Please ensure your speaker is turned on. The above video shows how hydrogen peroxide (H2O2) decomposes into water and oxygen gas. As mentioned in the video, manganese dioxide (MnO2) is used as a catalyst for this reaction.
By using the comment function of this post, answer the follow questions.
(a) Write a balance equation showing H2O2 decomposing into water and oxygen gas.
(b) By quoting relevant data from the data booklet, show how MnO2 functions as a catalyst.
hint 1: MnO2 can act as both an oxidising agent and a reducing agent.
hint 2: MnO2 is regenerated. This means MnO2 is first used in a redox reaction with H2O2 to form a Mn of a different oxidation state. And the Mn (of a different oxidation state) reacts wth H2O2 in a redox fashion to give u back your MnO2.
(c) Suggest the identity of the white fumes.
Note:
1. <-> is a sign (although inaccurate) that denotes the reaction is reversible.
2. Please leave down your name at the end of your entry. Failure to do so would imply that you have not submitted the homework assignment.
3. Mn(II) can be written as Mn^2+. And water can be written as H2O.
20 comments:
(a) 2H2O2(aq) --> 2H2O(l) + O2(g)
(b) The decomposition of H2O2 is a disproportionation reaction. It is redeuced to H2O and oxidised to O2.
When only H2O2 is present and MnO2 is added, H2O2 will be oxidised, hence:
O2 + 2H^+ + 2e- <-> H2O2 Eө=+0.68V
MnO2 + 4H^+ + 2e- <-> Mn^2+ + 2H2O Eө=+1.23V
Eөcell = +1.23 - (+0.68) = +0.55V, reaction is spontaneous.
Reduction of H2O2:
H2O2 + 2H^+ + 2e- <-> 2H2O Eө= +1.77V
MnO2 + 4H^+ + 2e- <-> Mn^2+ + 2H2O Eө=+1.23V
(Mn^2+ is the reducing agent)
Eөcell = +1.77 - (+1.23) = +0.54V, reaction is spontaneous. Mn^2+ is oxidised to form MnO2(regenerated).
(c)Oxygen gas
Angel
Melissa Wibowo
1SB4
a) 2H202aq)--> 2H20(l)+ O2(g)
b) MnO2 + 4H^+ + 2e^- <--> Mn^2+ + 2H2O E=1.52
Mn^2+ + 2e- <--> Mn E=-1.10
Firstly, MnO2 will be reduced to form Mn^2+
After that, Mn is oxidised as it is more favourable to be oxidised (E value is negative, equilibrium position shifts to the left), to become Mn^2+..
c) Oxygen
(a).H2O2 -> O2 + 2H2O
(b).MnO4^- + 4H^+ + 3e <-> MnO2 + 2H2O
E=+1.67V
Oxidation: MnO2 + 2H2O -> MnO4^- + 4H^+ + 3e
Reduction: MnO4^- + 4H^+ + 3e -> MnO2 + 2H2O
Since E value is positive, the reactions are spontaneous in standard conditions i.e. 298K the oxidation of MnO2 will occur.
The water molecules are from the decomposition of H2O2. A small amount of H2O2 will be decomposed to form water molecules without adding MnO2. Thus, MnO2 is being oxidised to form MnO4^- ions which shows that MnO2 act as a reducing agent. MnO4^- will then react with hydrogen ions (which comes from water molecules and reduction of MnO2) to form back MnO4+. Therefore, it can be seen that MnO2 is not being reduced or changed in amount which is a characteristic of a catalyst and it increases the rate of decomposition of H2O2. Hence, MnO2 functions as a catalyst.
(c).The white fumes are oxygen gas.
Name: Ng Xue Wen
Class: 1SB4
a) 2(H2O2) -> 2(H2O) + O2
b) Quoting from the data booklet,
MnO2 + 4(H+) + 2(e-) <-> Mn(2+) + 2(H2O) E* = +1.23 V [1]
H2O2 + 2(H+) + 2(e-) <-> 2(H2O) E* = +1.77 V [2]
O2 + 2(H+) + 2(e-) <-> H2O2 E* = +0.68 V [3]
Looking at the reaction [3] and [2],
H2O2 has less positive E* value, hence H2O2 would be oxidised to form O2. MnO2 would act as an oxidising agent, producing Mn(2+).
Looking at the reaction [1] and [2],
H2O2 has a more positive E* value and hence H2O2 would be reduced - giving H2O as a product.
In this reaction, Mn(2+) is oxidised back to MnO2.
Hence MnO2 is regenerated and it helps the oxidation and reduction of H2O2 to form both O2 and H2o.
