[Fish200398]

2MnO4 2- + A H2O2 + B H^{+} --> 2Mn2+ + D O_2 + E H_2O

what is A, B, D, E?

why i get 2 answers ?
2MnO4 2- +  1 H2O2 + 6 H^{+} --> 2Mn2+ +  3 O_2 +  4H_2O
2MnO4 2- + 5 H2O2 + 6 H^{+} --> 2Mn2+ + 5 O_2 + 8 H_2O

[AWK]

H₂O₂ may decompose independently of oxidation by permanganate. There is countless solution.

Check coefficient at oxygen in the first reaction

[Fish200398]

it said under acidic conditions

[AWK]

Your equation can be balanced in an infinite number of ways: this is a combination of two different reactions - oxidation and catalytic decomposition of hydrogen peroxide.

[Borek]

When balancing fails

[chenbeier]

2MnO4 2- + A H2O2 + B H^{+} --> 2Mn2+ + D O_2 + E H_2O

what is A, B, D, E?

why i get 2 answers ?
2MnO4 2- +  1 H2O2 + 6 H^{+} --> 2Mn2+ +  3 O_2 +  4H_2O
2MnO4 2- + 5 H2O2 + 6 H^{+} --> 2Mn2+ + 5 O_2 + 8 H_2O

Check the charges, both equations are wrong.

[AWK]

Helly, though through no fault of her own, encountered an extremely difficult and complicated reaction. The reaction can, of course, be "politely" balanced with coefficients 1, 2 and 4 on both sides of the equation, respectively.
But this reaction is formally the sum of two independent reactions, and therefore you can create an infinite number of solutions as a linear combination of both equations.
These are the reactions:
MnO₄^2- + 4H⁺ = Mn²⁺ + O₂ + 2H₂O
and
2H₂O₂ = O₂ + 2H₂O
The above reaction is catalyzed by manganese compounds.

With half the amount of acid, you can still write such a reaction
MnO₄^2- + 2H₂O₂ + 2H⁺ = Mn(OH)₂ + 2O₂ + 2H₂O
(and for this reaction of course, you can also write an infinite number of linear combinations with the catalytic decomposition of hydrogen peroxide)

[MNIO]

as you know by now the typical steps to balancing a redox reaction don't work well. 
  (1) identify the oxidiation states of every atom
  (2) determine which atoms are oxidized and which are reduced
  (3) write half reactions (including e's)
  (4) balance e's in half reactions
  (5) combine half reactions and cancel e's
  (6) rearrange as needed and add counter ions
  (7) add any remaining chemical species... then balance them

why doesn't it work well?

*** 1 **
on the left
  Mn in MnO4(2-) is +6
    O in MnO4(2-) is -2
    H in H2O2 is +1
    O in H2O2 is -1.. -1 not -2 !!!!
    H in H+ is of course +1

on the right
   Mn in Mn(2+) is +2
     O in O2 is 0
     H in H2O is +1
     O in H2O is -2

*** 2 ***
reduction half
.. Mn goes from +6 to +2 and is REDUCED (reduction is reduction in oxidation state)

clear so far right?  but what happens to the O??
  some O might go from -2 to 0 and be oxidized
  some O might go from -1 to 0 and be oxidized
  some O might go from -1 to -2 and be reduced
  some O might go from -2 to -2 and remain unchanged

so which is which?
  does the O in MnO4(2-) remain unchanged or get oxidized to 0?
  does the O in H2O2 get reduced to -2 or get oxidized to 0?

let's start by assuming the manganate ion oxidizes the peroxide oxygen atoms and the O in the manganate remains unchanged.  i.e.
  O in H2O2 goes from -1 to 0 and is oxidized
   O in MnO4(2-) remains unchanged

*** 3 ***
half reactions
  2 Mn(6+) + 8 e's ----> 2 Mn(2+)
  2 O(-1) ----> 2 O(0) + 2 e's

*** 4 ***
balancing e's
  2 Mn(6+) + 8 e's ----> 2 Mn(2+)
  8 O(-1) ----> 8 O(0) + 8 e's

*** 5 ***
combining and canceling e's
  2 Mn(6+) + 8 e's + 8 O(-1) ----> 2 Mn(2+) + 8 O(0) + 8 e's
then
  2 Mn(6+) + 8 O(-1) ----> 2 Mn(2+) + 8 O(0)

