Balance the following half-reaction: mno4-→ mno2(s)

1. the first species to balance is the redox active species, the element that is changing oxidation state. in this reaction, the redox active species, mn, is already balanced.
2. how many moles of h2o must there be on the reactant side of this half reaction

MnO₄⁻ + 2 H₂O + 3 e⁻ ⇒ MnO₂ + 4 OH⁻

2 moles of H₂O are on the reactant side.

Explanation:

Let´s consider the following half-reaction.

MnO₄⁻ ⇒ MnO₂

This reduction of permanganate to manganese dioxide happens in a neutral medium. The oxidation number of Mn changes from +7 to +4. When we are in a neutral medium, we can balance the reaction using the ion-electron method either as we were in acid or in basic medium. In this case, to satisfy the condition is the task (water on the reactant side) we will balance this half-reaction as if it were in basic media. Since Mn is already balanced, we have to balance O. We will add as many moles of water as excess of oxygen there is on 1 side, and the double of OH⁻ on the other side. Finally, we add as many electrons as we need to balance it electrically.

MnO₄⁻ + 2 H₂O + 3 e⁻ ⇒ MnO₂ + 4 OH⁻

answer: h+ its less than oh

explanation:

it's filtration, but this is an assumption so my apologies if i got this