Question

In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 solution is added from a buret to a colorless, acidified solution of H2O2(aq) in an Erlenmeyer flask. (Note: At the end point of the titration, the solution is a pale pink color.)

Answer 1

Answer:

The oxygen element in H2O2 is the specie that is reduced in H2O and oxidized into O2.

Explanation:

5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)

Oxidation is defined as the outright loss of electrons. Oxidation leads to an increase in an element's oxidation number.

If this reaction is broken down into reduction and oxidation halves

It is observed that, of the reactants above,

H202 becomes H2O and O2

MnO4- + H+ becomes Mn2+ and H2O

Oxidation number of Mn changes from +7 In MnO4- to +2 In Mn2+ (evidently reduction)

The Oxygen in MnO4- doesn't change oxidation numbers as its oxidation number stays at -2

Oxidation number of Oxygen changes from -1 in H2O2 to -2 In H2O and 0 in O2

The Hyrogen in H2O2 doesn't change oxidation numbers, itsoxidation number stays at +1 in H2O2 and H2O.

This indicates that H2O2 undergoes oxidation and reduction; more specifically, the oxygen element in H2O2 is the specie that is reduced in H2O and oxidized into O2.

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