Balance this equation using oxidation reduction numbers?

Question

Also, state which substances are being oxidised and which are being reduced. Then fully balance the equation
[MnO4-] +H2O2+[H+] => [Mn2+] +O2+H2O

Answer 1

H is always +1, except when it is 0 in H2 and -1 in hydrides. O is always -2, except when it is 0 in O2 and -1 in peroxides. H2O2 is a peroxide. So Mn starts out as +7, because +7 - 8 = -1 charge on MnO4-. Mn finishes as +2, the charge on Mn2+. O starts out as -1 and finishes as 0 in O2.

Draw a line over the top of the equation from MnO4- to Mn2+ with "+5e-" written in the middle. This means that MnO4- gains 5 electrons going from +7 to +2.

Draw a line under the equation from H2O2 to O2 with "-2e-" written in the middle. This means that each O loses 1 electron going from -1 to 0, and there are 2 O's on each side, so -2e-.

Electrons lost must equal electrons gained, so multiply both MnO4- and Mn2+ by 2, and multiply both H2O2 and O2 by 5. Now each loses or gains 10e-.

Next count the O's on each side. 2MnO4- has 8 O's and 5H2O2 has 10 O's, so there are 18O's on the left. 5O2 has 10 O's, so the H2O on the right must be 8H2O, so there can be 18 O's on the right. 8H2O has 16H's, so H+ on the left must be 6H+. That balances the equation.

Answer 2

2MnO4- + 5H2O2 + 6H+ --> 2Mn2+ + 5O2 + 8H2O

MnO4- is the oxidizing agent and it is reduced. (Oxidation number of Mn decreases from +7 to +2)

H2O2 is the reducing agent and it is oxidized. (Oxidation number of O increases from -1 to 0)

Answer 3

reduced: Permanganate ion
oxidized: Peroxide

Answer 4

2MnO4- + 5H2O2 + 6H+ --> 2Mn2+ + 5O2 + 8H2O

Answer 5

eh no..