Theoretical basics of Inorganic chemistry

Oxidation and reduction. In the two preceding sections, we described precipitation reactions (reactions producing a precipitate) and acid–base reactions (reactions involving proton transfer). Here we discuss the third major class of reactions, oxidation–reduction reactions, which are reactions involving a transfer of electrons from one species to another.

 

 

Figure 11.1 Reaction of iron with Cu²⁺(aq)

 

As a simple example of an oxidation–reduction reaction, let us look at what happens when you dip an iron nail into a blue solution of copper (II) sulfate (figure 11.1). What you see is that the iron nail becomes coated with a reddish-brown tinge of metallic copper. The molecular equation for this reaction is

 

Fe(s) + CuSO₄ (aq) → FeSO₄ (aq) + Cu (s)

 

The net ionic equation is

 

Fe(s) + Cu²⁺ (aq) → Fe²⁺ (aq) + Cu(s)

 

The electron-transfer aspect of the reaction is apparent from this equation. Note that each iron atom in the metal loses two electrons to form an iron (II) ion, and each copper (II) ion gains two electrons to form a copper atom in the metal. The net effect is that two electrons are transferred from each iron atom in the metal to each copper (II) ion. The concept of oxidation numbers was developed as a simple way of keeping track of electrons in a reaction. Using oxidation numbers, you can determine whether electrons have been transferred from one atom to another. If electrons have been transferred, an oxidation–reduction reaction has occurred [15, p.185].

 

Oxidation numbers. We define the oxidation number (or oxidation state) of an atom in a substance as the actual charge of the atom if it exists as a monatomic ion, or a hypothetical charge assigned to the atom in the substance by simple rules. An oxidation–reduction reaction is one in which one or more atoms change oxidation number, implying that there has been a transfer of electrons. Consider the combustion of calcium metal in oxygen gas (figure 11.2).

 

2Ca(s) + O₂ (g) → 2CaO(s)

 

This is an oxidation–reduction reaction. To see this, you assign oxidation numbers to the atoms in the equation and then note that the atoms change oxidation number during the reaction.

Since the oxidation number of an atom in an element is always zero, Ca and O in O₂ have oxidation numbers of zero. Another rule follows from the definition of oxidation number: The oxidation number of an atom that exists in a substance as a monatomic ion equals the charge on that ion. So the oxidation number of Ca in CaO is +2 (the charge on Ca²⁺), and the oxidation number of 

 

 

Figure 11.2 The burning of calcium metal in oxygen. The burning calcium emits a red-orange flame.

 

O in CaO is -2 (the charge on O^2-). To emphasize these oxidation numbers in an equation, we will write them above the atomic symbols in the formulas.

 

      _{o     o          +2 -2}

2Ca + O₂ → 2CaO

 

From this, you see that the Ca and O atoms change in oxidation number during the reaction. In effect, each calcium atom in the metal loses two electrons to form Ca²⁺ ions, and each oxygen atom in O₂ gains two electrons to form O^2- ions. The net result is a transfer of electrons from calcium to oxygen, so this reaction is an oxidation–reduction reaction. In other words, an oxidation–reduction reaction (or redox reaction) is a reaction in which electrons are transferred between species or in which atoms change oxidation number.

Note that calcium has gained in oxidation number from 0 to +2. (Each calcium atom loses two electrons.) We say that calcium has been oxidized. Oxygen, on the other hand, has decreased in oxidation number from 0 to -2. Each oxygen atom gains two electrons.

We say that oxygen has been reduced. An oxidation–reduction reaction always involves both oxidation (the loss of electrons) and reduction (the gain of electrons).Formerly, the term oxidation meant “reaction with oxygen.” The current definition greatly enlarges the meaning of this term. Consider the reaction of calcium metal with chlorine gas (figure 11.3); the reaction looks similar to the burning of calcium in oxygen. The chemical equation is

 

  _{o       o          +2 -2}

Ca + Cl₂ → CaCl₂

 

In this reaction, the calcium atom is oxidized, because it increases in oxidation number (from 0 to +2, as in the previous equation). Chlorine is reduced; it decreases in oxidation number from 0 to -1. This is clearly an oxidation–reduction reaction that does not involve oxygen [15, p. 186].

 

Figure 11.3 The burning of calcium metal in chlorine. The reaction appears similar to the burning

of calcium in oxygen.

