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I've researching electrolysis recently, and all of the examples I've seen use two different metals as the cathode and anode.

In my attempts to make a hydrogen producing cell (table salt used as the electrolyte in water) I've used two steel nails as both my cathode and anode. When I power it up, one side bubbles and the other stays normal. The side where the bubbling occurred, corrodes and makes the water murky with what I presume to be rust. The other side does nothing, not even get electroplated with the other side's metal.

My question is this: Does electrolysis work the same way when the cathode and anode are made of the same metal? It might be a stupid question, but I can't find anything that answers this question specifically.

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    $\begingroup$ Can there be a word in which the same letter occurs twice? Yes. Why? Because (1) we can't possibly think of a valid reason that would prevent it and (2) we have a working example. The same applies to your case. $\endgroup$ Dec 16 '20 at 5:33
  • $\begingroup$ You cannot electroplate anything from a solution that hardly contains the respective ion, and I'm pretty sure you cannot electroplate with iron at all. You need a positive standard potential, otherwise you always get hydrogen. $\endgroup$
    – Karl
    Dec 16 '20 at 7:59
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My question is this: Does electrolysis work the same way when the cathode and anode are made of the same metal? It might be a stupid question, but I can't find anything that answers this question specifically.

The battery or the power source does not care about the electrode material. Its job is to pump the electrons in the circuit. Now the problems originate from the chemistry rather. By definition, if a direct current passes through an electrolyte, electrolysis (=chemical decomposition) has to occur. Something has to reduce at the cathode and something has to oxidize at the anode. Here, we enter the competition, as shown below in the case of your simple aqueous NaCl set up.

At the anode (the electrode connected to the positive terminal of the DC supply), you have to think... (a) Will the electrode oxidize itself? Possible (b) Will the solvent i.e., water oxidize to oxygen? Possible (c) Will the chloride ion oxidize to chlorine? Possible (d) Na+ ions, they should be repelled (e) Can the electrode react with the electrolytic product? Very much possible

Similarly at the cathode (the electrode connected to the negative terminal of the DC supply), think about ... (a) Can the electrode reduce itself? If it is a metal, it cannot reduce further. (b) Will the solvent i.e., water reduce to hydrogen? Possible (c) Can the Na+ ions be reduced to metallic sodium? Possible, but not in plain water. (d) Cl- should be pushed away. (e) Can the electrode react with the electrolytic products? Very much possible

So you can see here, that depending on the electrode material, yes it can be oxidized if used an anode. It can also be oxidized indirectly, if the evolving chlorine begins to attack the metal, which happened in your case.

In any case, electrolytic cell does not care about the electrodes, the electrodes can certainly be the same but they have to be inert. People always choose inert electrodes so that the metallic cathode or anode do not interfere chemically during the electrolysis. Platinum is a good choice but it is too expensive for home experiments. Carbon/graphite electrode is another choice. As you have seen, stainless steel is not a good choice as an anode. It is readily oxidized by chlorine.

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If $\ce{NaCl}$ is electrolyzed between two iron electrodes, the iron is not modified at the cathode. The cathodic reaction is : $$\ce{2 H2O + 2 e- -> H2 + 2OH-}$$ So gaseous $\ce{H2}$ is produced in bubbles, and the solution becomes basic, because of the ions $\ce{OH-}$. The nature of the metal at the cathode does not matter. The reaction at the anode will be : $$\ce{Fe -> Fe^{2+} + 2 e-}$$ But after a while the ions produced at both electrodes interact and produce a dark precipitate $\ce{Fe(OH)2}$ , $$\ce{Fe^{2+} + 2 OH^- -> Fe(OH)2}$$ which is soon oxidized by air into a brown precipitate $\ce{Fe(OH)3}$ that looks like rust. $$\ce{4Fe(OH)2 + O2 + 2 H2O -> 4 Fe(OH)3}$$

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