This is how MnO2 acts as catalyst.
c) the white fumes is oxygen vapour.
Name: Jeanny
a)2H2O2(l)->2H2O(l)+O2(g)
b)MnO2 acts as an oxidising reagent and reacts with H2O2 to form Mn^2+, oxygen gas and water
MnO2(s)+H2O2(l)+2H^+(aq)<->Mn^2+(aq)+)2(g)+4H2O(l)
It then acts as a reducing agent and reacts with H2O2 again to form back MnO2
Mn^2+(aq)+H2O2(l)<->MnO4(s)+2H^+(aq)
C)Oxygen gas
YiPin
a) 2H2O2 -> 2H2O + O2
b) From the data booklet, MnO4 is seen to be able to undergo both oxidation and reduction.
MnO4^- + 4H^+ + 3e^- <-> MnO2 + 2H2O
and
MnO2 + 4H^+ + 2e^- <-> Mn^2+ + H2O
It reacts with H2O2 in two steps. Firstly, MnO2 oxidies H2O2 into water and oxygen gas with itself being reduced into Mn^2+ ions. The Mn^2+ ions is then oxidised by H202 to form MnO2 again.
c) Water Vapour
Joshua Toh.
a)2 H2O2 → 2 H2O + O2
b)The Eº of
MnO4- + 4H+ + 3e<-> MnO2 + 2H2O
is +1.67V
and
MnO2 + 4H+ + 3e <-> Mn2+ + 4H2O is +1.23v.
During the reaction, MnO2 would disproportionate into Mn2+ and MnO4-. To catalyze the H2O2 into Oxygen gas, MnO4- and H+ is used.
Once the reaction proceeds to the end, MnO4-,Mn2+,H2O and 4H+ would remain in the solution. However, they spontaneously react back to form the MnO2 used as the catalyst.
(Using the forward reaction of the first equation and the backward reaction of the second equation in part b)
MnO4- is reduced and Mn2+ is oxidised. Using
Eºcell=EºReduction hc - Eº oxidation HC
= +1.67- +1.23
= +0.44V
This is a positive value and the reaction is therfore spontaneous and the forward reaction proceeds to form back MnO2.
c) The white fumes could be gaseous H2O as the reaction is highly exothermic and would raise the temperature of the reaction above the boiling point of liquid water.
Daryl Toh 1sb4
(a) 2 H2O2 → 2 H2O + O2
(b) H2O2 + 2 H+ + 2 e <-> 2 H2O
E = +1.77V
MnO4- + 4H+ + 3e <-> MnO2 + 2H2O E= +1.67V
Ecell = +1.77 - (+1.67) = +0.10V
reaction is sponstaneous, H2O2 oxidises MnO2 to MnO4-.
O2 + 2H+ + 2e <-> H2O2 E = +0.68V
Ecell = +1.67 - (+0.68) = +0.99V
reaction is spontaneous, MnO4- reacts with H2O2 to form back the catalyst MnO2.
(c) The white fumes is oxygen.
Done by: Daniel 1SB4.
(a)2H2O2->O2+2H2O
(b) H2O2 is being oxidised to oxygen and water by MnO2. MnO2 is being reduced to Mn^2+. Then, the Mn^2+ will be oxidised by H2O2 to form MnO2 again while H2O2 is being reduced to water. In this way, the MnO2, which acts as the catalyst to both the reduction and oxidation of H2O2, is being regenerated. This shows that MnO2 will not be used up during the reaction. From the data booklet, the standard reduction potential of MnO2 to Mn^2+ is +1.23V and O2 to H2O2 is +0.68V. E-cell is +0.55V. Reaction is spontaneous. When Mn^2+ is oxidised to MnO2, the E value of it is +1.23V and E value for reduction of H2O2 to H2O is +1.77V. E-cell is +0.54V. Reaction is spontaneous. Since both E-cell values are positive and close to each other, i can deduce that MnO2 does not affect the reaction. As the reaction is non-spontaneous initially without addition of MnO2 and addition of it makes it spontaneous, i can conclude that MnO2 acts as catalyst that lowers the activation energy of the reaction and resulting it being spontaneous.