 

Oxidation-Number Rules. So far, we have used two rules for obtaining oxidation numbers: (1) the oxidation number of an atom in an element is zero, and (2) the oxidation number of an atom in a monatomic ion equals the charge on the ion. These and several other rules for assigning oxidation numbers are given in table 11.1.

In molecular substances, we use these rules to give the approximate charges on the atoms. Consider the molecule SO₂. Oxygen atoms tend to attract electrons, pulling them from other atoms (sulfur in the case of SO₂). As a result, an oxygen atom in SO₂ takes on a negative charge relative to the sulfur atom. The magnitude of the charge on an oxygen atom in a molecule is not a full -2 charge as in the O^2- ion. However, it is convenient to assign an oxidation number of -2 to oxygen in SO₂ (and in most other compounds of oxygen) to help us express the approximate charge distribution in the molecule. Rule 3 in table 11.1 says that an oxygen atom has an oxidation number of -2 in most of its compounds.

Rules 4 and 5 are similar in that they tell you what to expect for the oxidation number of certain elements in their compounds. Rule 4, for instance, says that hydrogen has an oxidation number of +1 in most of its compounds.

Rule 6 states that the sum of the oxidation numbers of the atoms in a compound is zero. This rule follows from the interpretation of oxidation numbers as (hypothetical) charges on the atoms. Because any compound is electrically neutral, the sum of the charges on its atoms must be zero. This rule is easily extended to ions: the sum of the oxidation numbers (hypothetical charges) of the atoms in a polyatomic ion equals the charge on the ion.

You can use Rule 6 to obtain the oxidation number of one atom in a compound or ion, if you know the oxidation numbers of the other atoms in the compound or ion. Consider the SO₂ molecule. According to Rule 6,

 

(Oxidation number of S) +2 × (oxidation number of O) = 0

 

Or

 

(Oxidation number of S) +2 × (-2) = 0.

 

Therefore,

 

Oxidation number of S (in SO₂) = -2 × (-2) = +4

 

Table 11.1

Rules for Assigning Oxidation Numbers

Rule	Applies to	Statement
1	Elements	The oxidation number of an atom in an element is zero.
2	Monatomic ions	The oxidation number of an atom in a monatomic ion equals the charge on the ion.
3	Oxygen	The oxidation number of oxygen is -2 in most of its compounds. (An exception is O in H2O2 and other peroxides, where the oxidation number is -1.)
4	Hydrogen	The oxidation number of hydrogen is +1 in most of its compounds. (The oxidation number of hydrogen is -1 in binary compounds with a metal, such as CaH2.)
5	Halogens	The oxidation number of fluorine is -1 in all of its compounds. Each of the other halogens (Cl, Br, I) has an oxidation number of -1 in binary compounds, except when the other element is another halogen above it in the periodic table or the other element is oxygen.
6	Compounds and ions	The sum of the oxidation numbers of the atoms in a compound is zero. The sum of the oxidation numbers of the atoms in a polyatomic ion equals the charge on the ion.

 

Describing Oxidation–Reduction Reactions. We use special terminology to describe oxidation–reduction reactions. To illustrate this, we will look again at the reaction of iron with copper (II) sulfate. The net ionic equation is

 

Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s)

 

We can write this reaction in terms of two half-reactions. A half-reaction is one of two parts of an oxidation–reduction reaction, one part of which involves a loss of electrons (or increase of oxidation number) and the other a gain of electrons (or decrease of oxidation number). The half-reactions for the preceding equation are

 

Fe(s) → Fe²⁺(aq) + 2e^- (electrons lost by Fe)

Cu²⁺(aq) + 2e^- → Cu(s) (electrons gained by Cu²⁺)

 

Oxidation is the half-reaction in which there is a loss of electrons by a species (or an increase of oxidation number of an atom).

 

Reduction is the half-reaction in which there is a gain of electrons by a species (or a decrease in the oxidation number of an atom).

 

Thus, the equation Fe(s) → Fe²⁺(aq) + 2e^- represents the oxidation half-reaction, and the equation Cu²⁺(aq) + 2e^- → Cu(s) represents the reduction half-reaction.

Recall that a species that is oxidized loses electrons (or contains an atom that increases in oxidation number) and a species that is reduced gains electrons (or contains an atom that decreases in oxidation number). An oxidizing agent is a species that oxidizes another species; it is itself reduced. Similarly, a reducing agent is a species that reduces another species; it is itself oxidized. In our example reaction, the copper (II) ion is the oxidizing agent, whereas iron metal is the reducing agent [15, p. 187].

