Question

10.13. Среди приведённых реакций указать реакции диспропор-цирования: 1) S + KOH  K2SO3 + K2S + H2S; 2) HCl + CrO3  CrCl + Cl2 + H2O; 3) Au2O3  Au + O2; 4) H2O + Cl2  HCl +HClO.

Answer 1

Среди приведённых реакций указать реакции диспропор-цирования:1) S + KOH  K2SO3 + K2S + H2S;

4) H2O + Cl2  HCl +HClO.

Answer 2

Reactions of Disproportionation

The reactions of disproportionation involve the simultaneous oxidation and reduction of a single species. In these reactions, the same element undergoes both oxidation and reduction, resulting in the formation of two different oxidation states of that element. Let's analyze the given reactions and identify the reactions of disproportionation:

1) S + KOH → K2SO3 + K2S + H2S - In this reaction, sulfur (S) is oxidized from an oxidation state of 0 in elemental sulfur to -2 in sulfide ions (H2S) and reduced from 0 to +4 in sulfite ions (K2SO3) and from 0 to +2 in elemental sulfur (K2S). Therefore, this reaction involves the disproportionation of sulfur.

2) HCl + CrO3 → CrCl3 + Cl2 + H2O - In this reaction, chlorine (Cl) is oxidized from an oxidation state of -1 in hydrochloric acid (HCl) to 0 in chlorine gas (Cl2) and reduced from +7 in chromate ions (CrO3) to +3 in chromium ions (CrCl3). Therefore, this reaction involves the disproportionation of chlorine.

3) Au2O3 → Au + O2 - In this reaction, gold (Au) is reduced from +3 in gold(III) oxide (Au2O3) to 0 in elemental gold (Au) and oxygen (O) is oxidized from -2 in gold(III) oxide (Au2O3) to 0 in molecular oxygen (O2). Therefore, this reaction involves the disproportionation of gold.

4) H2O + Cl2 → HCl + HClO - In this reaction, chlorine (Cl) is reduced from 0 in chlorine gas (Cl2) to -1 in hydrochloric acid (HCl) and oxidized from 0 to +1 in hypochlorous acid (HClO). Therefore, this reaction involves the disproportionation of chlorine.

To summarize, the reactions of disproportionation among the given reactions are: 1) S + KOH → K2SO3 + K2S + H2S 2) HCl + CrO3 → CrCl3 + Cl2 + H2O 3) Au2O3 → Au + O2 4) H2O + Cl2 → HCl + HClO

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