Question

5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g) 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.) Which element is being oxidized during the titration, and what is the element’s change in oxidation number?

Answer 1

Answer:

Oxygen in hydrogen peroxide oxidizes from -1 to 0.

Explanation:

Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.

The given reaction is shown below as:

$5 H_2O_2_{(aq)} + 2 MnO_4^-_{(aq)} + 6 H^+_{(aq)}\rightarrow 2 Mn^{2+}_{(aq)} + 8 H_2O_{(l)} + 5 O_2_{(g)}$

Manganese in $MnO_4^-$ has oxidation state of +7

Manganese in $Mn^{2+}$ has an oxidation state of +2

It reduces from +7 to +2

Oxygen in hydrogen peroxide has an oxidation state of -1.

Oxygen in molecular oxygen has an oxidation of 0.

Thus, oxygen in hydrogen peroxide oxidizes from -1 to 0.