 

Some Common Oxidation–Reduction Reactions. Many oxidation–reduction reactions can be described as one of the following:

1.      Combination reaction

2.      Decomposition reaction

3.      Displacement reaction

4.      Combustion reaction

We will describe examples of each of these in this section.

Combination Reactions A combination reaction is a reaction in which two substances combine to form a third substance. Note that not all combination reactions are oxidation–reduction reactions. However, the simplest cases are those in which two elements react to form a compound; these are clearly oxidation–reduction reactions.

 

2Na(s) + Cl₂ (g) → 2NaCl(s)

 

Antimony and chlorine also combine in a fiery reaction.

 

2Sb + 3Cl₂ → 2SbCl₃

 

Some combination reactions involve compounds as reactants and are not oxidation–reduction reactions. For example,

 

CaO (s) + SO₂ (g) → CaSO₃(s)

 

(If you check the oxidation numbers, you will see that this is not an oxidation–reduction reaction.)

 

Decomposition Reactions A decomposition reaction is a reaction in which a single compound reacts to give two or more substances. Often these reactions occur when the temperature is raised. In Chapter 1, we described the decomposition of mercury (II) oxide into its elements when the compound is heated. This is an oxidation–reduction reaction.

 

2HgO(s) → 2Hg (l) + O₂ (g)

 

Another example is the preparation of oxygen by heating potassium chlorate with manganese(IV) oxide as a catalyst.

 

2KClO₃(s) → 2KCl(s) + 3O₂ (g)

                                                                       _MnO2

 

In this reaction, a compound decomposes into another compound and an element; it also is an oxidation–reduction reaction.

Not all decomposition reactions are of the oxidation–reduction type. For example, calcium carbonate at high temperatures decomposes into calcium oxide and carbon dioxide.

 

CaCO₃(s) → CaO (s) + CO₂ (g)

 

Displacement Reactions.  A displacement reaction (also called a singlereplacement reaction) is a reaction in which an element reacts with a compound, displacing another element from it. Since these reactions involve an element and one of its compounds, these must be oxidation–reduction reactions. An example is the reaction that occurs when you dip a copper metal strip into a solution of silver nitrate.

 

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

 

From the molecular equation, it appears that copper displaces silver in silver nitrate, producing crystals of silver metal and a greenish-blue solution of copper (II) nitrate. The net ionic equation, however, shows that the reaction involves the transfer of electrons from copper metal to silver ion:

 

Cu(s) + 2Ag⁺ (aq) → Cu²⁺ (aq) + 2Ag(s)

 

When you dip a zinc metal strip into an acid solution, bubbles of hydrogen form on the metal and escape from the solution.

 

Zn(s) + 2HCl (aq) → ZnCl₂ (aq) + H₂ (g)

 

Zinc displaces hydrogen in the acid, producing zinc chloride solution and hydrogen gas. The net ionic equation is

 

Zn(s) + 2H⁺ (aq) → Zn²⁺ (aq) + H₂ (g)

 

Whether a reaction occurs between a given element and a monatomic ion depends on the relative ease with which the two species gain or lose electrons. Figure 11.4 shows the activity series of the elements, a listing of the elements in decreasing order of their ease of losing electrons during reactions in aqueous solution. The metals listed at the top are the strongest reducing agents (they lose electrons easily); those at the bottom, the weakest. A free element reacts with the monatomic ion of another element if the free element is above the other element in the activity series. The highlighted elements react slowly with liquid water, but readily with steam, to give H₂.

Consider this reaction:

 

2K(s) + 2H⁺ (aq) → 2K⁺ (aq) + H₂ (g)

 

You would expect this reaction to proceed as written, because potassium metal (K) is well above hydrogen in the activity series. In fact, potassium metal reacts violently with water, which contains only a very small percentage of H⁺ ions. Imagine the reaction of potassium metal with a strong acid like HCl!

 

Figure 11.4 Activity Series of the Elements

 

Combustion Reactions.  A combustion reaction is a reaction in which a substance reacts with oxygen, usually with the rapid release of heat to produce a flame. The products include one or more oxides. Oxygen changes oxidation number from 0 to -2, so combustions are oxidation–reduction reactions.

Organic compounds usually burn in oxygen or air to yield carbon dioxide. If the compound contains hydrogen (as most do), water is also a product. For instance, butane (C₄H₁₀) burns in air as follows:

 

2C₄H₁₀ (g) + 13O₂ (g) → 8CO₂ (g) + 10H₂O (g)

 

Many metals burn in air as well. Although chunks of iron do not burn readily in air, iron wool, which consists of fine strands of iron, does. The increased surface area of the metal in iron wool allows oxygen from air to react quickly with it.

