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In a voltaic pile with copper and zinc as the two metals and with aqueous sodium chloride (salt water) as the electrolyte, what happens to the electrons and ions and what reactions occur?

I understand that in a copper/zinc voltaic pile using sulfuric acid as the electrolyte, the zinc electrode decomposes into positive zinc ions (which dissolve into the electrolyte) and 2 electrons, which travel through the exterior wire to the copper electrode. At the copper electrode, the electrons are then accepted by positive hydrogen ions from the electrolyte solution which form hydrogen gas at the copper electrode.

The problem I see when this is adapted to a sodium chloride electrolyte is that the positive deposited ion would be pure sodium and it seems unlikely that sodium would form in its pure form both because zinc cannot reduce sodium (being lower in the reduction potential list) and because sodium is very hard to obtain in its pure form. Nonetheless, using a sodium chloride electrolyte for a voltaic pile seems to work frequently in many household experiments (according to my research),

So my question is: What is going on? Is there a new, different reaction? What happens to the ions and electrons?

Answer 1:

It's been a long time since I've been into chemistry, but I think I understand what is happening. If so, then you are correct about electro negativity (I think that is what you are referring to as "electron potential"): zinc will not reduce the sodium because zinc is more electronegative than sodium, but will reduce hydrogen because it is less electronegative than hydrogen. However, there *is* hydrogen in the system in an oxidized state: namely, the water the solution of sodium chloride is dissolved in. This will dissolve the zinc and produce hydrogen gas, exactly as if you had used sulphuric acid. The sodium chloride from what I can tell is just there to make the water electrically conducting - nothing will happen if the electricity has no way to get where the emf is sending it.


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