Reactions with double oxides

At reactions with double oxidesoccur when this class of oxides is placed in a container with one of the following substances:

- Water (H₂O)

- Inorganic Acids (HX)

- Inorganic bases (WOH)

It is noteworthy that the double oxides they result from an association between two different oxides of the same element, which always has two cations.

To determine the products that will be formed in the double oxide reactions,it is necessary to know the cations of the elements in these oxides and the anions formed from these cations, as shown below:

• Element iron

In double oxides, the element iron is in the form of iron II cation (Fe⁺²) and iron III (Fe⁺³). Furthermore, it is an element capable of originating two different anions: ferrate (FeO₄^-2) and ferrite (FeO₂^-1).

• lead element

In double oxides, the lead element is in the form of lead II cation (Pb⁺²) and lead IV (Pb⁺⁴). Furthermore, it is an element capable of originating two different anions: the plumbate (PbO₃^-2) and the lead (PbO₂^-2).

• Element tin

In double oxides, the tin element is in the form of tin II cation (Sn

⁺²) and tin IV (Sn⁺⁴). Furthermore, it is an element capable of originating two different anions: stannate (SnO₃^-2) and the tin (SnO₂^-2).

• Element manganese

In double oxides, the element manganese is in the form of manganese II cation (Sn⁺²) and manganese III (Mn⁺³). Furthermore, it is an element capable of originating two different anions: manganate (MnO₄^-2) and manganese (MnO₃^-2).

Reactions with double oxides and water

When one double oxide reacts with water, two inorganic bases (YOH) are formed, each presenting one of the different cations that form the oxide, as in the following general equation:

Y₃O₄ + H₂O → Y(OH)_The + Y(OH)_B

Note.: The indices a and b in the equation refer to the charge of each of the cations that form the double oxide.

For example, if double iron oxide (Fe₃O₄) is placed in the presence of water, there will be the formation of iron hydroxide II [Fe (OH)₂] and iron III hydroxide [Fe(OH)₃], since in this double oxide there is the presence of iron II cations (Fe⁺²) and iron III (Fe⁺³).

These products (the bases) are formed because in the double iron oxide (Fe₃O₄) there is the presence of iron II and iron III cations, which interact with the hydroxyls (OH^-1) of water and form the bases, as in the following balanced equation:

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Faith₃O₄ + 4H₂O → 1Fe(OH)₂ + 2Fe(OH)₃

Reactions with double oxides and bases

When one double oxide reacts with a base Inorganic (WOH), occurs the formation of two different inorganic salts and water. Salts are formed by the interaction between the base cation and the two anions formed by the metal of the oxide, as in the following equation:

Y₃O₄ + WOH → WYO_ç + WYO_d + H₂O

Note.: The indices c and d in the products refer to the different amounts of oxygen present in the anions formed by the metal of the oxide.

For example: if the double lead oxide (Pb₃O₄) react with calcium hydroxide [Ca(OH)₂, and calcium is a +2 charge cation, there will be the formation of calcium plumbate (CaPbO₃), which is the result of the interaction between Ca⁺² and PbO₃^-2, and calcium plumbite (CaPbO₂), which is the result of the interaction between Ca⁺² and PbO₂^-2.

These products are formed because the lead element forms the lead anions (PbO₂^-2) and plumbato (PbO₃^-2), which interact with the calcium cation (Ca⁺²) of calcium oxide. The following is the balanced equation of this process:

1Pb₃O₄ + 3Ca (OH)₂ → 1CaPbO₃ + 2CaPbO₂ + 3H₂O

Reactions with double oxides and acids

When one double oxide reacts with an inorganic acid (HX) occurs the formation of two different inorganic salts and water. Salts are formed by the interaction between each of the cations that form the double oxide and the anion (X^-) of acid, as in the following equation:

Y₃O₄ + HX → YX + YX + H₂O

If the double manganese oxide (Mn₃O₄) react with phosphoric acid (H₃DUST₄), there will be the formation of manganese phosphate II [Mn₃(DUST₄)₂], because of the interaction between the Mn⁺² and PO₄^-3, and manganese phosphate III (MnPO₄), resulting from the interaction between the Mn⁺³ and PO₄^-3.

These products are formed because in the double manganese oxide there is the presence of manganese cations II and manganese III, which interact with the phosphate anion of the acid, as in the balanced equation a follow:

3Mn₃O₄ + 8h₃DUST₄ → 1Mn₃(DUST₄)₂ + 6MnPO₄ + 12:00₂O

By Me. Diogo Lopes Dias

Would you like to reference this text in a school or academic work? Look:

DAYS, Diogo Lopes. "Reactions with double oxides"; Brazil School. Available in: https://brasilescola.uol.com.br/quimica/reacoes-com-oxidos-duplos.htm. Accessed on June 28, 2021.