 

4Fe(s) + 3O₂ (g) → 2Fe₂O₃(s) [15, p. 190]

 

Balancing Simple Oxidation–Reduction Equations Oxidation–reduction reactions can often be quite difficult to balance. Some are so complex in fact that chemists have written computer programs to accomplish the task. In this section, we will develop a method for balancing simple oxidation–reduction reactions that can later be generalized for far more complex reactions.

One of the advantages of using this technique for even simple reactions is that you focus on what makes oxidation–reduction reactions different from other reaction types.  

At first glance, the equation representing the reaction of zinc metal with silver (I) ions in solution might appear to be balanced.

 

Zn(s) + Ag⁺ (aq) → Zn²⁺ (aq) + Ag(s)

 

However, because a balanced chemical equation must have a charge balance as well as a mass balance, this equation is not balanced: it has a total charge of +1 for the reactants and +2 for the products. Let us apply the half-reaction method for balancing this equation [15, p. 192].

 

Half-Reaction Method Applied to Simple Oxidation–Reduction Equations. The half-reaction method consists of first separating the equation into two half-reactions, one for oxidation, the other for reduction. You balance each half-reaction, then combine them to obtain a balanced oxidation–reduction reaction. Here is an illustration of the process. First we identify the species being oxidized and reduced and assign the appropriate oxidation states.

 

Zn(s) + Ag⁺ (aq) → Zn²⁺ (aq) + Ag(s)

 

Next, write the half-reactions in an unbalanced form.

 

Zn → Zn²⁺ (oxidation)

Ag⁺ → Ag (reduction)

 

Next, balance the charge in each equation by adding electrons to the more positive side to create balanced half-reactions. Following this procedure, the balanced half-reactions are:

 

Zn → Zn²⁺ + 2e^- (oxidation half-reaction)

Ag⁺ + e^- → Ag (reduction half-reduction)

 

Note that the number of electrons that Zn loses during the oxidation process (two) exceeds the number of electrons gained by Ag⁺ during the reduction (one). Since, according to the reduction half-reaction, each Ag⁺ is capable of gaining only one electron, we need to double the amount of Ag⁺ in order for it to accept all of the electrons produced by Zn during oxidation.

To meet this goal and obtain the balanced oxidation–reduction reaction, we multiply each half-reaction by a factor (integer) so that when we add them together, the electrons cancel.

We multiply the first equation by 1 (the number of electrons in the second half-reaction) and multiply the second equation by 2 (the number of electrons in the first half-reaction).

 

1 × (Zn → Zn²⁺ + 2e^-)

 

The electrons cancel, which finally yields the balanced oxidation–reduction equation:

 

Zn(s) + 2Ag⁺ (aq) → Zn²⁺ (aq) + 2Ag(s)

 

The majority of the chemical reactions discussed in this chapter take place in solution. This is because the reaction between two solid reactants often proceeds very slowly or not at all.

In a solid, the molecules or ions in a crystal tend to occupy approximately fixed positions, so the chance of two molecules or ions coming together to react is small. In liquid solutions, reactant molecules are free to move throughout the liquid; therefore, reaction is much faster.

When you run reactions in liquid solutions, it is convenient to dispense the amounts of reactants by measuring out volumes of reactant solutions. In the next two sections, we will discuss calculations involved in making up solutions [15, p. 193].

 

Oxidizing and Reducing Agents. Chemists frequently use the terms oxidizing agent and reducing agent to describe certain of the reactants in redox reactions, as in statements like fluorine gas is a powerful oxidizing agent, or calcium metal is a good reducing agent. Let us briefly consider the meaning of these terms.

In a redox reaction, the substance that makes it possible for some other substance to be oxidized is called the oxidizing agent, or oxidant. In doing so, the oxidizing agent is itself reduced.

Similarly, the substance that causes some other substance to be reduced is called the reducing agent, or reductant. In the reaction, the reducing agent is itself oxidized or stated in other ways an oxidizing agent (oxidant):

·         causes another substance to be oxidized

·         contains an element whose oxidation state decreases in a redox reaction

·         gains electrons (electrons are found on the left side of its half-equation)

·         is reduced

·         a reducing agent (reductant)

·         causes another substance to be reduced

·         contains an element whose oxidation state increases in a redox reaction

·         loses electrons (electrons are found on the right side of its half-equation)

·         is oxidized

In general, a substance with an element in one of its highest possible oxidation states is an oxidizing agent. If the element is in one of its lowest possible oxidation states, the substance is a reducing agent.

