Which Of The Following Reactions Would Be Classified As Oxidation-reduction? Check All That Apply.

Oxidation-Reduction: A Second Way to Classify Reactions

Another way of classifying reactions separates them into only two groups: (1) those that do not involve a change in oxidation number but do result in a decrease in the number of ions in solution and (2) those that involve a transfer of electrons and changes in oxidation number.

Those that result in a decrease in the number of ions in solution are usually double-displacement reactions (Section 8.2D). In a neutralization reaction, hydrogen ion combines with hydroxide ion to form the covalent, un-ionized compound water, thus decreasing the number of ions in solution. In a precipitation reaction, the insoluble product removes ions from the solution. Other reactions in this category are those that form weak electrolytes (Section 7.5C), such as acetic acid or aqueous ammonia (ammonium hydroxide). Examples are the reaction between sodium acetate and hydrochloric acid

NaC₂H₃O₂(aq) + HCl(aq) NaCl(aq) + HC₂H₃O₂(aq)

and the reaction between sodium hydroxide and ammonium chloride

NaOH(aq) + NH₄Cl(aq) NaCl(aq) + NH₃(aq) + H₂O(l)

Those reactions that involve a transfer of electrons include combination, displacement, and decomposition reactions. Reactions that involve a transfer of electrons are known as oxidation-reduction or redox reactions. The reaction of sodium with chlorine is a redox reaction:

2 Na + Cl₂ 2 NaCl

Each chlorine atom gains an electron to form a chloride ion:

Each sodium atom loses an electron to form a sodium ion:

The element that loses electrons is oxidized. In the reaction of sodium with chlorine, sodium is oxidized. The element that gains electrons is reduced. In this reaction, chlorine is reduced.

Displacement reactions are usually oxidation-reduction reactions. A typical displacement reaction is that of copper with silver nitrate:

Cu(s) + 2 AgNO₃(aq) 2 Ag + Cu(NO₃)₂(aq)

In this reaction, the copper loses electrons (is oxidized):

Cu⁰ Cu²⁺ + 2 e^-

and the silver ion gains electrons (is reduced):

Ag⁺ + e^- Ag⁰

Oxidation without reduction is impossible. If an element in a reaction loses electrons, another element in the reaction must gain electrons.

A. Identifying Oxidation-Reduction Reactions
Oxidation numbers (Section 6.2A) have many uses, but the one that concerns us here is their role in determining whether or not a particular reaction involves oxidation-reduction. In an oxidation-reduction reaction, at least two oxidation numbers change. The element that is oxidized increases its oxidation number, and the element that is reduced decreases its oxidation number. In the reaction of sodium with chlorine, sodium atoms are oxidized to sodium ions; the oxidation number of sodium increases from 0 to +1. At the same time, chlorine is reduced to chloride ions; the oxidation number of chlorine decreases from 0 to -1. By assigning oxidation numbers to all the elements in the reactants and the products of a reaction, we can determine whether the reaction results in a change in oxidation number. If a change does occur, the reaction is an oxidation-reduction reaction. For example, consider the reaction between magnesium and oxygen:

Below each element in each substance in the equation, we have written its oxidation number. The oxidation number of magnesium increases from 0 to +2; magnesium is oxidized. The oxidation number of oxygen decreases from 0 to -2; oxygen is reduced. Thus, the reaction of magnesium with oxygen is an oxidation-reduction reaction.

A reaction that is not an oxidation-reduction reaction will cause no changes in oxidation numbers. Consider the reaction of sodium hydroxide with hydrochloric acid:

Below each element in the equation, we have written its oxidation number. Because each element has the same oxidation number in the products as it does in the reactants, we know that this neutralization reaction is not an oxidation-reduction reaction.