(c) The white fume is oxygen.
a) 2H2O2 -> 2H2O + O2
b) MnO2 + 4H+ + 2e <-> Mn2+ + 2H2O
MnO2 is first used in a redox reaction with H2O2 where MnO2 is reduced from Mn4+ to Mn2+ while H2O2 is oxidised from O- to O, hence forming O2. Mn2+ continues to react with H2O2 in a redox reaction where Mn2+ is oxidised to Mn4+ hence regenerating MnO2, while H2O2 is reduced from O- to O2-, hence forming water with the H+ ions produced. MnO2 is therefore a catalyst as it speeds up the decomposition of H2O2 through the redox reactions , liberating H2O and O2.
c) Oxygen gas
-Katrina
a) 2H2O2(l) <-> 2H2O(l) + O2(g)
b)
1.H202(l) + 2H^+(aq) + 2e -> 2H2O(l) E=+1.77
2.H202(l) -> O2(g) + 2H^+(aq) + 2e E=+0.68
3.MnO2(s) + 4H+(aq) + 2e -> Mn^2+(aq) + 2H2O(l) E=+1.23
4.2H20(l) -> O2(g) + 4H^+(aq) +4e E=+1.23
hydrgen peroxide decomposes to form water and oxygen. according to the half equations, the Ecell of the decompostion of hydrogen peroxide would be +1.77-(+0.68)=+1.09V. hence, the E value of the decompostion hydrogen peroxide would be +1.09V. thus, the E of manganese(II)oxide would be higher than that of hydrogen peroxide. this would make manganese(II)oxide a stonger oxidising agent than hydrogen peroxide. this means that hydrogen pereoxide would be oxidised and manganese(II)oxide would be reduced. Ecell=Ered-Eox=+1.77-(+1.09)=+0.68V. this would mean that the reaction would take place and hence manganese(II)oxide serves as a catalyst. manganese(II) ions would then react with the water form as seen in equation 4 to form oxygen gas and manganese(II)oxide again and the cycle repeats itself.
c)oxygen gas. hydrogen peroxide decomposes to form oxygen and water. it is likely that because of the high temperature, water vapour is present.
rui jie
Hi Mr Kwok, this is my assignment answers:
(a) 2H2O2(aq) -> 2H2O(l) + O2(g)
(b) MnO2 + 4H+ + 2e- <-> Mn^2+ + 2H2O E= +1.23
MnO4- + 8H+ + 5e- <-> Mn^2+ + 4H2O E= +1.52
MnO4- + 4H+ + 3e- <-> MnO2 + 2H2O E= +1.67
MnO4 functions as a catalyst when it undergoes a process of regeneration through a reaction. As the reaction proceeds with the more postive E value as the more favourable reaction, MnO4 is firstly being oxidised to MnO4- is the 3rd equation above, with oxidation number of Mn increasing from +4 to +7.
Afterwhich, it is being reduced by the H+ ions in the 2nd equation from +7 to +2. Mn^2+ reacts with H2O to produce MnO2 again in the first equation, thus MnO2 fuctions like this as a catalyst.
(c) Water Vapour.
From,
Sherlyn Koh
1SB4
(a) 2H202 -> 2H20 + O2
(b) From Data Booklet:
Mn02 + 4H^+ + 2e <-> Mn^+2 + 2H2O (1)
O2 + 2H^+ + 2e <-> H2O2 (2)
H2O2 + 2H^+ + 2e <-> 2H2O (3)
Since 2 moles of H2O2 is used(from part (a)), amount of H^+ in (2) dissociated is 4 moles. Therefore, the 4H^+ would react with the MnO2 to form Mn^+2 and 2H20.
More H2O is produced from (3) due to the dissociation of H2O2. Thus, the H20 produced would react with the Mn^+2 formed in (1) to form MnO2.
Hence, the amount of MnO2 does not change at the end of the reaction. This, MnO2 functions as a catalyst in this reaction.
(c) Oxygen gas.
a) 2H2O2 (aq) -> 2H2O (l) + O2 (g)
b)[MnO2 speeds up the decomposition of H2O2 by forming an intermediate by reacting with H2O2. This intermediate then reacts with more H2O2 to produce H2O and O2, and to regenerate the MnO2. MnO2 is not used up in the reaction.]
c) the white fumes may be toxiz fumes produced from the heating of the plastic bottles as the plastic reacts with the oxygen produced from the decomposition of the hydrogen peroxide.
-elizabeth
(a)2H2O2-> 2H2O + O2
(b)the redox reaction in this case is between MnO2 and H2O2
reduction half eqn - MnO2 + 4H^+ + 2e <-> Mn^2+ + 2H2O
oxidation half eqn - O2 + 2H^+ + 2e <-> H2O2
Ecell = 1.23-0.68=0.55V(>0 therefore reaction is spontaneous)
Combined equation - H2O2 + MnO2 + 2H^+ -> Mn^2+ + O2 + 2H2O
(c) It is oxygen gas
Hari Veluri
(a) 2H2O2-> 2H2O + O2
(b) MnO2 + 4H^+ + 2e- <-> Mn^2+ +2H2O
E= +1.23V
MnO2 reduces itself, making it an oxidising agent. Hence, it acts as an oxidising agent and oxidises H2O2 to produce H2O and O2. Since oxidation is already taking place in H2O2, addition of MnO2 speeds up the rate of oxidation due to it being an oxidising agent.
