PROBLEM: Aqueous Sodium Permanganate plus solid Zinc plus liquid water yields solid Manganese Dioxide plus solid Zinc Hydroxide. (Note: The reaction takes place in basic solution)

Step 1: Write the unbalanced formula equations

NaMnO4(aq) + Zn(s) + H2O(l) --------> MnO2(s) + Zn(OH)2(s)

Step 2: Identify the species that are oxidized and reduced. Start by labeling the oxidation #s.

Since Zn goes from 0 to +2 it lost e– and is oxidized.

Since Mn goes from +7 to +4 it gained e– and is reduced.

Step 3: Write and balance the oxidation half reaction. Note – since Zn(OH)2 is a solid it does

NOT dissociateZn(s) --------> Zn(OH)2

Add 2 H2O to balance the O mass2 H2O + Zn(s) --------> Zn(OH)2

2 H2O + Zn(s) --------> Zn(OH)2 + 2 H+ + 2 e–

Next add e– to balance the charge – since the left is 0, and the right is +2, add 2e– to the right

NaMnO4(aq) + Zn(s) + H2O(l) --------> MnO2(s) + Zn(OH)2(s) +4+1 –2–20 +1–2+7 +2 –2 +1

Now we will balance the equation as if it was in acid and then switch it to base after we have the net ionic equation

2 H2O + Zn(s) --------> Zn(OH)2 + 2 H+

Next balance the H mass by adding 2 H+

Step 5: Next make the oxidation and reduction half-reactions have the same # of e–

MnO4– --------> MnO2

Step 4: Write and balance the reduction half reaction

Balance the O by adding H2O MnO4

– --------> MnO2 + 2 H2O

2 H2O + Zn(s) --------> Zn(OH)2 + 2 H+ + 2 e– (X3)

3 e– + 4 H+ + MnO4– --------> MnO2 + 2 H2O (X2)

6 H2O + 3 Zn(s) --------> 3 Zn(OH)2 + 6 H+ + 6 e–

Step 6: Combine the two half-reactions together, to make the net ionic equation. Remember to cancel out the items on opposite sides of arrow (water, H+ ions, and e–). In this case all the e–, 6 H2O , and 6 H+.

2 H2O + 2 MnO4– + 3 Zn + 2 H+ --------> 2 Zn(OH)2 + MnO2

Step 7: Convert to base by adding an OH– ion for every H+ ion. But remember – to keep the equation balanced you must add to both sides.

2 H2O + 2 MnO4– + 3 Zn + 2 H+ + 2 OH– --------> 3 Zn(OH)2 + 2 MnO2 + 2 OH–

Step 8: OH– plus H+ makes water – combine and cancel out waters from both sides4 H2O + 2 MnO4

– + 3 Zn --------> 3 Zn(OH)2 + 2 MnO2 + 2 OH–

Balance the H by adding H+

4 H+ + MnO4– --------> MnO2 + 2 H2O

Balance the charge by adding e–, since the left side is +3 and the right is 0, add 3 e– to the left3 e– + 4 H+ + MnO4

– --------> MnO2 + 2 H2O

6 e– + 8 H+ + 2 MnO4– --------> 2 MnO2 + 4 H2O

Step 7: Add back spectator ions and combine with other ions to write complete compounds and the balanced overall equation. Combine Na+ with the anions.

4 H2O + 2 NaMnO4 + 3 Zn --------> 3 Zn(OH)2 + 2 MnO2 + 2 NaOH