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Advanced Placement Chemistry

Lesson Plan # 9

2 class periods

Oxidation-Reduction Reactions

In precipitation reactions cations and anions come together to form an insoluble ionic compound. In neutralization reactions H+ ions and OH- ions come together to form H2O molecules. Ions may also participate in a third kind of reaction. The ions can transfer electrons between them, producing an oxidation-reduction reaction.

Oxidation and Reduction

Have you ever noticed corrosion at the terminals of an automobile battery? What we call corrosion is the conversion of a metal into a metal compound by a reaction between the metal and some substance in its environment.

    • When a metal undergoes corrosion, it loses electrons and forms cations.
    • When an atom, ion, or molecule has become more positively charged (that is, when it has lost electrons), we say that it has been oxidized.
    • Loss of electrons by a substance is called oxidation.
    • The term oxidation is used because the first reactions of this sort to be studied thoroughly were reactions with oxygen.
    • When an atom, ion, or molecule has become more negatively charged (gained electrons), we say that it is reduced; the gain of electrons by a substance is called reduction.
    • When one reactant loses electrons, another reactant must gain them; the oxidation of one substance is always accompanied by the reduction of another as electrons are transferred between them.
    • Reactions in which electrons are transferred between reactants are called oxidation-reduction, or redox reactions.
    • Two good mnemonics for redox reactions: (1) LEO the lion says GER: Lose electrons oxidation, gain electrons reduction; and (2) OIL RIG: Oxidation involves loss of electrons, reduction involves gain of electrons.

Oxidation Numbers

Because electrons are not shown explicitly in chemical equations, we must do some work to determine whether a reaction involves oxidation and reduction. The concept of oxidation numbers (also called oxidation states) was devised as a simple way of keep track of electrons in reactions.

    • We define the oxidation number of an atom in a substance to be the actual charge of the atom if it is a monatomic ion; otherwise, it is the hypothetical charge assigned to the atom using a set of rules.
    • An oxidation-reduction reaction is one in which one or more atoms change oxidation numbers, implying the transfer of electrons.
    • Oxidation occurs when there is an increase in oxidation number, whereas reduction occurs when there is a reduction in oxidation number.
    • We use the following rules for assigning oxidation numbers.

    • For an atom in its elemental form, the oxidation number is always zero. Thus, each H atom in the H2 molecule has an oxidation number of 0, and each P atom in the P4 molecule has an oxidation number of 0.
    • For an monatomic ion, the oxidation number equals the charge on the ion. Thus K+ has an oxidation number of +1, S2- has an oxidation state of -2, and so forth.

    • The alkali metal ions (group 1A) always have a 1+ charge, and therefore the alkali metals always have an oxidation number of +1 in their compounds.
    • The alkaline earth metals (group 2A) are always +2.
    • Aluminum (group 3A) is always +3 in their compounds.

    • In writing oxidation numbers, we will write the sign before the number to distinguish them from the actual electronic charges, which we write the number first.
    • Nonmetals usually have negative oxidation numbers, although they can sometimes be positive:

    • The oxidation number of oxygen is usually -2 in both ionic and molecular compounds. The major exception is in compounds called peroxides, which contain the O22- ion, giving each oxygen an oxidation number of -1.
    • The oxidation number of hydrogen is +1 when bonded to nonmetals and -1 when bonded to metals.
    • The oxidation number of fluorine is -1 in all compounds. The other halogens have an oxidation number of -1 in most binary compounds. When combined with oxygen, as in oxyanions, they have positive oxidation states.

    • The sum of the oxidation number of all atoms in a neutral compound is zero. The sum of the oxidation numbers in a polyatomic ion equals the charge of the ion.

SAMPLE EXERCISE 4.6

Determine the oxidation state of sulfur in each of the following: (a) H2S; (b) S8; (c) SCl2; (d) Na2SO3; (e) SO42-.

Solution:

  1. Hydrogen has an oxidation number of +1. Because the H2S molecule is neutral, the sum of the oxidation number must equal zero. Letting x equal the oxidation of S, we have 2(+1) + x = 0. Thus, S has an oxidation number of -2.
  2. Because this is an elemental form of sulfur, the oxidation number of S is 0.
  3. Because this is a binary compound, we expect chlorine to have an oxidation number of -1. The sum of oxidation numbers must equal zero. Letting x equal the oxidation number of S, we have x + 2(-1) = 0. Consequently, the oxidation number of S must be +2.
  4. Sodium, an alkali metal, is always found in compounds with an oxidation number of +1. Oxygen has a common oxidation state of -2. Letting x equal the oxidation number of S, we have 2(+1) + x + 3(-2) = 0. Therefore, the oxidation number of S in this compound is +4.
  5. The oxidation state of O is -2. The sum of the oxidation numbers equals -2, the net charge of the SO42- ion. Thus we have x + 4(-2) = -2. From this relation we conclude that the oxidation number of S in this ion is +6.

