Zn oxidized 2e2. balance those atoms based on total electrons transferred
-1 → 5Zn+2 + N
5Zn
+
2NO
3
2
•
3. balance any other elements (other than H and O) - none
4. balance O by adding water as necessary
-1 → 5Zn+2 + N + 6H O
5Zn
+
2NO
3
2
2
•
5. balance H by adding H+1 as necessary
• 12H+1 + 5Zn + 2NO3-1 → 5Zn+2 + N2 + 6H2O
6. recheck by confirming that charge is balanced
• +10 = +10
25
25
-1
+2
Zn
Zn + NO3 →
+ N2 (in acidic)
Write out the two half reactions.
26
26
-1
+2
Zn
Zn + NO3 →
+ N2 (in acidic)
From time to time, AP will ask you to
write out the half reactions.
1. oxidation
• Zn → Zn+2 + 2e• note that charge balances in half reactions as well
2. reduction
• 2NO3-1 → N2
+ ions
this
may
involve
using
water
and
H
•
• 12H+1 + 2NO3-1 + 10e- → N2 + 6H2O
3. this corresponds with what we had determined by the
previous method
• N reduced 5e- gained per N, but the N2 requires 10etotal, Zn oxidized 2e- lost
27
27
In an basic solution
-1
Br2 + AsO2 →
-1
Br
+ AsO4
-3
1. Identify which elements are oxidized and reduced
2. balance those atoms based on total electrons
transferred
3. balance any other elements (other than H and O)
4. balance O by adding water as necessary
5. balance H by adding H+1 as necessary
6. if the reaction occurs in basic solution, convert to basic
by adding an equal number of OH-1 ions to both sides to
cancel out the H+1 ions
7. recheck by confirming that charge is balanced
28
28
-1
Br2 + AsO2 →
-1
Br
+ AsO4
-3
“pretend” in acid, then convert to basic
1.
2.
3.
4.
5.
6.
7.
Identify which elements are oxidized and reduced
0
+ +3, -2
→ -1
+
+5, -2
•
• Br reduced 1e-/atom, As oxidized 2ebalance those atoms based on total electrons transferred
• Br2 + AsO2-1 → 2Br-1 + AsO4-3
balance any other elements (other than H and O) - none
balance O by adding water as necessary
• 2H2O + Br2 + AsO2-1 → 2Br-1 + AsO4-3
balance H by adding H+1 as necessary
• 2H2O + Br2 + AsO2-1 → 2Br-1 + AsO4-3 + 4H+1
if the reaction occurs in basic solution, convert to by adding an equal number of
OH-1 ions to both sides to cancel out the H+1 ions
• 4OH-1 + 2H2O + Br2 + AsO2-1 → 2Br-1 + AsO4-3 + 4H+1 + 4OH-1
• 4OH-1 + 2H2O + Br2 + AsO2-1 → 2Br-1 + AsO4-3 + 4H2O
• 4OH-1 + Br2 + AsO2-1 → 2Br-1 + AsO4-3 + 2H2O
recheck by confirming that charge is balanced
• -5 = -5
29
29
In basic solution
-1
Br2 → BrO3 +
-1
Br
1. Identify which elements are oxidized and reduced
• This is a disproportionation reaction - the same
element is both oxidized and reduced.
2. balance those atoms based on total electrons transferred
3. balance any other elements (other than H and O)
4. balance O by adding water as necessary
5. balance H by adding H+1 as necessary
6. if the reaction occurs in basic solution, convert Rx to
basic by adding an equal number of OH-1 ions to both
sides to cancel out the H+1 ions
7. recheck by confirming that charge is balanced
30
30
-1
Br2 → BrO3 +
-1
Br
“pretend” in acid, then convert to basic
1.
2.
3.
4.
5.
6.
7.
Identify which elements are oxidized and reduced
0
→
+5, -2
+
-1
•
• Br reduced 1e-/atom, and Br oxidized 5e-/atom
balance those atoms based on total electrons transferred
• 6 Br2 → 2 BrO3-1 + 10Br-1
balance any other elements (other than H and O) - none
balance O by adding water as necessary
• 6 Br2 + 6 H2O → 2 BrO3-1 + 10Br-1
balance H by adding H+1 as necessary
• 6 Br2 + 6 H2O → 2 BrO3-1 + 10 Br-1 + 12 H+1
if the reaction occurs in basic solution, convert to by adding an equal number of
OH-1 ions to both sides to cancel out the H+1 ions
• 12OH-1 + 6 Br2 + 6H2O → 2 BrO3-1 + 10Br-1 + 12H+1 + 12OH-1
• 12OH-1 + 6 Br2 + 6H2O → 2 BrO3-1 + 10Br-1 + 12H2O
• 12OH-1 + 6 Br2 → 2 BrO3-1 + 10Br-1 + 6H2O
recheck by confirming that charge is balanced
• -12 = -12
31
31
-1
-1
Br
Br2 → BrO3 +
(in basic)
Write out the half reactions.
32
32
-1
-1
Br
Br2 → BrO3 +
(in basic)
Write out the half reactions.
1. oxidation
-1 + 12 H+1 + 10eBr
+
6
H
O
→
2
BrO
2
2
3
•
• which you can convert to basic if necessary.
2. reduction
-1
Br
+
2e→
2Br
2
•
3. this corresponds with what we had determined by the
previous method
• N reduced 5e- gained per N, but the N2 requires 10etotal, Zn oxidized 2e- lost
33
33
(in basic)
-1
-1
-1
CN + MnO4 → CNO + MnO2
1. Identify which elements are oxidized and reduced
• You may worry about what to do with the oxidation
numbers - consider N to be -3 and watch the C change
2. balance those atoms based on total electrons transferred
3. balance any other elements (other than H and O)
4. balance O by adding water as necessary
5. balance H by adding H+1 as necessary
6. if the reaction occurs in basic solution, convert to by
adding an equal number of OH-1 ions to both sides to
cancel out the H+1 ions
7. recheck by confirming that charge is balanced
34
34
(in basic)
-1
-1
-1
CN + MnO4 → CNO + MnO2
1.
•
•
•
2.
•
3.
4.
•
5.
•
6.
•
•
•
7.
•
Identify which elements are oxidized and reduced
You may worry about what to do with the oxidation numbers - consider N to be -3 and
watch the C change
+2, -3
+
+7, -2
→
+4, -3, -2
+
+4, -2
Mn reduced 3e-, and C oxidized 2ebalance those atoms based on total electrons transferred
3CN-1 + 2MnO4-1 → 3CNO-1 + 2MnO2
balance any other elements (other than H and O) - done
balance O by adding water as necessary
3CN-1 + 2MnO4-1 → 3CNO-1 + 2MnO2 + H2O
balance H by adding H+1 as necessary
3CN-1 + 2MnO4-1 + 2H+1 → 3CNO-1 + 2MnO2 + H2O
if the reaction occurs in basic solution, convert to by adding an equal number of OH-1 ions to
both sides to cancel out the H+1 ions
2OH-1 + 3CN-1 + 2MnO4-1 + 2H+1 → 3CNO-1 + 2MnO2 + H2O + 2OH-1
3CN-1 + 2MnO4-1 + 2H2O → 3CNO-1 + 2MnO2 + H2O + 2OH-1
3CN-1 + 2MnO4-1 + H2O → 3CNO-1 + 2MnO2 + 2OH-1
recheck by confirming that charge is balanced
35
-5 = -5
35