Question

Методом ионно-электронных уравнений подберите коэффициенты в следующих окислительно-восстановительных реакциях. Укажите окислитель и восстановитель. Какой их элементов окисляется , какой восстанавливается? MnO₂ + H₂SO₄ -> MnSO₄ + O₂ + H₂O

Answer 1

Ion-electron equations for the given redox reactions:

The given redox reaction is: MnO₂ + H₂SO₄ -> MnSO₄ + O₂ + H₂O

To balance this equation using the ion-electron method, we need to follow these steps:

1. Separate the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction. 2. Balance the atoms in each half-reaction, excluding oxygen and hydrogen. 3. Balance the oxygen atoms by adding water molecules (H₂O) to the side that needs more oxygen. 4. Balance the hydrogen atoms by adding hydrogen ions (H⁺) to the side that needs more hydrogen. 5. Balance the charge by adding electrons (e⁻) to the side that needs more charge. 6. Multiply each half-reaction by a coefficient to make the number of electrons equal in both half-reactions. 7. Add the two balanced half-reactions together and cancel out any common terms.

Let's go through these steps to balance the given redox reaction:

Balancing the oxidation half-reaction:

The oxidation half-reaction involves the conversion of MnO₂ to MnSO₄. The oxidation state of Mn changes from +4 to +2.

MnO₂ -> MnSO₄

To balance the atoms, we have:

Mn: 1 Mn on both sides O: 2 O on the left, 4 O on the right S: 0 S on the left, 1 S on the right

To balance the oxygen atoms, we add 2 water molecules (H₂O) to the right side:

MnO₂ -> MnSO₄ + 2H₂O

To balance the hydrogen atoms, we add 4 hydrogen ions (H⁺) to the left side:

MnO₂ + 4H⁺ -> MnSO₄ + 2H₂O

To balance the charge, we add 4 electrons (e⁻) to the left side:

MnO₂ + 4H⁺ + 4e⁻ -> MnSO₄ + 2H₂O

Balancing the reduction half-reaction:

The reduction half-reaction involves the conversion of H₂SO₄ to O₂ and H₂O. The oxidation state of S changes from +6 to +4.

H₂SO₄ -> O₂ + H₂O

To balance the atoms, we have:

H: 4 H on the left, 2 H on the right S: 1 S on the left, 0 S on the right O: 4 O on the left, 2 O on the right

To balance the oxygen atoms, we add 2 water molecules (H₂O) to the right side:

H₂SO₄ -> O₂ + 2H₂O

To balance the hydrogen atoms, we add 2 hydrogen ions (H⁺) to the left side:

H₂SO₄ + 2H⁺ -> O₂ + 2H₂O

To balance the charge, we add 6 electrons (e⁻) to the left side:

H₂SO₄ + 2H⁺ + 6e⁻ -> O₂ + 2H₂O

Balancing the overall reaction:

To balance the overall reaction, we need to make the number of electrons equal in both half-reactions. To do this, we multiply the oxidation half-reaction by 6 and the reduction half-reaction by 4:

6(MnO₂ + 4H⁺ + 4e⁻) -> 4(H₂SO₄ + 2H⁺ + 6e⁻) + 6(MnSO₄ + 2H₂O)

Simplifying the equation:

6MnO₂ + 24H⁺ + 24e⁻ -> 4H₂SO₄ + 8H⁺ + 24e⁻ + 6MnSO₄ + 12H₂O

Cancelling out common terms:

6MnO₂ + 24H⁺ -> 4H₂SO₄ + 6MnSO₄ + 12H₂O

Identifying the oxidizing and reducing agents:

In the balanced equation, the oxidizing agent is the species that gets reduced, while the reducing agent is the species that gets oxidized.

In this case, MnO₂ is the reducing agent because it loses electrons and undergoes oxidation, changing from an oxidation state of +4 to +2. H₂SO₄ is the oxidizing agent because it gains electrons and undergoes reduction, changing from an oxidation state of +6 to +4.

Therefore, MnO₂ is oxidized and H₂SO₄ is reduced in the given redox reaction.

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