Balancing Redox Equations
Reduction/Oxidation = Redox
A new iron bolt (Fe)
A rusted (oxidized) bolt
(rust is Fe2O3)
Iron (Fe) rusts (is oxidized) when it reacts with oxygen (O2) in the air.
2Fe + 3O2  Fe2O3 [iron(III) oxide]
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Redox Equations
Rusting is a an example of a redox reaction.
2Fe + 3O2  Fe2O3 [iron(III) oxide]
In redox reactions, one chemical is oxidized (e.g., Fe is oxidized) and another chemical is
reduced (e.g., O2 is reduced).
Oxidation and reduction always occur together. One cannot occur without the other.
In this reaction, when iron is oxidized, it gains oxygen.
This definition of oxidation (‘gain of oxygen’), while correct in this case, is inadequate,
because many redox reactions do not involve oxygen.
A better definition follows, but do not disregard the example of rusting. It is something
we all have observed and will help you understand and remember the broader
definitions.
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Redox Equations
0
0
2Fe + 3O2

lose 3 e’s, LEO
+3
-2
Fe2O3
[iron(III) oxide]
gain 2 e’s, GER
What are the oxidation numbers of iron and oxygen in the rust equation?
Definitions to Memorize:
Loss of Electrons = Oxidation
LEO the Lion says GER
Gain of Electrons = Reduction
To do redox chemistry, you must be able to determine which atoms have lost e’s (have
been oxidized) and which atoms have gained e’s (have been reduced).
Thus, iron has been oxidized by oxygen. Oxygen is an ‘oxidizing agent’.
and oxygen has been reduced by iron. Iron is a ‘reducing agent’.
3
Balancing Redox Equations by the Half-Reaction Method
The following rules describe the most versatile method for balancing redox equations ...
1. Break the eqn. into two ½-reactions, one for oxidation (oxid.) and one for reduction (red.)
2. Balance all atoms in each ½-reaction other than H and O
3. Balance O and H as follows ...
a) first balance O by adding H2O,
b) then balance H’s by adding H+
c) finally, balance the charges by adding sufficient electrons
e-'s are always added to the right side of the oxid. ½-reaction because oxid. = loss of e-'s
e-'s are always added to the left side of the red. ½-reaction because red. = gain of e-'s
4. Multiply each ½-reaction by a coefficient so that the same number of e-'s are transferred in
each ½-reaction. The number of e’s gained must equal the number of e’s lost.
5. Add the ½-reactions and cancel common terms (e.g., H2O and/or H+) from each side.
6. If the reaction is carried out in neutral or basic solution, add enough OH- to neutralize all
H+ (combine them to make H2O) and be sure to add the same quantity of OH- to both sides
to keep mass and charge balanced.
7. Cancel common terms (H2O and/or OH-) from each side and check for mass and charge
balance
4
+5
+4
Bromate and sulfite react as shown : BrO3- + SO3-2
Oxidation ½-reaction:
1.
3a) balance O w. H2O
3b) balance H w. H+
3c) balance charge w e’s
4.
Reduction ½-reaction:
1.
2. balance Br
3a) balance O w. H2O
3b) balance H w. H+
3c) balance charge w. e’s
+6
𝐻+ 0
Br2 + SO4-2
gain 5 e’s, GER
-2
𝐻+
lose 2 e’s, LEO
SO3
SO4-2
H2O + SO3-2  SO4-2
H2O + SO3-2  SO4-2 + 2H+
H2O + SO3-2  SO4-2 + 2H+ + 2e(H2O + SO3-2  SO4-2 + 2H+ + 2e-) × 5
-
BrO3-, the oxidizer is reduced
SO3-2, the reducer is oxidized
𝐻+
BrO3
Br2
2BrO3- Br2
2BrO3- Br2 + 6H2O
12H+ + 2BrO3-  Br2 + 6H2O
10e- + 12H+ + 2BrO3-  Br2 + 6H2O
O can only be balanced w. H2O
H can only be balanced w. H+
4. Multiply both ½-reactions by the appropriate coefficient so that both reactions involve the same number of e’s.
Find the lowest common multiple (LCM) of the e’s transferred (10 e’s is the LCM of 2e’s and 10e’s).
Combine the left sides of the ½-reactions and combine the right sides of the ½-reactions and reduce them.
