2. Oxidation-Reduction Reactions

All chemical reactions can be divided into two groups. In those of the first group, the oxidation state of all the elements in the reactants remains constant, while in those of the second group, the oxi­dation state of one or several elements changes.

We can exemplify the reactions of the first group by the neutralization reaction:

HCl + NaOH = NaCl + H₂O

An example of a reaction of the second group is the reaction of metal with an acid:

Zn + 2HC1 = ZnCl₂ + H₂↑.

If in the neutralization reaction no element changes its oxidation number, in the second example the oxidation number of the Zinc changes from 0 to +2, and of the Hydrogen from +1 to 0.

Reactions as a result of which the oxidation numbers of elements change are called oxidation-reduction (redox) reactions.

Oxidation-reduction reactions are of great importance in biological systems. Photosynthesis, breathing, digestion - all these are chains of oxidation-reduction reactions. In engineering, the significance of redox reactions is also great. For example, the entire metallurgical industry is based on oxidation-reduction pro­cesses during which metals are recovered from natural compounds.

Simple example of an oxidation-reduction reaction is the reac­tion of formation of an ionic compound from elementary substances, for example, the reaction of Sodium with Chlorine:

2Na + Cl₂ = 2NaCl.

This reaction, like any heterogeneous one, proceeds in several steps. In one of them, the Sodium atoms transform into positively charged ions; the oxidation number of Sodium changes from 0 to -1:

Na = Na⁺ + .

Such a process - the losing of electrons attended by an increase in the oxidation number of element - is called oxidation.

The Chlorine atoms, which transform into negatively charged ions, gain the electrons lost by Sodium; the oxidation number of Chlorine changes from 0 to -1:

Cl₂ + 2 = 2C1^-.

The gaining of electrons attended by a decrease in the oxidation number of an element is called reduction.

Hence, in the reaction considered above, Sodium is oxidized, and Chlorine is reduced.

Substance containing an element that is oxidized is called reducing agent, and substance containing an element that is reduced is called oxidizing agent (oxidant).

Consequently, in the above example, Sodium is a reducing agent, and Chlorine an oxidizing one.

Inspection of the oxidation-reduction equations shows that one molecule of Chlorine when reduced gains two electrons, while the oxidation of one Sodium atom is attended by the loss of one electron. The total number of electrons in the system in chemical reactions does not change: the number of electrons lost by the molecules (atoms, ions) of the reducing agent is equal to the number of electrons gained by the molecules (atoms, ions) of the oxidizing agent. Therefore, one molecule of Chlorine can oxidize two Sodium atoms.

3. Compiling Equations of Oxidation-Reduction Reactions

In previous section, we considered very simple example of redox reaction - the formation of a compound from two elementary substances. Usually, the equations of redox reactions are more complicated and it is often quite difficult task to balance them. We shall consider a few examples.

Example 1. The reaction between Hydrogen Iodide and concentrated Sulfuric acid. This reaction proceeds as follows:

HI + H₂SO₄ (conc) → I₂ + H₂S + H₂O.

If we calculate the oxidation number of each element in the reactants and in the products, we shall see that it changes both in Iodine and in Sulfur. In the Iodine of HI it is -1, and in free Iodine it is 0. The oxidation number of Sulfur, on the other hand, changes from +6 (in H₂SO₄) to -2 (in H₂S). Hence, the oxidation number of Iodine increases, and of Sulfur decreases. Con­sequently, Iodine is oxidized, and Sulfur is reduced.

The equation of Iodine oxidation has a simple form:

2I^- = I₂ + 2 (oxidation)

The equation of Sulfur reduction is more complicated because both the reactant (H₂SO₄ or SO₄^2-) and the product (H₂S) contain other elements in addition to Sulfur. In compiling this equation, we shall proceed from the fact that the reaction takes place in an acidic aqueous solution, and the ion SO₄^2- transforms into the molecule H₂S:

SO₄^2- ® H₂S.

The four Oxygen atoms liberated in this process should be combined with Hydrogen into four molecules of water. Eight Hydrogen ions are needed for this. In addition, two Hydrogen ions are needed to form a molecule of H₂S. Hence, ten Hydrogen ions should react with the SO₄^2- ion:

SO₄^2- + 10 H⁺ ®H₂S + 4H₂O.

The total charge on the ions in the left-hand side of this formula is eight elementary positive charges, while the right-hand side con­tains only uncharged particles. Since the total charge does not change in the course of the process, eight electrons also participate in the reduction process:

SO₄^2- + 10H⁺ + 8 ® H₂S + 4H₂O (reduction).

