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University of Pennsylvania Department of Chemistry
Chemistry 101 - Engineering Sections
Balancing Oxidation-Reduction (Redox) Reactions
I. The basic process: Oxidation-Reduction Reaction
A process which involves a "transfer of electrons (e- ) " when species react chemically to
form products.
II. Definition of terms:
Oxidation = when a species "loses" or supplies electrons to another species. This "supplying"
species is said to be oxidized. It is also referred to as the reducing agent or reductant.
Reduction = when a species accepts electrons from another species. This "accepting" species is
said to be reduced. It is also referred to as the oxidizing agent. Thus, the species that undergoes
oxidation is the reducing agent because it supplies the electrons to the species that is reduced.
Alternatively, the species that undergoes reduction is the oxidizing agent or oxidant
because it accepts electrons from the species that is oxidized.
Oxidation Number (O.N.) = a parameter which enables one to determine if a species is
oxidized or reduced as the species undergoes chemical reaction.
III. Oxidation Numbers (ON) [or Oxidation States (OS)]
Some rules for determining Oxidation Numbers (ON) (See the text, for a more
detailed list.)
1. Free elements: ON of each atom is 0 (zero).
Ex.: Zn(s) - ON of Zn is 0 ; H2(g) - ON of each H atom is 0 (zero) ;
S8(s) - ON of each S is 0 (zero).
2. Ions consisting of one (1) atom: ON = charge on ion.
Ex.: Cl- - ON of Cl is -1 ; Al+3 - ON of Al is +3 .
3. Hydrogen, in compounds with other elements: ON = +1, except in hydrides:
LiH, NaH, CaH2, etc. In these cases, ON = -1.
4. Oxygen, in compounds with other elements: ON = -2, except in peroxides:
H2O2 , Na2O2, etc. In these cases, ON = -1. In KO2 ON of O is -1/2.
5. Algebraic sum of ON's in a neutral compound must be zero (0) ; in a polyatomic ion
the sum of the ON's equals the charge on the ion.
Ex.: Li2O : ON of each Li = +1; ON of O = -2; sum = (2)(+1) + (-2) = 0.
H3PO4 : Since the molecule is neutral, sum of ON's = 0 (zero).
\ 0 = (3)(+1) + (4)(-2) + (ON of P) ; thus, ON of P = +5.
Show: ON of Mn in MnO4-1 is +7; and ON of Cr (each one), in Cr2O7-2 is +6.
6. ON of each F in F2 = 0 (zero). ON of F in compounds with other atoms = -1.
"Always"!! We will see why when we discuss atomic structure and chemical bonding.
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IV. Balancing Redox Equations:
A. Oxidation States Method
This method will be presented by way of an example. The sample reaction is the combustion
of methanol to carbon dioxide and water.
Unbalanced Reaction: CH3OH(l) + O2(g) ------> CO2(g) + H2O(l) .
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Step 1 - Determine if the reaction is a redox reaction.
Using the rules, we have oxidation state (OS) of H in methanol is +1 and the OS of O
is -2. So, the OS of C must be -2 by the sum rule. The OS of O in O2 is 0. In CO2
the OS of O is -2. From the sum rule, the OS of C is +4. The OS of O in H2O is -2 and the
OS of H in water is +1. Consolidating this information, we conclude the following:
• C : The OS changes from -2 in methanol to +4 in carbon dioxide. C is oxidized and
so methanol is the reducing agent or reductant.
• O : For each O, the OS changes from 0 in O2 to -2 in water (or carbon dioxide).
O is reduced and so O2 is the oxidizing agent or oxidant.
Thus, this is a redox reaction.
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Step 2 - Balance INCREASE in OS (of C (↑ of 6)) of oxidized species by the
DECREASE in OS of reduced species (of each O (Ø of 2 each)).
Need 3x as much O in O2 as have C in methanol to balance the change.
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CH3OH(l) + 2 O2(g) -----> CO2(g) + 2 H2O(l) . The 2 in front of H2O is
finally inserted to balance atoms of each element on both sides of the reaction.
B. Balancing Redox Reactions Using the Half-Reaction Method
This method will be presented by way of an example. This example will contain another
common situation - oxidation-reduction processes often occur in either (aqueous) acidic
conditions or (aqueous) basic conditions. The following example reaction will be the
oxidation of ethanol (C2H5OH) by dichromate anion (Cr2O7-2) in acidic aqueous solution.
The products are acetic acid (CH3CO2H) and the Cr+3 cation. The physical states of all
species will be included in the initial unbalanced reaction and in the final balanced reaction. To
avoid clutter we will leave out the physical states in the intermediate steps.
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Unbalanced Rxn: C2H5OH (aq) + Cr2O7-2(aq) -----> CH3CO2H (aq) + Cr+3(aq)
(orange-red)
(green)
Step 1 - What is oxidized/reduced?
Using the rules, we have oxidation state (OS) of H in ethanol and in acetic acid is +1. The OS
of O is -2 in ethanol, in the dichromate anion, and in acetic acid. So, the OS of each C
must be -2 in ethanol by the sum rule. Also from the sum rule, the OS of each Cr is +6 in the
dichromate anion. The OS of each carbon in the acetic acid is 0 by the sum rule. Finally,
the OS of Cr in the Cr+3 cation is +3. Consolidating this information, we conclude the
following:
• C : The OS changes from -2 for each C in ethanol to 0 in acetic acid. C is oxidized and
so ethanol is the reducing agent or reductant.
• Cr : For each Cr, the OS changes from +6 in Cr2O7-2 to +3 in the Cr+3 cation.
Cr is reduced and so Cr2O7-2 is the oxidizing agent or oxidant.
