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)

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72

the answer is barium and copper bromide