Write a balanced half-reaction for the oxidation of manganese ion (mn2 ) to solid manganese dioxide (mno2) in acidic aqueous solution. Be sure to add physical state symbols where appropriate.

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s) + 4H+(aq) + 2e-

Explanation:

Step 1: Data given

The oxidation number of manganese ion (Mn2+ ) is +2

The oxidation number of manganese dioxide  is +(MnO2)4

This means the oxidation number from Mn will go from +2 to +4, since it's increased, this is an oxidation reaction

Mn2+(aq)  ⇒ MnO2(s)

We have to balance both sides. Mn is already the same. But on the right side we have O atoms. T obalance both sides we have to add O atoms to the left side. This by adding 2x H2O

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s)

Now the amount of O atoms is balanced, but we have H- atoms at the left side. To balance we have to add 4 H atoms to the right side

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

Now the amount of atoms is balanced at both sides. We also have to check if the charge on both sides is the same.

Since the left side has a charge of +2, and right has a charge of +4, we have to add 2 electrons to balance this.

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s) + 4H+(aq) + 2e-

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

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