The oxidation state of the nitrogen in dinitrogen tetroxide is nearly the maximum value attainable, and hence N₂O₄ is generally (N₂O₄) an oxidizing agent. Conversely, the nitrogen atom in hydrazine (N₂H₄) is in nearly the lowest oxidation state, and hence hydrazine is generally a reducing agent. When these two liquid compounds are mixed, a vigorous reaction takes place:

  

N₂O₄ (l) + 2N₂H₄ (l) → 3N₂ (g) + 4H₂O (g)

 

In this reaction, is the oxidizing agent and is the reducing agent. This reaction releases so much energy that it is used in some rocket propulsion systems.

Certain substances in which the oxidation state of an element is between its highest and lowest possible values may act as oxidizing agents in some instances and reducing agents in others. For example, in the reaction of hydrazine with hydrogen to produce ammonia, hydrazine acts as an oxidizing agent.

 

N₂H₄ (l) + H₂ (g) → 2NH₃ (g)

 

Permanganate ion, is a versatile oxidizing agent that has many uses in the chemical laboratory. In the next section, we describe its use in the quantitative analysis of iron that is, the determination of the exact (quantitative) amount of iron in an iron-containing material. Ozone, a triatomic form of oxygen, is an oxidizing agent used in water purification, as in the oxidation of the organic compound phenol,

 

C₆H₅OH (aq) + 14O₃ (g) → 6CO₂ (g) + 3H₂O (l) + 14O₂ (g)

 

Aqueous sodium hypochlorite, NaOCl(aq), is a powerful oxidizing agent. It is the active ingredient in many liquid chlorine bleaches. The bleaching action of NaOCl(aq) is associated with the reduction of the OCl^- ion to Cl^- ;the electrons required for the reduction come from colored compounds in stains. The bleaching action of NaOCl(aq) is demonstrated.

Thiosulfate ion, S₂O₃^2-, is an important reducing agent. One of its industrial uses is as an antichlor to destroy residual chlorine from the bleaching of fibers.

 

S₂O₃^2- (aq) + 4Cl₂ (aq) + 5H₂O (l) → 2HSO^4- (aq) + 8H⁺ (aq) + 8Cl^- (aq)

 

Oxidizing and reducing agents also play important roles in biological systems - in photosynthesis (using solar energy to synthesize glucose), metabolism (oxidizing glucose), and the transport of oxygen [3, p. 199].

 

Electrolysis

(a)   Cells and electrolysis. In cells, oxidation-reduction reactions proceed spontaneously, and the chemical energy accompanying the chemical reactions is converted to electric energy. If voltage is applied to the cell from the direction reverse to the electromotive force, a chemical reaction that corresponds to the negative electromotive force is induced. In other words, reactions that do not occur spontaneously are now induced by the electric energy. This process is called electrolysis. The charging of a lead storage battery is an example of electrolysis.

 

The total reaction of the Daniell cell is as follows.

 

Zn + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu

 

Suppose a voltage higher than is applied in the direction reverse to the electromotive force, the reverse reaction takes place. Thus, zinc will deposit and copper will begin to dissolve.

 

Zn²⁺ (aq) + Cu → Zn + Cu²⁺ (aq)

 

Figure 11.5 shows a schematic representation of the chemical reaction which occurs when a reverse voltage is applied to the Daniel cell.

 

(b)   Faraday’s law of electrolysis. In the first half of the 19^th century, Faraday investigated the relation between the quantity of electricity which flows in a cell and the quantity of substances chemically changed at the electrodes during the electrolysis. He summarized the results in two laws in 1833.

 

Faraday’s law of electrolysis

1.      The quantity of substances produced at the electrodes is proportional to the quantity of electricity that flows in the cell.

2.      When a certain quantity of electricity flows in the cell, the number of moles of substances changed at the electrodes is constant regardless of the type of substance. For instance, the quantity of electricity necessary to deposit 1 mole of a monovalent metal is 96,485 C (Coulomb) regardless of the type of metal.

 

Faraday’s Laws of Electrolysis.

·         Mass of an element produced at an electrode is proportional to the amount of electrical charge Q passed through the liquid. If a current of I Amperes (A) is passed through an electrolyte solution for t seconds (s), we have (1)

 

Q = It (1)

 

where the units of Q is the Coulomb (C). Obviously, C = A × s.

