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There is a beaker in which there are two iron electrodes (there is no salt bridge between two beakers-there's just one beaker) and one iron electrode is being bubbled with a supply of oxygen from a pipe (causing it to rust and fall to the bottom of the beaker). There is a flow of current in an external circuit (which I think is because the oxygen prevents one electrode from dissociating into $\ce{Fe^2+}$ and $\ce{2e-}$). The one that is not being bubbled with oxygen is the anode, but I'm not sure why. I think it's because the other one can't dissociate since it's reacting with oxygen. Also, the rate of rusting increases when $\ce{NaCl}$ is added to the water. Why is that? The Setup

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  • $\begingroup$ It is not clear what are you asking. Could you point out questions? Or the points of interest. $\endgroup$ – Jaroslav Kotowski Nov 12 '15 at 14:16
  • $\begingroup$ @JaroslavKotowski Why is the anode electrode B and why does adding NaCl to the water increase the rate of rusting? Also I thought that electrolyte was needed for flow of electrons in the first place, so how is just water doing the trick? $\endgroup$ – Airdish Nov 12 '15 at 14:39
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If your diagram is correct (the notation of the polarity of the electrodes) then the anode should be the one that is not being pumped with oxygen, because the negative end of any electro-chemical cell is the anode.

Let's see how this happens.

First of all let us consider the ions and molecules at two electrodes the system.

We have $\ce{O2}$ at electrode A and $\ce{Fe}$ at B .

Now since the reduction potential of $\ce{O2}$ is greater than $\ce{Fe}$ a reaction takes place at the $\ce{O2}$ pumping electrode. This is a reduction:

$$\ce{O2 + 2H2O + 4e- -> 4OH-}$$

Now at the other electrode, an oxidising reaction takes place:

$$\ce{Fe -> Fe^3+ + 3e-}$$

So you can see there's an excess of electrons at the $\ce{Fe}$ electrode. Now these electrons travel across to the $\ce{O2}$ containing electrode since the cell reaction there needs electrons. This travel of electrons give rise to an electric current. You might know from your previous lessons that current flows from negative terminal to the positive terminal inside a cell. Hence the $\ce{Fe}$ electrode becomes the negative anode.

As to your second question why does $\ce{NaCl}$ increase corrosion, for all these reactions to take place electrons should be moved about. $\ce{NaCl}$, being a strong electrolyte, increases the conductivity of water. That means electrons are moved rapidly giving a rise in the rate of corrosion. And you have asked in the comments "How is just water doing the trick?". You should keep in mind that water exists as $\ce{H+}$ and $\ce{OH-}$ in the aqueous liquid state.

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    $\begingroup$ A point you didn't cover - In a basic solution you'd get iron hydroxide, not iron oxide. $\endgroup$ – MaxW Nov 12 '15 at 17:28
  • $\begingroup$ but OPs question clearly states that iron(III)oxide is formed $\endgroup$ – slhulk Nov 12 '15 at 17:30
  • $\begingroup$ @slhulk I thought that water only exists as H+ and OH- when it's acting as a solvent? Also at electrode A I thought that the reaction was simply between Fe and O2 (3O2 + 6e- = 6O2- at electrode A, forming rust). Why is water part of the equation to give rise to OH-? And isn't OH- already there as you say that's how water exists? $\endgroup$ – Airdish Nov 13 '15 at 5:04
  • $\begingroup$ Water exists as H+ & OH- in aq medium. The reaction at A is determined by the reduction potential of elements present there. O2 reduces to OH-. Notice the change of oxidation state from 0 to -2. To balance it water is necessary. $\endgroup$ – slhulk Nov 14 '15 at 13:58
  • $\begingroup$ Water exists as H+ & OH- in aq medium. The reaction at A is determined by the reduction potential of elements present there. O2 reduces to OH-. Notice the change of oxidation state from 0 to -2. To balance it water is necessary. $\endgroup$ – slhulk Nov 14 '15 at 13:58

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