Question

Запишите электронные уравнения окисления и восстановления, соответствующие реакции H2S + KMnO4 + HCl → S + MnCl2 + KCl + H2O окислитель восстановитель и уравняйте эту реакцию методом электронного баланса. Укажите окислитель и восстановитель

Answer 1

5H2S+2KMnO4+6HCl=5S+2MnCl2+2KCl+8H2O
S с.о -2 -2=Sс.о. 0*5
Mnс.о. +7+5 =Mm с.о. +2*2
Сера восста.
Mn окисл.

Answer 2

Electrochemical Equations for Oxidation and Reduction Reactions

The given reaction is: H2S + KMnO4 + HCl → S + MnCl2 + KCl + H2O

To determine the oxidation and reduction half-reactions, we need to identify the species that undergoes oxidation and reduction.

1. Oxidation Half-Reaction: - In this reaction, the sulfur in H2S is oxidized from an oxidation state of -2 to 0 in S. - The oxidation half-reaction can be written as: H2S → S + 2e- [[1]].

2. Reduction Half-Reaction: - In this reaction, the manganese in KMnO4 is reduced from an oxidation state of +7 to +2 in MnCl2. - The reduction half-reaction can be written as: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O [[2]].

Balancing the Reaction using the Electron Balance Method

To balance the overall reaction, we need to ensure that the number of electrons transferred in the oxidation and reduction half-reactions is equal.

1. Multiply the oxidation half-reaction by 5 to balance the electrons: - 5H2S → 5S + 10e-

2. Multiply the reduction half-reaction by 2 to balance the electrons: - 2MnO4- + 16H+ + 10e- → 2Mn2+ + 8H2O

3. Now, add the two half-reactions together: - 5H2S + 2MnO4- + 16H+ → 5S + 2Mn2+ + 8H2O

Identifying the Oxidizing and Reducing Agents

The oxidizing agent is the species that gets reduced, while the reducing agent is the species that gets oxidized.

1. In this reaction, the oxidizing agent is KMnO4, which is reduced from an oxidation state of +7 to +2 in MnCl2.

2. The reducing agent is H2S, which is oxidized from an oxidation state of -2 to 0 in S.

Therefore, the oxidizing agent is KMnO4, and the reducing agent is H2S.

Summary

The balanced reaction is: 5H2S + 2MnO4- + 16H+ → 5S + 2Mn2+ + 8H2O

The oxidation half-reaction is: H2S → S + 2e-

The reduction half-reaction is: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

The oxidizing agent is KMnO4, and the reducing agent is H2S.

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