Question

Встановити коефіцієнти йонно-електронним методом в рівнянні окисно-відновної реакції, вказати процеси окиснення та відновлення, речовини окисники та відновники: NaI+MnO₂+ H₂SO₄= I₂ + MnSO₄+Na₂SO₄+H₂O

Answer 1

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Answer 2

Ion-electron method for balancing redox reactions

The ion-electron method, also known as the half-reaction method, is commonly used to balance redox reactions. This method involves breaking down the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction. Each half-reaction represents the process of either losing or gaining electrons.

Let's balance the given redox reaction using the ion-electron method:

Reaction: NaI + MnO₂ + H₂SO₄ = I₂ + MnSO₄ + Na₂SO₄ + H₂O

Step 1: Identify the oxidation and reduction half-reactions

In the given reaction, sodium iodide (NaI) is oxidized to form iodine (I₂), while manganese dioxide (MnO₂) is reduced to form manganese sulfate (MnSO₄). Therefore, the oxidation half-reaction involves NaI, and the reduction half-reaction involves MnO₂.

Step 2: Balance the atoms other than hydrogen and oxygen

Let's start by balancing the atoms other than hydrogen and oxygen in each half-reaction.

Oxidation half-reaction: NaI → I₂ Reduction half-reaction: MnO₂ → MnSO₄

The number of iodine atoms is already balanced in the oxidation half-reaction. In the reduction half-reaction, there is one manganese atom on the left side and one on the right side, so it is already balanced.

Step 3: Balance the oxygen atoms

Next, we balance the oxygen atoms by adding water (H₂O) molecules to the half-reactions. The number of oxygen atoms in the reduction half-reaction is already balanced, so we only need to balance the oxidation half-reaction.

Oxidation half-reaction: NaI → I₂ + 2H₂O

Step 4: Balance the hydrogen atoms

Now, we balance the hydrogen atoms by adding hydrogen ions (H⁺) to the half-reactions. The number of hydrogen atoms in the oxidation half-reaction is already balanced, so we only need to balance the reduction half-reaction.

Reduction half-reaction: MnO₂ + 4H⁺ → MnSO₄ + 2H₂O

Step 5: Balance the charges

Next, we balance the charges in the half-reactions by adding electrons (e⁻). The number of electrons in the oxidation half-reaction should be equal to the number of electrons in the reduction half-reaction.

Oxidation half-reaction: 2NaI → I₂ + 2H₂O + 2e⁻ Reduction half-reaction: MnO₂ + 4H⁺ + 2e⁻ → MnSO₄ + 2H₂O

Step 6: Multiply the half-reactions

To make the number of electrons equal in both half-reactions, we multiply the oxidation half-reaction by 2.

Oxidation half-reaction: 4NaI → 2I₂ + 4H₂O + 4e⁻ Reduction half-reaction: MnO₂ + 4H⁺ + 2e⁻ → MnSO₄ + 2H₂O

Step 7: Combine the half-reactions

Finally, we combine the two half-reactions by adding them together. The electrons on both sides of the equation cancel out.

Overall balanced equation: 4NaI + MnO₂ + 4H⁺ → 2I₂ + MnSO₄ + 2H₂O + 4Na⁺

The balanced equation for the given redox reaction using the ion-electron method is:

4NaI + MnO₂ + 4H⁺ → 2I₂ + MnSO₄ + 2H₂O + 4Na⁺