Balancing Redox Equation...Help!?

Mn^2+ --> MnO4^- + 5e- ( now balance charge with water and H+

4 H2O + Mn^2+ ----------> MnO4^- + 5 e- + 8H+ (use 4 H2O on left to provide O for MnO4-)

H2O2 + 2 H+ + 2 e- --------> 2 H2O (least common multiple 2,5 is 10 so cross multiply to balance electrons, tyhen cancel like terms.

2 Mn^2+ + 5 H2O2 + 10 H+ + 8H2O ----> 2 MnO4^- + 10 H2O + 16 H+

2 Mn^2+ + 5 H2O2 -------> 6 H+ + 2 MnO4^- + 2 H2O (balanced atoms and charge)

Start with the two half-reactions and make sure the atoms that are changing oxidation state are balanced:

Mn2+ --> MnO4-

H2O2 --> 2 H2O

Next, balance oxygens by adding H2O

Mn2+ + 4 H2O --> MnO4-

H2O2 --> 2 H2O

Now, balance hydrogen by adding H+:

Mn2+ + 4 H2O --> MnO4- + 8 H+

2 H+ + H2O2 --> 2 H2O

Now, balance charges by adding e-:

Mn2+ + 4 H2O --> MnO4- + 8 H+ + 5 e-

2 H+ + 2 e- + H2O2 --> 2 H2O

Now, you multiply one or both equations by a small number to make the number of electrons the same in the two reactions:

2 Mn2+ + 8 H2O --> 2 MnO4- + 16 H+ + 10 e-

10 H+ + 10 e- + 5 H2O2 --> 10 H2O

Now, add the two equations, cancelling anything that is on both sides:

2 Mn2+ + 5 H2O2 --> 2 MnO4- + 2 H2O + 6 H+

And you're done...