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Step 2 - Write out each half-reaction (reductant & associated product(s) and oxidant &
associated product(s)). Deal with stoichiometry of species changing their OS.
Material balance each reaction first with respect to species undergoing OS change.
Oxidation:
C2H5OH -----> CH3CO2H (already balanced for C)
Reduction:
Cr2O7-2 -----> 2 Cr+3
(now - balanced for Cr)
Balance OS change by appropriate addition of electrons.
• Each C atom changes OS form -2 to 0. This gives rise to an overall loss of 4 electrons,
(i.e., (2)(-2) ---> 0). Thus we add 4 electrons to the products side of the oxidation
half-reaction.
• Each Cr atom changes OS form +6 to +3. This gives rise to an overall gain of 6
electrons, (i.e., (2)(+6) ---> (2)(+3)). Thus we add 6 electrons to the reactants side
of the reduction half-reaction.
Oxidation:
C2H5OH -----> CH3CO2H + 4 eReduction:
6 e- + Cr2O7-2 -----> 2 Cr+3
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Step 3 - Balance each half-reaction for:
(i) Charge
(ii) Atoms
• In acid: Add H+ & H2O as needed.
• In base: Add OH- & H2O as needed.
Oxidation: (i) Add 4 H+ to right-hand-side to balance charge.
(ii) Add H2O to left-hand-side to balance mass [H's & O's].
[A]
C2H5OH + H2O -----> CH3CO2H + 4 e- + 4 H+
(Net charge of 0 on each side of reaction [A].)
Reduction: (i) Add 14 H+ to left-hand-side to balance charge.
(ii) Add 7 H2O to right-hand-side to balance mass [H's & O's].
[B]
6 e- + Cr2O7-2 + 14 H+ -----> 2 Cr+3 + 7 H2O
(Net charge of +6 on each side of reaction [B].)
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Step 4 - Multiply half-reactions by the smallest integers to balance the # of
electrons ... then add reactions and simplify.
Here, to balance # of electrons: [A] x 3 &
[B] x 2 Add and simplify.
-----> 3 CH3CO2H + 12 e- + 12 H+
[A] x 3:
3 C2H5OH + 3 H2O
[B] x 2:
12 e- + 2 Cr2O7-2 + 28 H+ -----> 4 Cr+3 + 14 H2O
3 C2H5OH(aq) + 2 Cr2O7-2(aq) + 16 H+ (aq) ----->3 CH3CO2H(aq) + 4 Cr+3(aq) + 11 H2O(l)
Double-check for yourself that atoms and net charge are balanced on each side of the reaction.
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Another Example - Disproportionation Reactions
A redox reaction where just a single substance acts as both oxidant and reductant.
The example is the disproportionation of the aqueous manganate anion (MnO4-2) into
solid manganese (IV) oxide (MnO2) and the aqueous permangante anion (MnO4-) in
aqueous acid solution. This time we will display the steps without detailed comment.
Unbalanced Reaction: MnO4-2(aq) -----> MnO2(s) + MnO4-(aq) .
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Step 1 - What is oxidized/reduced?
The OS of O is -2 throughout. From the sum rule, we have the following.
Mn in MnO4-2 has an OS of +6, the Mn in MnO2 has an OS of +4, and the Mn in
MnO4- has an OS of +7. As revealed above, Mn is both oxided and reduced.
• MnO4-2 / MnO4- : The OS of Mn changes from+6 to +7. Mn is oxidized in this case.
• MnO4-2 / MnO2 : The OS of Mn changes from+6 to +4. Mn is reduced in this case.
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Step 2 - Write out each half-reaction. Deal with stoichiometry of species changing
their OS. Material balance each reaction first with respect to species undergoing
OS change.
Oxidation:
MnO4-2 -----> MnO4(already balanced for Mn)
-2
Reduction:
MnO4 -----> MnO2
(already balanced for Mn)
Balance OS change by appropriate addition of electrons.
• In the oxidation half-reaction, the change in OS of Mn is +1, i,e., an overall loss of 1
electron. Thus we add 1 electron to the products side of the oxidation half-reaction.
• In the reduction half-reaction, the change in OS of Mn is -2, i,e., an overall loss of 2
electrons. Thus we add 2 electrons to the reactants side of the reduction half-reaction.
Oxidation:
Reduction:
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MnO4-2 -----> MnO4- + e2 e- + MnO4-2 -----> MnO2
Step 3 - Balance each half-reaction for:
(i) Charge
(ii) Atoms
• In acid: Add H+ & H2O as needed.
Oxidation: (i) No H+ needed to balance charge.
(ii) No H2O needed to balance mass [H's & O's].
[A]
MnO4-2 -----> MnO4- + e(Net charge of -2 on each side of reaction [A].)
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Step 3 - Continued:
Reduction: (i) Add 4 H+ to left-hand-side to balance charge.
(ii) Add 2 H2O to right-hand-side to balance mass [H's & O's].
[B]
2 e- + MnO4-2 + 4 H+ -----> MnO2 + 2 H2O
(Net charge of zero (0) on each side of reaction [B].)
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Step 4 - Multiply half-reactions by the smallest integers to balance the # of
electrons ... then add reactions and simplify.
Here, to balance # of electrons: [A] x 2 &
[B] - No change. Add and simplify.
[A] x 2:
2 MnO4-2
[B] unchanged: 2 e- + MnO4-2 + 4 H+
-----> 2 MnO4- + 2 e-----> MnO2 + 2 H2O
3 MnO4-2(aq) + 4 H+ (aq) -----> 2 MnO4-(aq) + MnO2(s) + 2 H2O(l)
Double-check for yourself that atoms and net charge are balanced on each side of the reaction.