R EDOX R EACTIONS Chapter 20

S ECTION 20.1

R EDOX R EACTIONS A chemical rxn in which electrons are transferred from one atom to another is called an oxidation- reduction reaction, or redox rxn. In a redox reaction, the loss of electrons from atoms of a substance is called oxidation. The gain of electrons is called reduction.

C HANGES IN O XIDATION N UMBER You already learned that the oxidation number of an atom is equal to the number of electrons gained or lost in an ionic compound. Oxidation increases an atom’s oxidation number; reduction decreases the oxidation number.

C OMPLEMENTARY P ROCESSES Oxidation and reduction are complementary processes that always occur together. The substance that is reduced in a redox reaction is the oxidizing agent. The substance that is oxidized in a redox rxn is the reducing agent.

E XAMPLE Redox rxn are not limited to reactions in which atoms change to ions or vice versa. For example in the rxn, identify the oxidizing and reducing agent H 2 (g) + Cl 2 (g)  2HCl (g)

P RACTICE P ROBLEMS 1. Identify what is oxidized and what is reduced. Also identify the oxidizing agent and the reducing agent. a. Zn + Ni 2+  Ni + Zn 2+ b. 2I - + Br 2  I 2 + 2Br - c. 2NO  N 2 + O 2 d. 2H 2 + S 2  2H 2 S

D ETERMINING O XIDATION N UMBERS Use these rules to determine the oxidation numbers in redox reactions. 1. The oxidation number of an uncombined atom is zero. 2. The oxidation number of a monatomic ion is equal to the charge on the ion. 3. The oxidation number of the more electronegative atom in a molecule or complex ion is the same as the charge it would have if it were an ion. 4. The most oxidation number of oxygen in compounds is always -2, except in peroxides, where it is -1. When bonded to fluorine the oxidation number of oxygen is 2+.

E XAMPLE Determine the oxidation number of each element in the following compound and ion. a. SrCO 3 b. Cr 2 O 7 2-

P RACTICE P ROBLEMS 2. Determine the oxidation number of the boldface element in each of these compounds. a. Li 2 Si O 3 b. Al 4 C 3 c. Ca H 2 d. Be Se O 4 e. K 2 Ge F 6 f. Al( Cl O 3 ) 3 3. Determine the oxidation number of the boldface element in each of these ions. a. P O 4 3- b. Hg 2 2+ c. H S O 4 - d. Pt Cl 6 2- e. Pu O 2 + f. Te O 3 2-

R EACTIONS AS R EDOX R XN Not all chemical rxns are redox rxns. Most double-displacement rxns are not redox rxns, because there is not transfer of electrons between atoms. Combustion and single-replacement rxns are always redox rxns. Many synthesis and decomposition rxns are redox rxns as well.

S ECTION 20.3

B ALANCING R EDOX R EACTION When balancing redox rxns in addition to balancing the # of each atom, the number of electrons gained or lost must also be balanced. The method used is balancing half-reactions. A half-reaction shows the oxidation half of the rxn, and the reduction half of the rxn in each half-rxn. One half-rxn shows the oxidation and the other shows the reduction.

B ALANCING H ALF -R EACTIONS The steps to balance a redox half-reaction are as follows: 1. Write the net ionic equation for the rxn, omitting spectator ions. 2. Write the oxidation and reduction half-reactions for the net ionic equation. 3. Balance the atoms and charges in each half- reaction. 4. Adjust the coefficients so that the # of electrons lost in the oxidation equals the # of electrons gained in reduction. 5. Add the balanced half-reactions and return spectator ions.

E XAMPLE Use the half-reaction method to balance this equation. K 2 Cr 2 O 7 (aq) + HCl (aq)  CrCl 3 (aq) + KCl (aq) + Cl 2 (g)

P RACTICE Use the half-reaction method to balance. a. I 2 (s) + H 2 SO 3 (aq)  I - (aq) + HSO 4 - (aq) b. Fe 2+ (aq) + MnO 4 - (aq)  Fe 3+ (aq) + Mn 2+ (aq) c. Zn (s) + Cr 2 O 7 2- (aq)  Zn 2+ (aq) + Cr 3+ (aq) d. IO 3 - (aq) + I - (aq)  I 2 (s)