Question

CI2+J2+H2O=HJO3+HCE Окислительно-восстоновительная рекция

Answer 1

5Cl2+J2+6H2O=2HJO3+10HCl
2Clo ----+2-----Cl- окисл.5
Jo------ -5e ------J5+восст.2

Answer 2

Oxidation-Reduction Reaction: CI2 + J2 + H2O = HJO3 + HCE

The given chemical equation represents an oxidation-reduction reaction. In this reaction, chlorine (Cl2) and iodine (I2) react with water (H2O) to produce periodic acid (HJO3) and hydrochloric acid (HCE). Let's break down the reaction and discuss the oxidation and reduction processes involved.

The balanced chemical equation for the reaction is as follows:

CI2 + J2 + H2O → HJO3 + HCE

To understand the oxidation and reduction processes, we need to assign oxidation numbers to the elements involved in the reaction. The oxidation number is a positive or negative number assigned to an atom to indicate its degree of oxidation or reduction in a compound or ion.

In this reaction: - Chlorine (Cl) has an oxidation number of 0 in Cl2. - Iodine (I) has an oxidation number of 0 in I2. - Hydrogen (H) has an oxidation number of +1 in H2O. - Oxygen (O) has an oxidation number of -2 in H2O. - Iodine (I) has an oxidation number of +5 in HJO3. - Chlorine (Cl) has an oxidation number of -1 in HCE.

Now, let's analyze the oxidation and reduction processes involved in the reaction:

1. Oxidation: - Chlorine (Cl) is oxidized from an oxidation number of 0 in Cl2 to -1 in HCE. - Iodine (I) is oxidized from an oxidation number of 0 in I2 to +5 in HJO3.

2. Reduction: - Hydrogen (H) is reduced from an oxidation number of +1 in H2O to 0 in HCE. - Oxygen (O) is reduced from an oxidation number of -2 in H2O to -1 in HJO3.

The reaction can be understood as follows: - Chlorine (Cl2) is oxidized to form hydrochloric acid (HCE), where chlorine gains an electron and its oxidation number decreases from 0 to -1. - Iodine (I2) is oxidized to form periodic acid (HJO3), where iodine loses electrons and its oxidation number increases from 0 to +5. - Water (H2O) is reduced, where hydrogen gains an electron and its oxidation number decreases from +1 to 0, and oxygen gains electrons and its oxidation number decreases from -2 to -1.

Please note that the reaction you provided is not a commonly known or widely discussed reaction. The specific reaction conditions, such as temperature, pressure, and catalysts, may affect the reaction kinetics and product formation. If you have any further questions or need more information, feel free to ask!