Example

For the following reactions decide: (1) Is it an oxidation-reduction? (2) If so, which element is oxidized and which element is reduced?

a. Bromine can be prepared by bubbling chlorine gas through a solution of sodium bromide. The equation for this reaction is:

b. If you blow through a straw into limewater, the solution becomes milky. In chemical terms, if carbon dioxide is bubbled through a solution of calcium hydroxide in water a milky white precipitate of calcium carbonate forms:

Solution

Reaction a

1. Write the oxidation number under each element in the equation:

2. Do any elements change oxiation number? Yes, both chlorine and bromine do. Therefore, this reaction is oxidation-reduction.

3. The oxidation number of chlorine changes from 0 to -1: chlorine is reduced. The oxidation number of bromine changes from -1 to 0: bromine is oxidized.

Reaction b

Below each element we have written its oxidation number. None of these nubers changed during the reaction: the reaction is not an oxidation-reduction.

In an oxidation-reduction reaction, the substance that gains electrons is the oxidizing agent. The substance that loses electrons is the reducing agent. In the reaction of magnesium with oxygen,

2 Mg(s)
0
loses electrons
is oxidized to Mg²⁺
is the reducing agent

O₂(g)
0
gains electrons
is reduced to 2 O^2-
is the oxidizing agent

Mg
+2,

O
-2

Example

In the reaction of sodium with chlorine to form sodium chloride, which substance is the oxidizing agent? Which is the reducing agent?

Solution

Write the equation for the reaction and assign oxidation numbers:

Because chlorine changes oxidation number form 0 to -1, it is reduced: it is the oxidizing agent. Because sodium changes oxidation number from 0 to +1, it is oxidized; it is the reducing agent.

The characteristics of oxidation and reduction are summarized in Table 8.4. Chapter 14 covers oxidation-reduction in greater depth. Our discussion here is merely an introduction.

**TABLE 8.4** Characteristics of oxidation-reduction reactions
Substance oxidized	Substance reduced
loses electrons	gains electrons
attains a more positive oxidation number	attains a more negative oxidation number
is the reducing agent	is the oxidizing agent

B. Combustion Reactions
Combustion reactions are a special type of oxidation-reduction reaction. We correctly associate combustion reactions with burning. In the usual combustion reaction, the elements in the reacting compound combine with oxygen to form oxides as in the combustion of propane to form carbon dioxide and water:

2 C₃H₈(l) + 10 O₂(g) 6 CO₂(g) + 8 H₂O(l)

Figure 8.6 shows the combustion of gasoline in oxygen.

The above reactions take place when an adequate supply of oxygen is present. In the absence of an adequate supply of oxygen, carbon monoxide may be formed instead of carbon dioxide.

2 C₃H₈(l) + 7 O₂(g) 6 CO₂(g) + 8 H₂O(l)

FIGURE 8.6 Combustion as illustrated by burning gasoline storage tanks. The combustion of gasoline or other petroleum products is usually accompanied by yellow flames and dense black smoke.

Example

Write the balanced equations for the complete combustion in oxygen of:

a. sulfur to form sulfur dioxide

b. butane, C₄H₁₀

c. ethyl alcohol, C₂H₅OH

Solution

Note that the physical states of these substances are not given. They are therefore omitted from the equations.

a. Write the formulas of the reactants, sulfur and oxygen, and of the product, SO₂.

b. Butane contains only carbon and hydrogen and the combustion is complete, so the products are carbon dioxide, and water. The unbalanced equation is:

Four atoms of carbon on the left give four molecules of carbon dioxide on the right. Ten atoms of hydrogen on the left give 5 molecules of water on the right which requires thirteen atoms of oxygen (6 1/2 molecules) on the left.

To write the equation using only whole numbers of molecules, we must multiply through by 2 to get:

c. Ethyl alcohol contains carbon, hydrogen, and oxygen. The producs of complete combustion will be carbon dioxide and water.

Initial balancing gives two atoms of carbon on the left, two molecules of carbon dioxide on the right; six atoms of oxygen on the right, 3 1/2 molecules of oxygen on the left:

We multiply by 2 to clear the fraction and give the balanced equation.