In the above example, the ratio of the number of electrons participating in the reduction process to that liberated in oxidation is 4:1. To obtain the overall equation of the reaction, we have to take this ratio into account when summing the equations of the reduction and oxidation processes, i.e. multiply the reduction equation by four. It is customary practice to indicate the required multipliers to the right of a vertical line when writing the equations:

2I- = I2 + 2 + SO42- + 10H+ + 8 ® H2S + 4H2O	4 1
SO42- + 8 I- + 10H+ = 4 I2 + H2S + 4H2O

The reaction equation obtained can also be written in the molecular form:

H₂SO₄ + 8HI = 4I₂ + H₂S + H₂O.

Example 2. Reaction of Aluminium with Potassium Nitrate in a basic solution. The skeleton equation is:

KNO₃ + Al + KOH + H₂O ® NH₃ + KAlO₂.

Here the oxidation number changes in Nitrogen and Aluminium. Metallic Aluminium (its oxidation number is zero) transforms into the ion A1O₂^-, in which the oxidation number of the Aluminium is +3. To compile the oxidation equation, we shall proceed from the scheme:

Al ® AlO₂^-.

In a basic solution, the OH^- ion is the source of the Oxygen needed for this process to occur. Four hydroxide ions are required to combine one Aluminium atom into AlO₂^- ion:

Al + 4 OH^- ® AlO₂^- + 2H₂O.

The left-hand side of the formula contains four negative charges, and the right-hand side only one. Hence, in the course of the process, three electrons are given up:

Al + 4 OH^- = AlO₂^- + 2H₂O + 3 (oxidation).

To obtain the reduction equation, we shall proceed from the scheme:

NO₃^- ® NH₃.

Here in the course of the process, the Nitrogen atoms lose Oxygen atoms and is combined with Hydrogen atoms. In a basic solution, it is possible with the participation of water molecules. Three water molecules are needed to be combined with three Oxygen atoms, and three more water molecules to form a molecule of NH₃:

NO₃^- + 6H₂O ® NH₃ + 9 OH^-.

The total charge of the right-hand side of the formula is nine negative charges, and of the left-hand side - one. Consequently, eight electrons participate in the process:

NO₃^- + 6H₂O + 8 = NH₃ + 9 OH^- (reduction).

The ratio between the number of electrons liberated in oxidation and the number of electrons gained in reduction is 3:8 in the given example. Hence, to obtain the overall equation of the reaction, we must summate the equations of the oxidation and reduction processes, multiplying the first of them by 8 and the second by 3:

+ Al + 4 OH- = AlO2- + 2H2O + 3 NO3- + 6H2O + 8 = NH3 + 9 OH-	8 3
8 Al + 3 NO3- + 5 OH- = 8 AlO2- + 3 NH3

or in the molecular form:

8 Al + 3 KNO₃ + 5 KOH + 2 H₂O = 8 KAlO₂ + 3 NH₃.

Example 3. Catalytic oxidation of ammonia. This reaction is used in the production of Nitric acid. It is conducted at a temperature of about 750⁰C. The skeleton equation is:

NH₃ + O₂ → NO + H₂O.

The condensation of water vapor (steam) at 750⁰C is impossible. We shall therefore not write the equations of the oxidation and reduction processes as we did for reactions proceeding in an aqueous solution - with the participation of water molecules, Hydrogen or hydroxide ions. We shall only count the number of electrons participating in the oxidation and reduction. We shall take into account that the increase in the oxidation number of an element is equal to the number of lost electrons, and the decrease equals the number of gained electrons.

According to the skeleton equation, the oxidation number changes both in the Nitrogen and in the Oxygen. In the former, it grows from - 3 to +2, and in the latter, it diminishes from 0 to -2. Let us write these changes as formulas, indicating the oxidation numbers of the relevant elements as superscripts. To avoid confusion with the charge of an ion, we shall use Roman numerals for this purpose:

N^-III → N^+II + 5 (oxidation);

O₂⁰ + 4 → 2 O^-II (reduction).

The ratio of the number of electrons gained in reduction to the number of electrons lost in oxidation is 4:5. Hence, five molecules of Oxygen can oxidize four molecules of Ammonia:

4 NH₃ + 5 O₂ = 4 NO + 6 H₂O.

The reaction equations in the above three examples were compiled in a definite sequence. It can also be adhered to in other cases when compiling the equations of redox reactions. The sequence of oper­ations is as follows:

1. Draw up a skeleton equation indicating the reactants and products.

2. Determine the oxidation numbers of the elements in the sub­stances of the right-hand and left-hand sides of the equation; indicate the elements whose oxidation number changes.

3. Draw up the reduction and oxidation equations; find the ratio of the number of electrons gained in reduction to that lost in oxida­tion.

4. Sum the oxidation and reduction equations with account taken of the ratio of the number of electrons found in point 3.