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Topic: Redox Reactions  (Read 3785 times)

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Offline Incendium

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Redox Reactions
« on: April 07, 2013, 12:59:46 AM »
Can Someone please help me with this question please?

Which one of the reactions below, to reduce a metal oxide, is likely to occur? Why?
A: PbO + H2 ---> Pb + H2O
B: Fe2O3 + 2 Al ---> Al2O3 + 2Fe
C: MgO + CO ---> Mg + CO2
D: 2CuO + C ---> 2 Cu + CO2

Thanks in Advance.

Offline Needaask

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Re: Redox Reactions
« Reply #1 on: April 07, 2013, 03:13:33 AM »
Can Someone please help me with this question please?

Which one of the reactions below, to reduce a metal oxide, is likely to occur? Why?
A: PbO + H2 ---> Pb + H2O
B: Fe2O3 + 2 Al ---> Al2O3 + 2Fe
C: MgO + CO ---> Mg + CO2
D: 2CuO + C ---> 2 Cu + CO2

Thanks in Advance.

I would say D.

For B, they are both solids so their activation energy is extremely high.
For C, MgO is extracted by electrolysis so I don't think that reaction will take place at all.
For A and D, both reaction will occur but looking at the reactivity series Cu is lower down than Pb and furthermore C is a better reducing agent than H2 as evidenced by H2's inability to reduce ZnO while C is capable of doing so. So I think the reaction that's most likely to occur is D.

Offline Borek

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Re: Redox Reactions
« Reply #2 on: April 07, 2013, 04:44:35 AM »
I would say D

Let's see...

Quote
For B, they are both solids so their activation energy is extremely high.

Same about D, yet you said D is possible.

Not to mention the fact B is a well known termite reaction.

Quote
For C, MgO is extracted by electrolysis so I don't think that reaction will take place at all.

Electrolysis of what?

I can think of three obvious methods of producing MgO - neither uses electrolysis.

B & C have commercial applications, I don't see any particular reason why A & D should be impossible.
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Offline Needaask

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Re: Redox Reactions
« Reply #3 on: April 07, 2013, 04:50:43 AM »
I would say D

Let's see...

Quote
For B, they are both solids so their activation energy is extremely high.

Same about D, yet you said D is possible.

Not to mention the fact B is a well known termite reaction.

Quote
For C, MgO is extracted by electrolysis so I don't think that reaction will take place at all.

Electrolysis of what?

I can think of three obvious methods of producing MgO - neither uses electrolysis.

B & C have commercial applications, I don't see any particular reason why A & D should be impossible.

Oops for C I meant to get Mg from the MgO i would have to electrolyse molten MgO. Oh that's right B is used in railway tracks I think. But comparing A and D i still think that D occurs more easily right? So I guess we would compare B and D for this question?

Edit: I thought of a crazy unorthodox method by comparing the difference between the metal oxide and the reducing agent in the reactivity series. So the more positive the difference the more likely the reaction? Other than that I'm out of ideas..
« Last Edit: April 07, 2013, 05:28:07 AM by Needaask »

Offline Needaask

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Re: Redox Reactions
« Reply #4 on: April 07, 2013, 09:11:46 AM »
sorry I think that we can't really use electrolysis to say that it that reaction would take more energy. The only reason why electrolysis is used for them is that they are cheaper right? We can actually also use carbon but we would have to supply a lot more energy to it. So in this case my new explanation for this is that copper in copper oxide is the least reactive. So it's the easiest for it to be reduced. So now we have to compare the oxidation of the reducing agent. Am I right?

Offline Arkcon

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Re: Redox Reactions
« Reply #5 on: April 07, 2013, 09:37:01 AM »
The only reason why electrolysis is used for them is that they are cheaper right? We can actually also use carbon but we would have to supply a lot more energy to it.

Can you explain this better?  What do you mean "put more energy into it?"  Exactly how, say if I've mixed carbon with a compound that can't be reduced, will I "put more energy into" the reaction, and make it go forward?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Needaask

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Re: Redox Reactions
« Reply #6 on: April 07, 2013, 09:54:37 AM »
The only reason why electrolysis is used for them is that they are cheaper right? We can actually also use carbon but we would have to supply a lot more energy to it.

