balance redox equation

Question

In titration, H2O2 solution + Manganate (VII) ion
H2O2 --> O2 + 2H+
O.N of H in H2O2 = +2
O.N of O in H2O2 = -1
O.N of O2 =0
O.N of H = 1
又oxidation,又reduction,即係點?

Answer 1

The reducing agent of the reaction is H2O2. The half equation for the oxidation of H2O­2 is:
H2O2(aq) → 2H+(aq) + O2(aq) + 2e-
Oxidation number of H in H2O2 and in H+ = +1
Hence, there is no change in oxidation number of H.
Oxidation number of O in H2O2 = -1
Oxidation number of O in O2 = 0
Hence, there is an increase in oxidation number of O. Therefore, H2O2 undergoes oxidation.

The oxidizing agent of the reaction in MnO4-, named as manganate(VII) ion (not manganese(VII) ion). The redox reaction usually occurs in an acidic medium. The half equation for the reduction of acidified MnO4- ion is
MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O
Oxidation number of H in H+ and in H2O = +1
Oxidation number of O in MnO4- and in H2O = -2
Hence, there is no change in oxidation numbers of H and O.
Oxidation number of Mn in MnO4- = +7
Oxidation number of Mn in Mn2+ = +2
Hence, there is a decrease in oxidation number of Mn. Therefore, MnO4- undergoes reduction.

Overall equation:
2MnO4-(aq) + 6H+(aq) + 5H2O­2(aq) → 2Mn2+(aq) + 8H2O + 5O2(aq)

2007-09-29 20:07:04 補充：
There is NO such ion called "manganese(VII) ion". However, "manganese(VII)" usually refers to all compounds of Mn with oxidation number 7, such as manganate(VII) ion, manganese(VII) oxide, etc.

2007-09-29 20:08:05 補充：
There is NO such ion called "manganese(VII) ion", "Mn^ 7". However, "manganese(VII)" usually refers to all compounds of Mn with oxidation number 7, such as manganate(VII) ion, manganese(VII) oxide, etc.

2007-09-29 20:09:39 補充：
Sorry, the positive signs could not be shown, due to fault of Yahoo.