Write a balanced half-reaction for the reduction of solid manganese dioxide MnO2 to manganese ion Mn 2 in acidic aqueous solution. Be sure to add physical state symbols where appropriate.

2 e⁻ + 4 H⁺(aq) + MnO₂(s) → Mn²⁺(aq) + 2 H₂O(l)

Explanation:

Let's write the half-reaction for the reduction of solid manganese dioxide MnO₂ to manganese ion Mn²⁺ in acidic aqueous solution.

MnO₂(s) → Mn²⁺(aq)

To balance the oxygen atoms, we will add 2 moles of water on the right side. To compensate for the hydrogen atoms added, we will add 4 hydrogen ions on the left side.

4 H⁺(aq) + MnO₂(s) → Mn²⁺(aq) + 2 H₂O(l)

Finally, we will balance the charges by adding 2 electrons on the left side.

2 e⁻ + 4 H⁺(aq) + MnO₂(s) → Mn²⁺(aq) + 2 H₂O(l)

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).