REDOX REACTIONS (honors)
Oxidation Number: A number assigned to an element in a molecule based on the distribution of
electrons in that molecule. The same element can have very different properties in different
The oxidation number of any uncombined element is 0 Ox. # of Na(s) is 0
The ox. # of a monatomic ion equals the charge of the ion Ox. # of Cl- is –1
The more electronegative element in a binary compound is Ox. # of O in NO is –2
assigned the # equal to the charge it would have if it were an ion
The ox. # of fluorine in a compound is always –1 Ox. # of F in LiF is –1
Oxygen has an ox. # of –2 unless it is combined with F (when it is Ox. # of O in NO2 is –2
+2), or it is in a peroxide like H2O2 (when it is –1) Ox. # of O in OF2 is +2
Ox. # of O in Na2O2 is –1
The ox. # of H in most compounds is +1 unless it is combined with Ox. # of H in H2O is +1
a metal, in which case it is -1 Ox. # of H in LiH is –1
In compounds, Group 1 & 2 elements and Al have ox. #s of +1,+2, Ox. # of Ca in CaCO3 is +2
The sum of ox. # of all atoms in a neutral compound is 0 Ox. # of C in CaCO3 is +4
The sum of the ox. # of all atoms in a polyatomic ion equals the Ox. # of P in H2PO4- in +5
charge of the ion
*See worksheets for more practice/examples
Oxidation is a reaction in which the atoms or ions of an element experience an increase in
oxidation state. An increase in oxidation state means that the oxidation number becomes more
positive. Therefore, the species has lost electrons.
Ex: Na Na+ + e-
Reduction is a reaction in which the oxidation state of an element decreases. A decrease in
oxidation sate means that the oxidation number becomes more negative. Therefore, the species
has gained electrons.
Ex: Cl2 + 2e- 2Cl-
Since oxidation is the losing of electrons and reduction is the gaining of electrons, they have to
occur simultaneously and the number of electrons produced in oxidation must equal the number of
electrons acquired in reduction. Any chemical process in which elements undergo changes in
oxidation number is an oxidation-reduction reaction, or redox reaction for short. The part of the
reaction involving oxidation or reduction alone can be written as a half-reaction. The overall
equation for a redox reaction is the sum of the two half-reactions.
Ex: 3Cu 3Cu2+ + 6e- (oxidation half-reaction)
+5 -2 +1 +2 -2 +1 -2
2NO3- + 6e- + 8H+ 2NO + 4H2O (reduction half-reaction)
0 +5 +2 +2
3Cu + 2NO3- + 8H+ 3Cu2+ + 2NO + 4H2O (redox reaction)
A reducing agent is a substance that has the potential to cause another substance to be reduced.
They lose electrons. An oxidizing agent is a substance that has the potential to cause another
substance to be oxidized. They gain electrons.
Term Change in Oxidation # Change in electrons
Oxidized In a positive direction Loss of electrons
Reduced In a negative direction Gain of electrons
Oxidizing Agent In a negative direction Gain of electrons
Reducing Agent In a positive direction Loss of electrons
Half-Reaction Balancing Method for Redox Reactions
Oxidation-reduction reactions in aqueous solution are often complicated, and are therefore often
difficult to balance by simple inspection. The steps below outline a special technique for balancing
redox reactions, called the half-reaction method.
If the reaction takes place in an acidic solution:
1. Write separate equations for the oxidation and reduction reactions. These are the half
2. For each half reaction:
a. Balance all elements except hydrogen and oxygen (the major species)
b. Balance oxygen using H2O
c. Balance hydrogen using H+
d. Balance the charge using electrons mnemonic: MAJOR OH –
3. If necessary, multiply one or both half reactions by an integer to equalize the number of
electrons transferred between the two half reactions.
4. Add the half reactions, and cancel identical species.
5. CHECK to see that both atoms and charge are balanced.
Example: H+(aq) + Cr2O7 2-(aq) + C2H5OH (l) Cr3+(aq) + CO2 (g) + H2O (l)
If the reaction takes place in a basic solution:
Follow steps 1 – 5 above as if H+ ions were present. Then:
6. Add OH- in the same amount as H+ to each side of the equation. (We want to eliminate H+ by
adding OH- and forming H2O.)
7. Form H2O molecules by combining H+ and OH- and eliminate the # of water molecules that
appear on both sides of the equation.
Example: Ag (s) + CN-(aq) + O2 (g) Ag(CN)2 –(aq) + H2O (l)