Question

Для реакции FeSO4+H2SO4+KMnO4-->Fe2(SO4)3+K2SO4+MnSO4+H2O составьте электронные ураврения. Расставьте коэффициенты Укажите окислитель и восстановитель

Answer 1

10FeSO4+8H2SO4+2KMnO4 = 5Fe2(SO4)3+K2SO4+ 2MnSO4+ 8H2O
2Fe(+2)-2e=2Fe(+3) 5 в-ль, ок-ие
Mn(+7) +5e=Mn(+2) 2. ок-ль, в-ие

Answer 2

Reaction and Electron Equations

The reaction you provided is:

FeSO4 + H2SO4 + KMnO4 --> Fe2(SO4)3 + K2SO4 + MnSO4 + H2O

To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Here is the balanced equation:

6 FeSO4 + 2 KMnO4 + 8 H2SO4 --> 3 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O

Now, let's write the electron equations for the oxidation and reduction half-reactions.

Oxidation Half-Reaction

In this reaction, the iron (Fe) in FeSO4 is oxidized from an oxidation state of +2 to +3. The oxidation half-reaction can be written as follows:

Fe --> Fe3+ + 3e-

Reduction Half-Reaction

In this reaction, the manganese (Mn) in KMnO4 is reduced from an oxidation state of +7 to +2. The reduction half-reaction can be written as follows:

MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O

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.

In this reaction: - The oxidizing agent is KMnO4 because it causes the reduction of Mn from +7 to +2. - The reducing agent is FeSO4 because it causes the oxidation of Fe from +2 to +3.

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