These examples illustrate that the oxidation number of given element depends on the compound in which it occurs. The oxidation numbers of sulfur, as seen in these examples, range from -2 to +6.

Oxidation of Metals by Acids and Salts

The reaction of a metal with either an acid or a metal salt conforms to the following general pattern:

A + BX AX + B

Example:

Zn(s) + 2HBr(aq) ZnBr2(aq) + H2(g)

This reaction is called displacement reactions because the ion in solution is displaced or replaced through oxidation of an element.

SAMPLE EXERCISE 4.7

  1. Write the balanced molecular and net ionic equations for the reaction of aluminum with hydrobromic acid.
  2. What element is oxidized in the reaction?

Solution:

  1. The given reactants are a metal (Al) and an acid (HBr). To write the chemical equation, we must predict the products. Metals react with acids to form a salt of the metal and elemental hydrogen, H2(g). The salt will contain the cation formed from the metal and the anion of the acid. Aluminum forms Al3+ ions; the anion from hydrobromic acid is Br-, so the salt is AlBr3:
  2. 2Al(s) + 6HBr(aq) 2AlBr3(aq) + 3H2(g)

    Both HBr and AlBr3 are soluble strong electrolytes. Thus the complete ionic equation is

    2Al(s) + 6H+(aq) + 6Br-(aq) 2Al3+(aq) +6Br-(aq) + 3H2(g)

    Because Br- is a spectator ion, the net ionic equation is

    2Al(s) + 6H+(aq) 2Al3+(aq) + 3H2(g)

  3. Because the oxidation number of Al increases, going from 0 to +3, it is oxidized. The H+ is reduced going from an oxidation state of +1 to 0.

The Activity Series

Can we predict whether a certain metal will be oxidized either by an acid or by a particular salt? This question is of practical importance as well as chemical interest. When a metal is oxidized, it appears to be eaten away as it reacts to form various compounds. Extensive oxidation can lead to the failure of metal machinery parts or the deterioration of metal structures.

Different metals vary in the ease with which they are oxidized. A list of metals arranged in order of decreasing ease of oxidation is called an activity series. The metals at the top of the table are most easily oxidized; that is, they react most readily to form compounds. Notice that the alkali metals and alkaline earth metals are at the top. They are called the active metals. The metals at the bottom of the activity series are very stable and form compounds less readily. Notice also that the transition elements from groups 8B and 1B are at the bottom of the list. These metals, which are used in making coins and jewelry, are called noble metals because of their low reactivity.

Activity Series of Metals in Aqueous Solution

Metal

Oxidation Reaction

Lithium

Li(s) Li+(aq) + e-

Potassium

K(s) K+(aq) + e-

Barium

Ba(s) Ba2(aq) + 2e-

Calcium

Ca(s) Ca2+(aq) + 2e-

Sodium

Na(s) Na+(aq) + e-

Magnesium

Mg(s) Mg2+(aq) + 2e-

Aluminum

Al(s) Al3+(aq) + 3e-

Manganese

Mn(s) Mn2+(aq) + 2e-

Zinc

Zn(s) Zn2+(aq) + 2e-

Chromium

Cr(s) Cr3+(aq) + 3e-

Iron

Fe(s) Fe2+(aq) + 2e-

Cobalt

Co(s) Co2+(aq) + 2e-

Nickel

Ni(s) Ni2+(aq) + 2e-

Tin

Sn(s) Sn2+(aq) + 2e-

Lead

Pb(s) Pb2+(aq) + 2e-

Hydrogen

H2(g) 2H+(aq) + 2e-

Copper

Cu(s) Cu2+(aq) + 2e-

Silver

Ag(s) Ag+(aq) + e-

Mercury

Hg(l) Hg2+(aq) + 2e-

Platinum

Pt(s) Pt2+(aq) + 2e-

Gold

Au(s) Au3+(aq) + 3e-

SAMPLE EXERCISE 4.8

Will an aqueous solution of iron(II) chloride oxidize magnesium metal? If so, write the balanced molecular and net ionic equations for the reaction.

Solution:

We are given two substances - an aqueous salt, FeCl2, and a metal, Mg - and asked if they react with each other. To answer this question, we use the activity series. Because Mg is above Fe in the table, we predict that the reaction will occur. The Mg metal will be oxidized to form the magnesium cation, and the iron cation in the salt will be reduced to Fe metal.

To write the equations for the reactions, we must remember the charges on common ions. Magnesium is always present in compounds as Mg2+; the chloride ion is Cl-. The magnesium compound formed in the reaction is MgCl2:

Mg(s) + FeCl2(aq) MgCl2(aq) + Fe(s)

Both FeCl2 and MgCl2 are soluble strong electrolytes and can be written in ionic form. When we do so, we find that Cl- is a spectator ion in the reaction. The net ionic equation is

Mg(s) + Fe2+(aq) Mg2+(aq) + Fe(s)

As the net ionic equation clearly shows, Mg is oxidized and Fe2+ is reduced in this reaction.