Can you explain this better?  What do you mean "put more energy into it?"  Exactly how, say if I've mixed carbon with a compound that can't be reduced, will I "put more energy into" the reaction, and make it go forward?

Hmm I was thinking if K2O+C->K+CO if we apply Hess's law K2O(s)->2K+(g)+O2-(g), C(s)->C(g) then O2-(g)+C(g)->CO(g)+2e, 2K+(g)+2e->2K(g) and K(g)->K(s). So I don't really see why carbon is exclusively used for zinc oxides to copper (II) oxide. Perhaps we should continue this discussion at the new post I made so that we keep the 2 questions here separate?
« Last Edit: April 07, 2013, 10:27:06 AM by Needaask »

Offline Arkcon

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Re: Redox Reactions
« Reply #7 on: April 07, 2013, 10:02:04 AM »
True enough, but still, you've gone and quoted Hess's Law here as well, but still -- in practical terms, how will the "energy" be put into the reaction?  Do you mean simply thermal energy?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Needaask

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Re: Redox Reactions
« Reply #8 on: April 07, 2013, 11:08:08 AM »
True enough, but still, you've gone and quoted Hess's Law here as well, but still -- in practical terms, how will the "energy" be put into the reaction?  Do you mean simply thermal energy?

Like for  K2O(s)->2K+(g) +O2-(g),  C(s)->C(g) energy is absorbed and in O2-(g)+C(g)->CO(g)+2e, 2K+(g)+2e->2K(g) and  K(g)->K(s) energy is being released?

But actually I'm not too sure where the energy goes.. Do some of these reaction have an activation energy? Because in boiling/solidification I don't think there's an activation energy right?

And for my O2-(g)+C(g)->CO(g)+2e it should be simplified to O2-(g)->O+2e and C+O->CO so I'm not sure if these 2 reactions here have an activation energy too.

I'm thinking this step by step process should be split into the bond breaking part of the reaction (which adds up to activation energy) and the bond forming part of the reaction (which adds up to the reverse activation energy).

So I would guess that energy is being put in to the bond breaking components of the reaction like this: http://postimg.org/image/6ufkwqa9d/

That's why I don't see why I can't use carbon for these reactions. Didn't carbon only increase the activation energy (cos i have to turn it into a gas) and also give out some energy (when it reformed CO). So thinking about this now if i were to just decompose 2K2O->4K+O2, then my activation energy would be smaller as now bond breaking or activation energy would just be: 2K2O(s)->4K+(g)+2O2-(g), 2O2-(g)->2O(g)+4e and bond forming would be: 4K+(g)+4e->4K(g), 4K(g)->4K(s) and 2O(g)->O2(g). So in this reaction won't the activation energy be smaller?

Hmm this is pretty confusing now.. I don't see the need for carbon besides it giving out energy with a reaction with an oxygen atom. Could you explain my misconceptions here? Thanks :)

Offline Incendium

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Re: Redox Reactions
« Reply #9 on: April 07, 2013, 12:18:15 PM »
Which one is least likely to occur?

Offline Needaask

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Re: Redox Reactions
« Reply #10 on: April 07, 2013, 09:54:31 PM »
Which one is least likely to occur?

Hi are you comparing the thermal decomposition vs the reduction by carbon? I would guess that reduction by carbon is more likely to occur. But I think that that reaction has a higher activation energy but lower enthalpy change than the decomposition reaction. So perhaps that's why carbon is added - to lower the enthalpy change of the reaction?

But was I right in that Hess's Law breakdown where certain reactions within it which only involves bond breaking adds up to the activation energy and those reactions which involve bond forming add up to the reverse activation energy?

I'm not too sure about this because in this case there is no trace of the collision theory.. so I'm not very sure about how using this Hess's Law shows us the activation energy as well the collision theory.
« Last Edit: April 07, 2013, 11:21:43 PM by Needaask »

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