4.4 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. The corrosion shown in Figure 4.12 results from the reaction of battery acid, H2SO4, with the metal clamp.

When a metal undergoes corrosion, it loses electrons and forms cations. For example, calcium is vigorously attacked by acids to form calcium ions, Ca2+:

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[4.24]

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. Thus, Ca, which has no net charge, is oxidized (undergoes oxidation) in Equation 4.24, forming Ca2+.

The term oxidation is used because the first reactions of this sort to be studied thoroughly were reactions with oxygen. Many metals react directly with O2 in air to form metal oxides. In these reactions the metal loses electrons to oxygen, forming an ionic compound of the metal ion and oxide ion. For example, when calcium metal is exposed to air, the bright metallic surface of the metal tarnishes as CaO forms:

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[4.25]

The analogous reaction of iron with oxygen in the presence of water is responsible for the formation of rust, an oxide of iron.

As Ca is oxidized in Equation 4.25, oxygen is transformed from neutral O2 to two O2– ions. 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, as shown in Figure 4.13. Reactions in which electrons are transferred between reactants are called oxidation-reduction, or redox reactions.

Figure 4.13 Oxidation is the loss of electrons by a substance; reduction is the gain of electrons by a substance. Oxidation of one substance is always accompanied by reduction of another.

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 keeping 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:

  1. 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.
  2. For any 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. Similarly, the alkaline earth metals (group 2A) are always +2, and 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 with the number first.)
  3. Nonmetals usually have negative oxidation numbers, although they can sometimes be positive:
    1. 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.
    2. The oxidation number of hydrogen is +1 when bonded to nonmetals and -1 when bonded to metals.
    3. 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.
  4. The sum of the oxidation numbers 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. For example, in the hydronium ion, H3O+, the oxidation number of each hydrogen is +1 and that of oxygen is -2. Thus the sum of the oxidation numbers is 3(+1) + (-2) = +1, which equals the net charge of the ion. This rule is very useful in obtaining the oxidation number of one atom in a compound or ion if you know the oxidation numbers of the other atoms, as illustrated in Sample Exercise 4.6.

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 (a) Hydrogen has an oxidation number of +1 (rule 3b). Because the H2S molecule is neutral, the sum of the oxidation numbers must equal zero (rule 4). Letting x equal the oxidation number of S, we have 2(+1) + x = 0. Thus, S has an oxidation number of -2.

(b) Because this is an elemental form of sulfur, the oxidation number of S is 0 (rule 1).

(c) Because this is a binary compound, we expect chlorine to have an oxidation number of -1 (rule 3c). The sum of the oxidation numbers must equal zero (rule 4). Letting x equal the oxidation number of S, we have x +2(-1) = 0. Consequently, the oxidation number of S must be +2.

(d) Sodium, an alkali metal, is always found in compounds with an oxidation number of +1 (rule 2). Oxygen has a common oxidation state of -2 (rule 3a). 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.

(e) The oxidation state of O is -2 (rule 3a). The sum of the oxidation numbers equals -2, the net charge of the SO42– ion (rule 4). 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 a 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.

PRACTICE EXERCISE

What is the oxidation state of the boldfaced element in each of the following: (a) P2O5; (b) NaH; (c) Cr2O72–; (d) SnBr4; (e) BaO2? Answers: (a) +5; (b) -1; (c) +6; (d) +4; (e) -1

Oxidation of Metals by Acids and Salts

There are many kinds of redox reactions. For example, combustion reactions (Section 3.2) are redox reactions because elemental oxygen is converted to compounds of oxygen. In this chapter we consider the redox reactions between metals and either acids or salts. In Chapter 20 we will examine more complex kinds of redox reactions.

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

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[4.26]

Examples:

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

Many metals undergo displacement reactions with acids, producing salts and hydrogen gas. For example, magnesium metal reacts with hydrochloric acid (see Figure 4.14) to form magnesium chloride and hydrogen gas. To show that oxidation and reduction have occurred, the oxidation number for each atom is shown below the chemical equation for this reaction:

AAAUWNUO

[4.27]

Notice that the oxidation number of Mg changes from 0 to +2. The increase in the oxidation number indicates that the atom has lost electrons and has therefore been oxidized. The H+ ion of the acid decreases in oxidation number from +1 to 0, indicating that this ion has gained electrons and has therefore been reduced. The Cl ion maintains an oxidation number of -1 and is a spectator ion in the reaction resulting in the following net ionic equation:

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[4.28]

Metals can also be oxidized by aqueous solutions of various salts. For example, iron metal is oxidized to Fe2+ by aqueous solutions of Ni2+, such as Ni(NO3)2(aq):

Molecular equation:

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[4.29]

Net ionic equation:

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[4.30]

Notice that the oxidation of Fe to form Fe2+ in this reaction is accompanied by the reduction of Ni2+ to Ni. Remember: Whenever one substance is oxidized, some other substance must be reduced.

SAMPLE EXERCISE 4.7

(a) Write the balanced molecular and net ionic equations for the reaction of aluminum with hydrobromic acid. (b) What element is oxidized in the reaction?

SOLUTION (a) 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:

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

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

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(b) 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.

PRACTICE EXERCISE

(a) Write the balanced molecular and net ionic equations for the reaction between magnesium and cobalt(II) sulfate. (b) What is oxidized and what is reduced in the reaction? Answers: (a) Mg(s) + CoSO4(aq) MgSO4(aq) + Co(s); Mg(s) + Co2+(aq) Mg2+(aq) + Co(s); (b) Mg is oxidized and Co2+ is reduced

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. For example, Equation 4.29 informs us that it would be unwise to store a solution of nickel nitrate in an iron container because the solution would dissolve the container. 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. For example, zinc metal is oxidized by aqueous solutions of copper(II), but silver metal is not. We conclude that zinc loses electrons more readily than does silver; that is, zinc is easier to oxidize than silver.

A list of metals arranged in order of decreasing ease of oxidation is called an activity series. Table 4.5 gives the activity series in aqueous solution for many of the most common metals. Hydrogen is also included in the table. 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.

The activity series can be used to predict the outcome of reactions between metals and either metal salts or acids. Any metal on the list can be oxidized by the ions of elements below it. For example, copper is above silver in the series. Thus, copper metal will be oxidized by silver ions, as pictured in Figure 4.15:

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[4.31]

Once again we note that the oxidation of copper is accompanied by a reduction reaction, namely, the reduction of silver ions to silver metal. The silver metal is evident on the surface of the copper wires in Figure 4.15(b) and (c). The copper(II) nitrate produces a blue color in the solution, which is most evident in part (c).

Only those metals above hydrogen in the activity series are able to react with acids to form H2. For example, Ni reacts with HCl(aq) to form H2:

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[4.32]

Because elements below hydrogen in the activity series are not oxidized by H+, Cu does not react with HCl(aq). Interestingly, copper does react with nitric acid, as shown earlier in Figure 1.14. However, this reaction is not a simple oxidation of copper by the H+ ions of the acid. Instead, the metal is oxidized to Cu(II) by the nitrate ion of the acid, accompanied by the formation of brown nitrogen dioxide, NO2(g):

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[4.33]

You might wonder what substance is being reduced as copper is oxidized in Equation 4.33. In this case the NO2 results from the reduction of NO3. We will examine reactions of this type in more detail in Chapter 20.

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, Table 4.5. 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 reaction, 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:

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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

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

PRACTICE EXERCISE

Which of the following metals will be oxidized by Pb(NO3)2: Zn, Cu, Fe? Answer: Zn and Fe