Hi Penny, the best way to do these is by the method of half-reactions where you first break down the reaction into the unbalanced oxidation and reduction half-reactions. Then you balance each half-reaction separately by the following procedure:
First do steps 1-4 to each half-reaction:
1. Balance the elements other than O and H.
2. Balance the oxygens by adding H2Os to the appropriate side.
3. Balance the hydrogens by adding H+ to the appropriate side.
4. Balance the charge by adding electrons (e�?/SUP>) to the appropriate side.
Then you are ready to do steps 5 & 6:
5. Balance the electrons lost and gained by multiplying one or both half-reactions through by an integer (so the electrons cancel when the half-reactions are added together).
6. Add the half-reactions together and combine like terms, if necessary.
By default, this gives the balanced reaction in acidic aqueous solution (since H+ ions are used to balance the hydrogens). To convert to basic solution, if asked for, simply add the same number of hydroxide ions, OH�?/SUP>, as you have H+ ions in the balanced reaction in acidic solution, to both sides of the equation (so it remains balanced). Convert each H+ + OH�?/SUP> to H2O and simplify the number of water molecules in the reaction to get the final reaction in basic solution. You could also do this to each balanced half-reaction before adding them together in step 6 if you prefer.
Here's #2 as an example:
+7 �? +3 �? +4 �? �?+1 +7 �?
MnO4�?/SUP> + ClO2�?/SUP> ---> MnO2 + OH�?/SUP> + ClO4�?/SUP> (basic solution)
I've added the oxidation states above the elements to aid in telling what is becoming oxidized (losing electrons) and what is becoming reduced (gaining electrons).
Since Cl in ClO2�?/SUP> goes from an oxidation state of +3 to +7, it must be losing 4 electrons, so ClO2�?/SUP> becomes oxidized. That automatically means that the other reactant, MnO4�?/SUP>, is becoming reduced, which it is, since Mn goes from +7 to +4 in the reaction by gaining 3 electrons.
Oxidation ClO2�?nbsp; ---> ClO4�?/SUP>
Half-Rxn
Reduction MnO4�?nbsp; ---> MnO2
Half-Rxn
Step 1: OK. Elements other than O and H are balanced.
Step 2: Balance the oxygens with H2O.
Oxidation ClO2�? + 2 H2O ---> ClO4�?/SUP>
Half-Rxn
Reduction MnO4�?/SUP> ---> MnO2 + 2 H2O
Half-Rxn
Step 3: Balance the H's with H+ ions.
Oxidation ClO2�?/SUP> + 2 H2O ---> ClO4�? + 4 H+
Half-Rxn
Reduction MnO4�?/SUP> + 4 H+ ---> MnO2 + 2 H2O
Half-Rxn
Step 4: Balance the charge on each side with electrons, e�?/SUP>.
Oxidation ClO2�?/SUP> + 2 H2O ---> ClO4�?/SUP> + 4 H+ + 4 e�?/SUP>
Half-Rxn (each side has a �? ion charge sum)
Reduction MnO4�? + 4 H+ + 3 e�?/SUP> ---> MnO2 + 2 H2O
Half-Rxn (each side has a zero ion charge sum)
Step 5: To balance the electrons lost and gained, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 4.
Oxidation 3 ClO2�? + 6 H2O ---> 3 ClO4�?/SUP> + 12 H+ + 12 e�?/SUP>
Half-Rxn
Reduction 4 MnO4�?/SUP> + 16 H+ + 12 e�?/SUP> ---> 4 MnO2 + 8 H2O
Half-Rxn
Step 6: Add the two half-reactions together and simplify.
3 ClO2�? + 6 H2O + 4 MnO4�?/SUP> + 16 H+ + 12 e�?nbsp; ---> 3 ClO4�? + 12 H+ + 12 e�?/SUP> + 4 MnO2 + 8 H2O
3 ClO2�?/SUP> + 4 MnO4�? + 4 H+ ---> 3 ClO4�?/SUP> + 4 MnO2 + 2 H2O
Check that the reaction is still element balanced and especially charge balanced! This is the balanced reaction in acidic solution.
To convert to basic solution, add 4 OH�?/SUP> to each side since we have 4 H+ ions to "neutralize" (forming 4 H2O).
3 ClO2�?/SUP> + 4 MnO4�?/SUP> + 4 H+ + 4 OH�?nbsp; ---> 3 ClO4�?/SUP> + 4 MnO2 + 2 H2O + 4 OH�?/SUP>
3 ClO2�?/SUP> + 4 MnO4�?/SUP> + 4 H2O ---> 3 ClO4�?/SUP> + 4 MnO2 + 2 H2O + 4 OH�?/SUP>
3 ClO2�?/SUP> + 4 MnO4�?/SUP> + 2 H2O ---> 3 ClO4�?/SUP> + 4 MnO2 + 4 OH�?/SUP>
Check to be sure the reaction is still element and charge balanced. Whew! This is the balanced reaction in basic solution. 
Patience is the key! Go slow, follow the steps, and stay on the alert for careless errors!
Steve
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