5H2O + 5SO3-2 + 10e- + 12H+ + 2BrO3-  5SO4-2 + 10H+ + Br2 + 6H2O + 10e5SO3-2 + 2H+ + 2BrO3-  5SO4-2 + Br2 + H2O
5
+7
Permanganate reacts w. hydrogen peroxide:
Oxidation ½-reaction:
1.
3b) balance H w. H+
3c) balance charge w e’s
6. neutralize H+ w. OH-
Reduction ½-reaction:
1.
3a) balance O w. H2O
3b) balance H w. H+
3c) balance charge w. e’s
6. neutralize H+ w. OH7. cancel H2O
MnO4-
-1
+ H2O2
𝑂𝐻 − +4
0
MnO2 + O2
𝑂𝐻 −
gain 3 e’s, GER
H2O2
O2
H2O2  O2 + 2H+
H2O2  O2 + 2H+ + 2e2OH- + H2O2  O2 + 2H2O + 2e(2OH- + H2O2  O2 + 2H2O + 2e-) × 3
-
𝑂𝐻 −
MnO4
MnO2
MnO4-  MnO2 + 2H2O
4H+ + MnO4-  MnO2 + 2H2O
3e- + 4H+ + MnO4-  MnO2 + 2H2O
3e- + 4H2O + MnO4-  MnO2 + 2H2O + 4OH3e- + 2H2O + MnO4-  MnO2 + 4OH(3e- + 2H2O + MnO4-  MnO2 + 4OH-) × 2
lose 1 e’s, LEO
e’s are always on the right side
of an oxidation ½-reaction
MnO4-, the oxidizer is reduced
H2O2, the reducer is oxidized
e’s are always on the left side
of a reduction ½-reaction
Multiply both ½-reactions by the appropriate coefficient so that both reactions involve the same number of e’s.
Find the lowest common multiple (LCM) of the e’s transferred (6 e’s is the LCM of 2e’s and 3e’s)
6OH- + 3H2O2 + 6e- + 4H2O + 2MnO4-  2MnO2 + 8OH- + 3O2 + 6H2O + 6e3H2O2 + 2MnO4-  2MnO2 + 2OH- + 3O2 + 2H2O
6
+6
Dichromate and methane react as shown :
Cr2O7-2
Oxidation ½-reaction:
1.
2. balance C’s
3a) balance O w. H2O
3b) balance H w. H+
3c) balance charge w e’s
6) neutralize H+ w. OH7) cancel H2O
gain 3 e’s, GER
Reduction ½-reaction:
1.
2. balance Cr
3a) balance O w. H2O
3b) balance H w. H+
3c) balance charge w. e’s
6. neutralize H+ w. OH7. cancel H2O
𝑂𝐻 −
-3
+ C2H6
+4
+3
CO2 + Cr+3
lose 7 e’s, LEO
C2H6
CO2
C2H6  2CO2
4H2O + C2H6  2CO2
4H2O + C2H6  2CO2 + 14H+
4H2O + C2H6  2CO2 + 14H+ + 14e4H2O + C2H6 + 14OH-  2CO2 + 14H2O + 14eC2H6 + 14OH-  2CO2 + 10H2O + 14e-2
𝑂𝐻 −
Cr2O7-2, the oxidizer is reduced
C2H6, the reducer is oxidized
𝑂𝐻 −
Cr2O7
Cr+3
Cr2O7-2  2Cr+3
Cr2O7-2  2Cr+3 + 7H2O
14H+ + Cr2O7-2  2Cr+3 + 7H2O
6e- + 14H+ + Cr2O7-2  2Cr+3 + 7H2O
6e- + 14H2O + Cr2O7-2  2Cr+3 + 7H2O + 14OH6e- + 7H2O + Cr2O7-2  2Cr+3 + 14OH7
+6
Cr2O7-2 +
gain 3 e’s, GER
Oxidation ½-reaction:
C2H6 + 14OH-  2CO2 + 10H2O + 14e-
-3
C2H6
𝑶𝑯− +4
lose 7 e’s, LEO
+3
CO2 + Cr+3 continued:
Reduction ½-reaction:
7H2O + Cr2O7-2 + 6e-  2Cr+3 + 14OH-
Multiply both ½-reactions by the appropriate coefficient so that both reactions involve the same number of e’s.