*** 6 ***
rearranging and adding counter ions as necessary
  2 Mn(6+) + 8 O(-1) ----> 2 Mn(2+) + 8 O(0)
  2 MnO4(2-) + 4 H2O2 ---> 2 Mn(2+) + 4 O2

*** 7 ***
adding additional chemicals
  2 MnO4(2-) + 4 H2O2 + __ H(+) ---> 2 Mn(2+) + 4 O2 + __ H2O
balancing O's
  2 MnO4(2-) + 4 H2O2 + __ H(+) ---> 2 Mn(2+) + 4 O2 + 8 H2O
balancing H's
  2 MnO4(2-) + 4 H2O2 + 8 H(+) ---> 2 Mn(2+) + 4 O2 + 8 H2O

*********
let's check to see if that's balanced
  left
     2 Mn
     2*4 + 4*2 = 16 O's
     4*2 + 8 = 16 H's
  right
     2 Mn
     4*2 + 8*1 = 16 O's
     8*2 = 16 H's
  electrons
     2 Mn's gain 4 e's each = 8 e's
     8 O's lose 1 e each = 8 e's

so
    2 MnO4(2-) + 4 H2O2 + 8 H(+) ---> 2 Mn(2+) + 4 O2 + 8 H2O
is certainly balanced.

the only issue is, that balancing job simplifies to
  1 MnO4(2-) + 2 H2O2 + 4 H(+) ---> 1 Mn(2+) + 2 O2 + 4 H2O

and we didn't need to start with a coefficient of 2 for MnO4(2-) and Mn(2+)
meaning something went wrong... right?

let's see if we can sort it out.
*********
*********
so.. either
   (A) I have the oxidation stuff wrong
   (B) the problem has a mistake in the charge of MnO4 ion.

let's see if (B) fixes the problem

*********
*********
if the reaction is really.
  2 MnO4(1-) + A H2O2 + B H(+) --> 2 Mn(2+) + D O2 + E H2O
which is a well known reaction where the "permanganate ion" oxidizes the peroxide, then going through the steps quickly

  Mn goes from +7 to +2 and is reduced
  O (in the H2O2) goes from -1 to 0 and is oxidized

half reactions
  2 Mn(7+) + 10 e's ---> 2 Mn(2+)
  2 O(-1) ---> 2 O(0) + 2 e's
balancing
  2 Mn(7+) + 10 e's ---> 2 Mn(2+)
  10 O(-1) ---> 10 O(0) + 10 e's
combining and canceling e's
  2 Mn(7+) + 10 O(-) ---> 2 Mn(2+) + 10 O(0)
adding counter ions and rearranging
  2 MnO4(-) + 5 H2O2 ---> 2 Mn(2+) + 5 O2
adding additional species and balancing
  2 MnO4(-) + 5 H2O2 + __ H(+) ---> 2 Mn(2+) + 5 O2 + __ H2O
  2 MnO4(-) + 5 H2O2 + __ H(+) ---> 2 Mn(2+) + 5 O2 + 8 H2O
  2 MnO4(-) + 5 H2O2 + 6 H(+) ---> 2 Mn(2+) + 5 O2 + 8 H2O

and all is well.

**********
my guess is you have a typo on that manganate / permanganate ion.. should be -1 not -2

[Fish200398]

yes, it was typo. thank you MNIO and all

[AWK]

Even for MnO₄^- The basic equation with coefficients 2, 5 and 6 on the left side of the equation is only a nice theory.
Catalytic decomposition of hydrogen peroxide with the manganese compound is very fast.

[Borek]

Even for MnO₄^- The basic equation with coefficients 2, 5 and 6 on the left side of the equation is only a nice theory.

I wouldn't dismiss it so easily. In low pH it is correct/stable/reproducible enough for the permanganate titration to be a known and valid method of quantitative hydrogen peroxide determination, listed in most analytical chemistry books.

[AWK]

Even for MnO₄^- The basic equation with coefficients 2, 5 and 6 on the left side of the equation is only a nice theory.

I wouldn't dismiss it so easily. In low pH it is correct/stable/reproducible enough for the permanganate titration to be a known and valid method of quantitative hydrogen peroxide determination, listed in most analytical chemistry books.

I agree, but  - in analytical procedures where an excess of sulfuric acid is used and acid is added before titration.