# Why does oxidation of the electrode take place in a solution?

What makes the electrode atoms start to ionize? Perhaps because of the solutes in the solvent?

So probably if you place it in distilled water it will not ionize, right?

Follow up question: If the solutes in the water are the reason of ionizing the electrode, Do they rip off the electrons from the electrodes atoms? (Most of websites illustrate that both electrons (on the electrode) and ions (in solution) contribute to the voltage of the electrode).

So why do I use electrode potentials if I have lets say a bowl made purely of $\ce{Zn}$ atoms and it contains lets say Copper ions. Even though I only have positive ions of copper. Does the copper ions generate the same voltage by their own without electrons?

## 1 Answer

Sorry to be so catty but the overall line of questioning indicates a profound lack of understanding. I can't explain all about of redox reactions in a few paragraphs to let me try to answer some of the questions in a general way and hope that those answers will provide some focus.

What makes the electrode atoms start to ionize? Perhaps because of the solutes in the solvent?

Every redox reaction has both a reduction reaction and an oxidation reaction.

I'm going to assume two electrodes in the solution. Thus in a redox reaction one electrode will have an oxidation reaction and the other will have a reduction reaction. The two electrodes will be connected via a wire through which electrons flow. Without the wire no reaction will occur.

So probably if you place it in distilled water it will not ionize, right?

There are two aspects to the answer to this question.

First distilled water doesn't conduct well. So even with our two electrodes in the distilled water we wouldn't see much current flow through the wire connecting the two electrodes. In this case it isn't the wire limiting current flow but the counter flow of ions through the distilled water.

Second chemists chose "electrodes" that will be inert in the solution being used. For example if a sodium metal "electrode" was dipped into water then a violent reaction would occur spontaneously on the contact surface between the electrode and the water. The products would be hydrogen gas and sodium hydroxide. So a sodium metal electrode in an aqueous solution just won't work.

Follow up question: If the solutes in the water are the reason of ionizing the electrode, Do they rip off the electrons from the electrodes atoms? (Most of websites illustrate that both electrons (on the electrode) and ions (in solution) contribute to the voltage of the electrode).

An overall redox reaction can be written as two half-cell reactions. The gist is that we can use electrons to balance the formula. This makes it easy to balance an overall redox reaction since we just have to balance the electrons.

So for an oxidation half-cell for a zinc electrode we have:
$\ce{Zn -> Zn^{2+} + 2e^-}$

For a reduction half-cell for a copper electrode we have
$\ce{Cu^{2+} + 2e^- -> Cu}$

Since we have two electrons in both half cells we can just add the equations as is to get:

$\ce{Cu^{2+} + Zn -> Cu + Zn^{2+}}$

Now there is a quirk in this particular reaction. If the copper ions in solution, which are to be reduced, come in contact with the Zn electrode then the copper ions would plate on the Zn electrode. So a membrane of some sort must be used as a barrier which keeps the copper ions away from the Zn electrode but still allows anions to flow through the barrier. Of course if we used ordinary glass as the barrier then neither anions nor cations could flow through the barrier and the battery formed by the two half-cells wouldn't work.

So why do I use electrode potentials if I have lets say a bowl made purely of Zn atoms and it contains lets say Copper ions. Even though I only have positive ions of copper. Does the copper ions generate the same voltage by their own without electrons?

Well a bowl can't be pure Zn and have Cu too.

So let me go to a hypothetical situation. I haven't checked Cu/Zn alloys (which would be possible but complicates the general idea here). So let's just assume a Cu/Zn alloy. When metals or alloys become solid they tend to form grains which are interlocked together by their 3D surfaces rather than one homogeneous material. The grains will vary in chemical composition. So the different grains have different electromotive potentials because of their compositional differences. Thus corrosion will occur. Some grains will dissolve and other grains will plate the dissolved ions back to metallic atoms.

Now let me clarify that corrosion is a very complicated subject for which I have made a gross generality. The actual corrosion reactions will depend on the metals in contact, and the exact composition of the solution with which they are in contact.

• ._., I didn't mean a galvanic cell. I just meant a half cell by it self. If you put an electrode in it, it starts to create ions and create a potential. chemguide.co.uk/physical/redoxeqia/introduction.html – Biker Sep 25 '16 at 18:01
• That is an entirely different question. Can I assume that your confusion is with the first three figures shown for the magnesium electrode? – MaxW Sep 25 '16 at 18:30
• Yes let me break the questions apart: Suppose you have a half cell and You place Mg electrode inside. Explain why: 1) Why does Mg atoms form ions in the water? Where do the electrons get the energy from to escape Mg atoms pull? 2)I have a problem with electrons being left on the electrode and not carried away ( if I say by a solvent). 3) I will post a picture of What I want – Biker Sep 25 '16 at 18:51
• Waiting for your picture, but let me make some comments about this. (1) This opens a can of worms. For all practical purposes you can ignore such dissolution at what I perceive your understanding level. (2) In order to measure any sort of electrode potential then you'd need two electrodes in the solution. (3) There are two phases here. One is solid the other liquid. The phase boundary is sort of like them being separated by a "force field" from the Starship Enterprise. ;-) So the ions or electrons have to somehow "jump" that barrier. – MaxW Sep 25 '16 at 20:20
• 1) I would like that can of worms for a better understanding .-., 2) i.imgur.com/yZDoOQl.png So at the top you can see a electrode disassociating in water forming copper ions and leaving electrons on the electrode(according to the website and you still didnt explain why :C) So from a physics point of view both electrons and ions contribute to the field generated between the ions and electrons thus should have a bigger voltage than (at the bottom) just copper ions their voltage should be a bit smaller than in the electrode. But we still interpret the copper ions and zin ions as E-pote – Biker Sep 26 '16 at 2:22