Chem: Energy changes
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1. I want to know what is produced when you burn hydrocarbons. Is it mostly carbon dioxide or carbon monoxide?
2. Electrial energy is produced by a simple cell as a result of
A the formation of covalent bonds between atoms
B the formation of negative ions from atoms
C the positive and negative ions coming together
D the transfer of electrons from a more reactive to less reactive element
My initial guess was D, and I switched to B later on thinking that release of energy involved exothermic process. How wrong was I. Actual answer's D. I need some explanation on this, preferably in basic terms. and tell me why the answer can't be B too. thanks.
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Students please feel free help your good brother bonkysleuth! I'll moderate silently, ie. if the discussion becomes misleading or errorneous, I (or my fellow Chem moderator, TinTin) will step in to say so.
This forum is for students to discuss and help each other learn. Carry on.
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For Qn 1, that will depends on the amount of oxygen that is avaliable.
In perfect equation, there should be all carbon dioxide. When the amount of oxygen is not enough, carbon monoxide will be created instead on carbon dioxide. The amount of carbon monoxide will depend on the amount oxygen present.
For example, methane only releases one carbon dioxide (CO2) for two water molecules (H2O) when burnt.
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Hi there!
I don't know how to explain how a current is produced when electrons are transferred...
But I am quite sure that simple cells involve 2 metals... So I don't see where the negative ions come from...
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A couple of comments...
Do not confuse electric current (which is the rate of flow of charged particles, eg. electrons, ions, etc) with energy (eg. heat energy).
In forming an anion from an atom, a quantifiable process known as electron affinity, depending on the element, energy may be released (in the form of heat), but no current flows.
Again, bonkysleuth, do not confuse "electric current" with "energy". They may be related (ie. current flow that can be used to do work is electrical energy), but specifically in the context of this MCQ qn, they are two completely distinct things that you seem to have confused.
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Hi. I myself am confused about why the transfer of electrons from a more reactive element to less reactive element can result in the production of electrical energy. Would someone please explain? Chanced upon this question in the TYS too.
Do reply asap because I want to understand the concept before submitting my TYS booklet to my teacher. And thanks bonkysleuth for raising this question up!
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Originally posted by anpanman:
Hi. I myself am confused about why the transfer of electrons from a more reactive element to less reactive element can result in the production of electrical energy. Would someone please explain? Chanced upon this question in the TYS too.
Do reply asap because I want to understand the concept before submitting my TYS booklet to my teacher. And thanks bonkysleuth for raising this question up!
A more reactive metal (eg. Na) wants to oxidize itself (ie. throw away it's valence electrons) to a more stable cationic state.But someone has to accept the electrons, or else the reactive metal (eg. Na) will not be allowed to anyhow throw away its electrons.
But no worries, a less reactive metal cation (eg. Cu2+) is here, willing to (or rather, 'bullied' into) accepting the electrons. So if Cu2+ accepts the electrons at the cathode, and Na throws away the electrons at the anode, then
[reduction at cathode] Cu2+ + 2e- ---> Cu
[oxidation at anode] 2Na ---> 2Na+ + 2e-
Now, you realize that the electrons cannot magically 'teleport' from the anode to the cathode. They have to travel or flow within a metal wire (that connects anode to cathode).
The rate of flow, of these electrons, in the external wire, is what we call "current".
Comprehende liao? anpanman and bonkysleuth?
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Originally posted by UltimaOnline:
A more reactive metal (eg. Na) wants to oxidize itself (ie. throw away it's valence electrons) to a more stable cationic state.But someone has to accept the electrons, or else the reactive metal (eg. Na) will not be allowed to anyhow throw away its electrons.
But no worries, a less reactive metal cation (eg. Cu2+) is here, willing to (or rather, 'bullied' into) accepting the electrons. So if Cu2+ accepts the electrons at the cathode, and Na throws away the electrons at the anode, then
[reduction at cathode] Cu2+ + 2e- ---> Cu
[oxidation at anode] 2Na ---> 2Na+ + 2e-
Now, you realize that the electrons cannot magically 'teleport' from the anode to the cathode. They have to travel or flow within a metal wire (that connects anode to cathode).
The rate of flow, of these electrons, in the external wire, is what we call "current".
Comprehende liao? anpanman and bonkysleuth?
Totally understood. ^^ Thank you! -
Incomplete combustion produces soot(unburnt carbon).