Chem 171
Oxidation Numbers
Balancing Redox Equations by the Method of Balancing Half Reactions
To assign an oxidation number (Nox):
 The oxidation number of an elemental substance is zero. examples: Na, Cl2, Ne, Fe
 The oxidation number for a monatomic ion is equal to the charge on that ion. examples: Na+, Cl—, Fe2+, Fe3+
 The sum of the oxidation numbers of all atoms in a species must equal the total charge on that species. examples: for
CO2, N oxC + 2•N oxO = 0; for CO32—, N oxC + 3•N oxO = —2
 Fluorine in polyatomic species is assigned Nox = —1.
 Hydrogen in polyatomic species is assigned Nox = +1 with nonmetals (covalent or molecular compounds) and Nox = —1
with metals (ionic compounds - hydride ion is H—).
 Oxygen in polyatomic species is usually assigned Nox = —2. Exceptions include compounds of O with F (assign F an
N ox of —1 first, then determine Nox of O); peroxides where oxygen is present as the peroxide anion, O22—, with N ox = —1
(example: H2O2); superoxides where oxygen is present as the superoxie anion, O2—, with N ox = —1/2 (example: KO2).
Steps for balancing equations for reactions taking place in acidic medium:
Consider the reaction between permanganate ion and oxalic acid to form manganese (II) ion and carbon dioxide. Write the
overall balanced redox equation for this reaction.
1.
Write the skeletal equation for the reaction; identify what is oxidized and reduced.
MnO4— + H2C 2O4  Mn2+ + CO2
C is oxidized, +3  +4 ∴ H2C 2O4 is the reducing agent
Mn is reduced, +7  +2 ∴ MnO4 is the oxidizing agent
2.
Write the skeletal half reactions for the oxidation and reduction.
oxidation: H2C 2O4  CO2
reduction: MnO4—  Mn2+
3.
Balance elements other than H and O.
oxidation: H2C 2O4  2 CO2
reduction: MnO4—  Mn2+
4.
Balanced O by adding H2O.
oxidation: H2C 2O4  2 CO2
reduction: MnO4—  Mn2++ 4 H2O
5.
Balance H by adding H+.
oxidation: H2C 2O4  2 CO2 + 2 H+
reduction: 8 H+ + MnO4—  Mn2++ 4 H2O
6.
Balance charge by adding electrons (e—).
oxidation: H2C 2O4  2 CO2 + 2 H+ + 2 e—
reduction: 5 e— + 8 H+ + MnO4—  Mn2++ 4 H2O
* At this point the individual half reactions are balanced *
7.
Prepare to add the half reacitons together by multiplying by an appropriate factor so that the electrons cancel out.
5 x [ oxidation: H2C 2O4  2 CO2 + 2 H+ + 2 e—]
2 x [reduction: 5 e— + 8 H+ + MnO4—  Mn2++ 4 H2O ]
oxidation: 5 H2C 2O4  10 CO2 + 10 H+ + 10 e—
reduction: 10 e— + 16 H+ + 2 MnO4—  2 Mn2++ 8 H2O
8.
Add 1/2 reactions together, simplify (cancel species that are the same on both sides of the equation) ; add physical states.
5 H2C 2O4 + 10 e— + 16 H+ + 2 MnO4—  10 CO2 + 10 H+ + 10 e— + 2 Mn2+ + 8 H2O
cancel e—, reduce H+
5 H2C 2O4 (aq) + 6 H+ (aq) + 2 MnO4— (aq)  10 CO2 (g) + 2 Mn2+ (aq) + 8 H2O (l)
Steps for balancing equations for reactions taking place in basic medium:
Consider the reaction between dichlorine heptaoxide gas and hydrogen peroxide to form chlorite ion and oxygen gas.
1.
Write the skeletal equation for the reaction; identify what is oxidized and reduced.
Cl2O7 + H2O2  ClO2— + O2
O is oxidized, —1  0 ∴ H2O2 is the reducing agent
Cl is reduced, +7  +3 ∴ Cl2O7 is the oxidizing agent
2.
Write the skeletal half reactions for the oxidation and reduction.
oxidation: H2O2  O2
reduction: Cl2O7  ClO2—
3.
Balance elements other than H and O.
oxidation: H2O2  O2
reduction: Cl2O7  2 ClO2—
4.
Balanced O by adding H2O.
oxidation: H2O2  O2
reduction: Cl2O7  2 ClO2— + 3 H2O
5.
Balance H by adding H+. Then, add that same number of OH— on both sides of the half reaction. Simplify by
recognizing that H+ + OH— = H2O
oxidation: 2 OH— + H2O2  O2 + 2 H+ + 2 OH—
reduction: 6 OH— + 6 H+ + Cl2O7  2 ClO2— + 3 H2O + 6 OH—
oxidation: 2 OH— + H2O2  O2 + 2 H2O
reduction: 6 H2O + Cl2O7  2 ClO2— + 3 H2O + 6 OH—
6.
Balance charge by adding electrons (e—).
oxidation: 2 OH— + H2O2  O2 +2 H2O + 2 e—
reduction: 8 e— + 6 H2O + Cl2O7  2 ClO2— + 3 H2O + 6 OH—
* At this point the individual half reactions are balanced *
7.
Prepare to add the half reacitons together by multiplying by an appropriate factor so that the electrons cancel out.
4 x [ oxidation: 2 OH— + H2O2  O2 +2 H2O + 2 e— ]
1 x [reduction: 8 e— + 6 H2O + Cl2O7  2 ClO2— + 3 H2O + 6 OH— ]
oxidation: 8 OH— + 4 H2O2  4 O2 +8 H2O + 8 e—
reduction: 8 e— + 6 H2O + Cl2O7  2 ClO2— + 3 H2O + 6 OH—
8.
Add 1/2 reactions together, simplify (cancel species that are the same on both sides of the equation) ; add physical states.
8 OH— + 4 H2O2 + 8 e— + 6 H2O + Cl2O7  4 O2 + 8 H2O + 8 e— + 2 ClO2— + 3 H2O + 6 OH—
cancel e—, reduce H2O and OH—
2 OH— (aq) + 4 H2O2 (aq) + Cl2O7 (g)  4 O2 (g) + 2 ClO2— (aq) + 5 H2O (l)