I had some time today and drew this pic in paint to help explain my question. The questions are bit wordy but I'll really appreciate if you can answer atleast one of them
1. What is the purpose of the salt bridge?
Is it because when electrons flow Cu side becomes positive and KI side becomes negative. To avoid this NO3- goes to the anode to try to regain the original negative charge and K+ vise versa. What happens if there was no salt bridge? Would the current flow both ways. Is the reason why this doesn't happen is because only Cu produces electrons.So electrons must flow from anode to cathode. Can reactants flow through. I don't think so because salt bridge is not a tube. It is paper in most cases isn't it.
I would state it a bit different to that (but similar basic ideas)
An electric current flows because of a potential difference across a conductor. The set up of the cell creates the p.d. (more on that later)
An electric current requires a charge carrier in the medium
In the metal wire the charge carrier is only electrons
In the solutions any mobile, dissolved ion can carry charge and will physically move to or from the electrodes depending on charge
So as well ions in solution, ions are needed in the salt bridge to complete the circuit.
Now, I have never actually seen a description of what actually happens at the interface between the salt bridge and the two solutions. I have thought about it and found nothing - so I forget about it until someone asks about the salt bridge. Sorry can't help on that bit
2. How does this really work?
Do the electrons from Cu go to the Cu electrode and then through the wire to the carbon electrode. Then Iodine recieves these electrons. Is this what happens.
The diagram in the following posthttp://www.chemicalforums.com/index.php?topic=26468.msg99853#msg99853
Gives a really good image of what is happening at the electrode/solution boundary - basically what you are describing goes on
3. My dislike towards reduction
Ok the reaction in KI is I2 + 2e- -------- 2I- . In the KI flask there is only I- present. However the equation implies I2 is originally present in the flask. Is it just the nature of the equation. If what happens is I2 gains electrons. Where is I2 to begin with there is only I- in the flask.
This is interesting. As you say for this reaction to occur there needs to be a source of I2
- which there isn't in this situation.
But is this the only reaction that can take place?
Another possible would be K+
But I guess this deposition of potassium would not take place because the equlibrium would lie well over the the left due to the relatively high, positive electrode potential of K+
What about other ions present in solution? Could something else in the KI solution be reacting (reducing) at the cathode? How could they behave and what are the electrode potentials?
Is the carbon significant? Is it just conducting electricity or reacting? I'm not sure. I did read something once about using carbon electrodes in a cell (and I think that's what is in standard batteries) but I can't find the link
Most cell setups I have studied have the electrode being the same material as the +ve ion in the salt solution. I suppose this makes the discussion and voltage calculation easier.
4. I thought oxidation and reduction can not happen separately. I mean in all redox equations balanced there have been 2 half equations. One for oxidation and one for reduction. I thought for oxidation to occur reduction must also occur and reverse.
Yes. Yes. Yes.
Also when we use electrochemical series to work out the cathode and anode we treat this as though oxidant and reductant are present in the same flask.
You have the key to the answer in your question.
The cell is set up in such a way to separate the redox two processes into two flasks. The setup uses the pd to drive a current through the wire and hence produce a usable and useful electric current when you complete the circuit. The whole purpose of doing this activity (apart from setting electrochemistry questions)
Indeed, as you write, if you mixed the reactants in the same container, the reactions would occur, but you wouldn't derive any usable electric current