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)

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

answer: i have one answer

acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (h+) ions

explanation:

hope this and sorry i could only come up with one answer!