The mass of element produced is proportional to the equivalent weight of the element.

 

·         The concept of “equivalent weights” is no longer recommended by IUPAC. However, it remains a useful concept in this context. One way to think about equivalent weight in the context of electrolysis is as the ratio of the molar mass (M) of the substance to the number of moles of electrons n that need to be added or removed to neutralize it. That is (2),

 

E = M/n (2)

 

Figure 11.5 Electrolysis. A reaction reverse to that in the Daniell cell takes place. Zinc deposits and copper dissolves [31, p. 15]

 

To illustrate, the equivalent weights E of a few elements are:

 

Reaction;                             M (g mol–1);       n;           E (g equiv–1)

H⁺(aq) + e^- → ½H2(g)        1.008                     1            1.008

Fe²⁺(aq) + 2e^- → Fe(s)       55.847                    2            27.924

Fe³⁺(aq) + 3e^- → Fe(s)       55.847                    3            18.616

Cl^-(aq) → ½Cl₂(g) + e^-       35.453                    1            35.453

 

The two laws can be combined into a convenient form by introducing the quantity known as the Faraday (F), which is the charge carried by one mole of electrons:

 

F = 1.6022 × 10–19 C × 6.0220 × 1023 mol^–1 = 96,485 C mol^–1.

 

Since one mole of electrons corresponds to one equivalent of the element, we may also think of the units of the F as C equiv^–1 [31, p.16].

 

(c)    Industrially important electrolyses. The 1st electrolysis attempted was the hydrolysis of water (1800). Davy immediately followed and successfully isolated alkali and alkaline earth metals. Even now electrolysis is used to produce various metals. Electrolysis is particularly useful for the production of metals with high ionization tendencies (e.g., aluminum).

The industrial production of aluminum by electrolysis was achieved in 1886 independently by the American inventor Charles Martin Hall (1863-1914) and the French inventor Paul Louis Toussaint Héroult (1863-1914) at the same time. The success of this electrolysis was due to the use of molten Na₃AlF₆ as the solvent of the ore (aluminum oxide; alumina Al₂O₃).

As a requirement for successful electrolysis, ions can migrate to the electrodes. An obvious way to give mobility to ions is to use its aqueous solution. However, in the case of electrolysis of alumina, an aqueous solution is inadequate because water is more readily reduced than the aluminum ion as shown below.

 

Al₃⁺ + 3e^- → Al normal electrode potential = -1.662 V

2H₂O +2e^- → H₂ + 2OH^- normal electrode potential = -0.828 V

 

Another method would be the use of molten salt. The trouble is that the melting point of Al₂O₃ is as high as 2050 ºC, and electrolysis at such a high temperature is not realistic. However, the melting point of a mixture of Al₂O₃ and Na₃AlF₆ is ca. 1000 ºC, and this temperature is easy to attain.

The details of the procedure are as follows: the ore, bauxite, contains various metal oxides as impurities. The ore is treated with alkali, and only amphoteric aluminum oxide dissolves. Insoluble materials are filtered off, and carbon dioxide is blown through the filtrate to cause hydrolysis (of salts). Alumina is deposited.

 

Al₂O₃(s) + 2OH^-(aq) → 2AlO^2- (aq) + H₂O(l)

2CO₂ + 2AlO₂^-(aq) + (n+1)H₂O(l) → 2HCO₃^- (aq) + Al₂O₃ × nH₂O(s)

 

The alumina thus obtained is mixed with Na3AlF6 and then molten salt electrolysis of the mixture is carried out. The reactions in the electrolytic cell are complicated. It is likely that initially alumina reacts with Na₃AlF₆ and then electrolytic reactions take place.

 

Al₂O₃ + 4AlF₆^3- → 3Al₂OF₆^2- + 6F^-

 

   The electrode reactions are as follows.

Negative electrode: 2Al₂OF₆^2- + 12F^- + C → 4AlF₆^3- + CO₂ + 4e^-

Positive electrode: AlF₆^3- + 3e^- → Al + 6F^-

Total reaction: 2Al₂O3 + 3C → 4Al + 3CO₂^-

 

The purity of aluminum obtained by this procedure is ca. 99.55 %. Aluminum is used as such and as alloys with other metals. The properties are excellent and, in addition, the price is modest. However, it must be remembered that the production of aluminum requires a tremendous amount of electricity [31, p. 17].