order of reaction of the decomposition of hydrogen peroxide

Question

potassium manganate(VII), hydrogen peroxide are reacting in a conical flask. At different time intervals, 10cm3 of reaction mixture is withdrawn, add to conical flask with 10cm3 H2SO4(stopwatch is started at this instant) and is then titrated with dilute K2MnO4 solution.

1. write a balance chemical equation to show the reaction between hydrogen peroxide and potassium permanganate.

2. what is the use of the manganese(IV)oxide?

3. explain why it is necessary to add the 10.0ml sample of the reaction mixture into sulphuric acid before titration. write balanced chemical equation to show any reaction.

4. write a balanced chemical equation to show the reaction occurred in titration.

5. what is the meaning of (V0 ml of potassium manganate(VII) solution are required for the oxidation of hydrogen peroxide at the beginning, and Vt ml are required after time t, then the concentration of hydrogen peroxide at these times are proportional to V0and Vt respectively.).

V0 represent what? and Vt represent what?

Answer 1

1.
A very small amount of potassium manganate(VII) is add to the hydrogen peroxide to produce manganese(IV) oxide, MnO2. Only a negligible proportion of hydrogen peroxide is used when potassium permanganate is completely reacted. The equation is :
2MnO4-(aq) + 3H2O2(aq) → 2MnO2(s) + 3O2(g) + 2H2O(l) + 2OH
-(aq)

=====
2.
Manganese(IV) oxide is used as the catalyst for the decomposition of hydrogen peroxide.
2H2O2(aq) → 2H2O(l) + O2(g) .... (catalyst: MnO2)

=====
3.
Excess H2SO4 is used to quench the reaction. In an acidic medium, the oxidizing power of manganese(VI) oxide is strong enough to undergo redox reaction with hydrogen peroxide.
MnO2(s) + H2O2(aq) + 2H+(aq) → Mn2+(s) + O2(g) + 2H2O(l)
In the absence of MnO2 catalyst, the decomposition of H2O2 becomes extremely slow and is practically stopped. Note that only negligible hydrogen peroxide is used in the above reaction.

=====
4.
2MnO4-(aq) + 5H2O2(aq) + 6H+(aq) → 2Mn2+(s) + 5O2(g) + 8H2O(l)
Note that the equation above is different from that in Q.1, because the reaction in Q.1 is done in a neutral medium while the above reaction is done in an acidic medium.

=====
5.
Vo : At the beginning (before the decomposition starts), 10 cm3 of the reaction mixture is withdrawing and the quenched with excess H2SO4. The resultant mixture is titrated with KMnO4 solution. It is found that Vo mL (not ml) of KMnO4 solution is added at the end point.

Vt : At time t, some hydrogen peroxide is decomposed whose concentration is lowered. 10 cm3 of the reaction mixture is withdrawing and the quenched with excess H2SO4. The resultant mixture is titrated with KMnO4 solution. It is found that Vt mL (Vt < Vo) of KMnO4 solution is added at the end point.

The higher the concentration of H2O2, the more volume of KMnO4 solution is needed. Therefore, the concentration of H2O2 is directly proportional to the volume of KMnO4 solution added.
At time = 0, volume of KMnO4 solution = Vo mL
At time = t, volume of KMnO4 solution = Vt mL