Find the lowest common multiple (LCM) of the e’s transferred ( 42 is the LCM’s of 14e’s and 6e’s)
(C2H6 + 14OH-  2CO2 + 10H2O + 14e-) × 3
(7H2O + Cr2O7-2 + 6e-  2Cr+3 + 14OH-) × 7
Combine the two ½-reactions:
3C2H6 + 42OH- 49H2O + 7Cr2O7-2 + 42e-  14Cr+3 + 98OH- + 6CO2 + 30H2O + 42e3C2H6 + 19H2O + 7Cr2O7-2  14Cr+3 + 56OH- + 6CO2
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Oxidation Number Method for Balancing Redox Equations
+5
+4
BrO3- + SO3-2
+6
𝐻+ 0
Br2 + SO4-2
gain 5 e’s, GER
lose 2 e’s, LEO
+7
MnO4-
-1
+ H2O2
𝑂𝐻 − +4
MnO2 + O2
gain 3 e’s, GER
+6
Cr2O7-2
lose 1 e’s, LEO
-3
+ C2H6
gain 3 e’s, GER
0
𝑶𝑯−
+4
+3
CO2 + Cr+3
lose 7 e’s, LEO
each Br atom in BrO3- gains 5 e’s.
each S atom in SO3-2 loses 2 e’s, 10e’s is the LCM
5 e’s × 2BrO3- = 10e’s gained
2 e’s × 5SO3-2 = 10e’s lost
2BrO3- + 5SO3-2  Br2 + 5SO4-2
each Mn atom in MnO4- gains 3 e’s.
each O atom in H2O2 loses 1 e’s or
each H2O2 lose 2 e’s, 6e’s is the LCM
3 e’s × 2MnO4- = 6e’s gained
2 e’s × 3H2O2 = 6e’s lost
2MnO4- + 3H2O2  2MnO2 + 3O2
each Cr atom in Cr2O7-2 gains 3 e’s or
each Cr2O7-2 gains 6 e’s
each C atom in C2H6 loses 7 e’s or
each C2H6 lose 14 e’s, 42e’s is the LCM
6 e’s × 7Cr2O7-2 = 42e’s gained
14e’s × 3C2H6 = 42e’s lost
7Cr2O7-2+ 3C2H6  14Cr+3 + 6CO2
9
A Difficult Redox Equation to Balance
+7
MnO4-
-8/5
+ C5H8
gain 3 e’s, GER
𝑂𝐻 −
+4
+4
MnO2 + CO2
lose 8/5 e’s, LEO
Equations like the one above are not uncommon in organic chemistry, where carbon may have fractional
oxidation numbers.
Although this can be balanced by the oxidation number method (with a bit of mental arithmetic),
most students will likely have more success using the Half-Reaction method.
The balanced equation is shown below. Can you obtain this answer?
28MnO4- + 3C5H8 + 2H2O
𝑶𝑯−
28MnO2 + 15CO2 + 28 OH-
10
Practice is the best teacher
11
Activity Series of Metals and Oxidation Potential (V = emf)
These very reactive
metals will be oxidized by
H2O and liberate H2 gas.
A positive oxidation potential means
that a reaction is favorable
(spontaneous) compared to
hydrogen and will produce the emf
shown (under standard conditions).
These moderately
reactive metals are not
oxidized by H2O, but are
oxidized by dilute acids
(H2SO4, HCl, etc.) and
liberate H2 gas.
A negative oxidation potential
means that a reaction is unfavorable
compared to hydrogen and will
require that a emf greater than the
value shown be applied to force the
reaction to occur as written.
These inert (‘noble’)
metals are unreactive.
They are not oxidized by
dilute acids.
The oxidation potential of hydrogen
is set to zero and used as a
reference for other redox reactions.
Reversing an oxidation reaction
yields a reduction reaction with the
same magnitude of emf, but with
opposite sign.
metal  metal cation + e’s Potential (V)
Li Li+ + 1e
K  K++ 1e
Na  Na+ + 1e
Mg  Mg+2 + 2e
Al  Al+3 +3e
Zn  Zn+2 + 2e
Fe  Fe+2 + 2e
Ni  Ni+2 + 2e
Sn  Sn+2 + 2e
Pb  Pb+2 + 2e
½ H2  H+ + e
Cu  Cu+2 + 2e
Hg  Hg+2 + 2e
Ag  Ag+ + e
Pt  Pt+2 + 2e
Au  Au+3 + 3e
+3.05
+2.924
+2.71
+2.36
+1.7
+0.76
+0.41
+0.23
+0.14
+0.13
0.00
-0.34
-0.85
-0.80
-1.20
-1.42
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