(c)The white fumes are a mixture of Oxygen gas and steam.
-jiayi
(a)
2H2O2 -> 2H2O + O2
(b)
When MnO2 is added to H2O2 solution, a redox reaction occurs.
MnO2 is reduced to Mn^2+ according to the equation:
MnO2 + 4H^+ -> Mn^2+ + 2H2O
The Mn^2+ ions formed are then oxidised as follow:
Mn^2+ + 4H2O -> MnO^4- + 8H^+ + 5e^-
After that, the MnO^4- ions undergo reduction again to form back MnO2:
MnO^4- + 4H^+ + 3e^- -> MnO2 + 2H2O
As seen from the above equations, MnO2 has been regenerated at the end of
the reaction.
Therefore, it functions as a catalyst.
(c)
Oxygen gas.
Quynh Phuong
a)2H2O2(aq)-> 2H2O(l) + O2(g)
b)Reduction half eqns:
MnO2 + 4H^+ + 2e <-> Mn^2+ + 2H2O
E(std)= +1.23V--------eqn1
MnO4^- + 4H^+ + 3e <-> MnO2 + 2H2O
E(std)= +1.67V--------eqn2
H2O2 + 2H^+ + 2e <-> 2H2O
E(std)= +1.77V---------eqn3
Oxidation half eqn:
MnO4^- + 8H^+ + 5e <-> Mn^2+ + 4H2O
E(std)= +1.52V-----------eqn4
O2 + 2H^+ + 2e <-> H2O2
E(std)= +0.68V-----------eqn5
Steps:
1)eqn 1 + eqn 5:
MnO2 + 4H^+ + H2O2 + 2e -> O2 + 2H^+ + Mn^2+ + 2H2O + 2e
Therefore,
MnO2 + 2H^+ +H2O2 -> O2 + Mn^2+ + 2H20 E(std cell)=(+1.23)-(+0.68)= +0.55V hence step 1 is spontaneous.
2)eqn 3 + eqn 4
Mn^2+ + 4H2O + H2O2 + 2H^+ + 2e -> MnO4^- + 8H^+ + 2H2O + 5e
Therefore,
Mn^2+ + 2H2O + H2O2 -> MnO4^- + 6H^+ +3e E(std cell)=(+1.77)-(+1.52)= +0.25V hence step 2 is spontaneous as well.
3)eqn 2
MnO4^- +4H^+ + 3e -> MnO2 + 2H2O
E(std)= +1.67V hence step 3 is spontaneous too.
Overall eqn: 2H2O -> 2H2O + O2
Since all the steps are spontaneous, it shows that MnO2 provides an alternative pathway for H2O2 to decompose with lower activation energy such that the decomposition can occur at standard conditions. Hence MnO2 works as a catalyst.
c) The identity of the white fumes could be steam, since the oxygen produced is a colourless gas.
Name: Sze Swai Ming
(a) 2H2O2 + 2H2O + O2
(b) MnO4 function as a homogenous catalyst where it takes part in the reaction by forming an intermediate with one of the reactants.
MnO2 + 2H2O -> MNO4 ־ + 4(H^+) + 3e ־ ___(x2)
H2O2 + 2(H^+) + 2e ־ -> 2H2O ___(x3)
2MnO2 + 3H2O2 -> 2MnO4 ־ + (2H^+) + 2H2O ___(1)
Thus, MnO4 ־ is the intermediate.
It reacts with more (H^+) to form MnO2, therefore the catalyst is regenerated
MnO4 ־ +4(H^+) + 3e ־ ->MnO2 + 2H2O___(x2)
H2O2 O2 + 2(H^+) + 2e ־ ____(x3)
2MnO4 ־ + 2(H^+) + 3H2O2 -> 2MnO2 + 4H2O + 3O2 ______(2)
Overall Equation: 6H2O2 -> 6H2O + 3O2
2H2O2 2H2O + O2
(c) The white fumes seen is water vapour that has been heated in the reaction when it is water.
done by: bing chang
a) 2H2O2 -> 2H2O + O2
b)
MnO2 is oxidised to Mn^2+:
MnO2 + 4H^+ -> Mn^2+ + 2H2O
Mn^2+ is then reduced to Mno2 and H2O is formed
MnO^4- + 4H^+ + 3e^- -> MnO2 + 2H2O
as MnO2 is regenerated at the end of reaction, it is a catalyst.
alvin
c)oxygen
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