Redox reaction help anyone

Ok, so I'm doing some homework for chem. I've been trying to understand half-reaction and the method, but I just can't understand it.


Use the half reaction method to write balanced ionic equations for each reaction. All are basic solutions!

a) MnO4-(aq) + ClO2-(aq) --> MnO2(s) + ClO4-(aq)
*The furthest I can get in this one is to make them in ionic form, which won't help me too much.

Would anyone be able to take the time and help with this question? Thank you so much... I'd really like to learn how to do this!!

Step 0: Assign oxidation numbers

MnO4-(aq) + ClO2-(aq) --> MnO2(s) + ClO4-(aq)
+7 -2 --------- +3 -2 ---------- +4 -2 ------- +7 -2

Step 1: Write unbalanced half-reaction equations

ClO2-(aq) -----------> ClO4-(aq) Oxidation
MnO4-(aq) ----------> MnO2(s) Reduction

Step 2: Balance all elements except for O and H; in this case they are already balanced

ClO2-(aq) -----------> ClO4-(aq)
MnO4-(aq) ----------> MnO2(s)

Step 3: Balance O by adding H2O for each O

ClO2-(aq) + 2H2O(l) -----------> ClO4-(aq)
MnO4-(aq) ----------> MnO2(s) + 2H2O(l)

Step 4: Balance H by adding H+ for each H

ClO2-(aq) + 2H2O(l) -----------> ClO4-(aq) + 4H+
MnO4-(aq) + 4H+ ----------> MnO2(s) + 2H2O(l)

Step 5: For each H add OH on both side of the equation

ClO2-(aq) + 2H2O(l) + 4OH -----------> ClO4-(aq) + 4H+ + 4OH
MnO4-(aq) + 4H+ + 4OH ----------> MnO2(s) + 2H2O(l) + 4OH

the H and OH combines to become H2O, so you have

ClO2-(aq) + 2H2O(l) + 4OH -----------> ClO4-(aq) + 4H2O
MnO4-(aq) + 4H2O ----------> MnO2(s) + 2H2O(l) + 4OH

Step 6: Balance the charges, and to balance the half-reaction charges, multiply the oxidation half-reaction by 3, and the reduction half-reaction by 4

3 [ ClO2-(aq) + 2H2O(l) + 4OH -----------> ClO4-(aq) + 4H2O + 4e- ]
4 [ MnO4-(aq) + 4H2O +3e- ----------> MnO2(s) + 2H2O(l) + 4OH ]