Use standard cell potentials to show that a catalyst can work

[resurgance2001]

Homework Statement

Use standard electrode potentials to show that MnO2 can work as a catalyst to break down hydrogen peroxide.

Relevant Equations

Equation with higher potential goes forward. Reverse equation with lower potential and add the two together to get the overall reaction (cancelling spectator ions) and add potentials together to get the overall potential.

My solution in shown on the attached files. The overall reaction between Mn02 is 0.81 Volts

The overall reaction which shows the reformation of the MnO2 catalyst is .27 Volts. The first reaction is more positive which shows that the MnO2 can work as a catalyst.

 

[mjc123]

What attached files?

 

[resurgance2001]

Homework Statement:: Use standard electrode potentials to show that MnO2 can work as a catalyst to break down hydrogen peroxide.
Relevant Equations:: Equation with higher potential goes forward. Reverse equation with lower potential and add the two together to get the overall reaction (cancelling spectator ions) and add potentials together to get the overall potential.

My solution in shown on the attached files. The overall reaction between Mn02 is 0.81 Volts

The overall reaction which shows the reformation of the MnO2 catalyst is .27 Volts. The first reaction is more positive which shows that the MnO2 can work as a catalyst.

What attached files?

Sorry - thought I had. Thanks

 

[mjc123]

Your comment on the last page is not quite right. Reaction 1 does not show the reaction which the question says is catalysed. That reaction is not simply the decomposition of H₂O₂ to O₂, but the disproportionation to O₂ and H₂O, in which some of the O from H₂O₂ is oxidised and some reduced. (In reaction 1 it all goes to O₂, and the O in H₂O all comes from the MnO₂ - see half-reactions 1 and 2.) The disproportionation is the combination of reactions 1 and 2. The electrode potentials show that MnO₂ can oxidise H₂O₂ to O₂, and Mn²⁺ can reduce H₂O₂ to H₂O (regenerating the MnO₂).

 

[resurgance2001]

Your comment on the last page is not quite right. Reaction 1 does not show the reaction which the question says is catalysed. That reaction is not simply the decomposition of H₂O₂ to O₂, but the disproportionation to O₂ and H₂O, in which some of the O from H₂O₂ is oxidised and some reduced. (In reaction 1 it all goes to O₂, and the O in H₂O all comes from the MnO₂ - see half-reactions 1 and 2.) The disproportionation is the combination of reactions 1 and 2. The electrode potentials show that MnO₂ can oxidise H₂O₂ to O₂, and Mn²⁺ can reduce H₂O₂ to H₂O (regenerating the MnO₂).

Hi - thanks for your answer. I will be honest I still don’t quite get how the electrode potentials are showing this. The calculations I did on the first two pages are ok, is that correct. I stop don’t quite get how electrode potentials are actually showing this. I feel I need to see another similar question or see a model answer. I did have a model answer but it did not really make sense to me! You are saying that the reaction potentials she this because they both sum to positive values? Is that correct? I have searched for similar questions and even thought of trying to do the same thing using potassium iodide if I could find the standard electrode potential for that. Hmm I feel confused.

 

[mjc123]

E°(I₂/I^-) = +0.54 V. https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)
This is an interesting one. The E°s show that under acidic conditions, I^- can reduce H₂O₂, but I₂ cannot oxidise it. So it does not cataslyse the disproportionation. If you add KI solution to an acidic solution of H₂O₂ you get a precipitate of I₂. However, if you add a little KI to an alkaline solution of H₂O₂, you get disproportionation, with rapid exothermic evolution of O₂. Why is this? Can you apply the Nernst equation to half-reactions 1 and 3 and see what the potentials would be at pH 14?

 

[resurgance2001]

I have also just received this answer from someone. Is this answer correct? Or is it way off the mark? It sounds reasonable to me but I am not experienced enough to be certain it is correct.

 

[mjc123]

I'm afraid this is complete nonsense.
Where do they get 2.86 V? Are they adding the two electrode potentials and getting it wrong? You should subtract them.
They then go on to compare the cell (full reaction) potential with an electrode (half-reaction) potential, which is not legitimate. (Ask them to write out the equation for the reaction with E_net = 1.35 V.)
Looking at your thread title, you may have the same confusion.

 

[resurgance2001]

I'm afraid this is complete nonsense.
Where do they get 2.86 V? Are they adding the two electrode potentials and getting it wrong? You should subtract them.
They then go on to compare the cell (full reaction) potential with an electrode (half-reaction) potential, which is not legitimate. (Ask them to write out the equation for the reaction with E_net = 1.35 V.)
Looking at your thread title, you may have the same confusion.

 

[resurgance2001]

I'm afraid this is complete nonsense.
Where do they get 2.86 V? Are they adding the two electrode potentials and getting it wrong? You should subtract them.
They then go on to compare the cell (full reaction) potential with an electrode (half-reaction) potential, which is not legitimate. (Ask them to write out the equation for the reaction with E_net = 1.35 V.)
Looking at your thread title, you may have the same confusion.

This is the answer I got originally from a really experienced A level chemistry teacher. This was the reason for my post because I don’t understand his answer. I am going to try writing out my own (revised) answer again. Unfortunately, I can’t find the original exam the question came from and so I can’t see the mark scheme and I can find no other similar questions in the textbook. I am not sure how you arrived at 1.35 Volts, so I need to look at that. I don’t see how there can be a ‘single’ overall potential because the three half equations can’t just be added. Looking at the first two equations the first one has to be reversed. Then looking at the second and third equation the second one has to be reversed. So you end up with two apparently independent ionic equations - although they are obvious not independent. There are no examples I can find that are similar to this.