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In lecture I was taught that, in a galvanic cell at standard conditions, platinum is often used as an inert electrode when the species being oxidized or reduced exists only in solution, because platinum is "inert," which means it is very stable at its oxidation state and resists reduction or oxidation. This concept confuses me, because "Reduction" means to lose charge, a.k.a. gain an electron(s). When the electrode has an electron passing through it in order to be transferred to another material, isn't it being reduced, albeit for a very miniscule amount of time?

For example, consider a galvanic cell with platinum electrodes and an iron wire. If we were to examine the cell at an instant and examine the platinum electrodes/wire, we would see a charge of -1 on some of the platinum atoms and iron wire, correct?

What I'd like to imagine is that, while platinum is capable of having a negative charge because of the room for electrons in its d-block, the species being reduced in the half-cell will always have a higher affinity for accepting the electron due to its higher standard reduction potential. This still leaves me confused on why platinum is willing to accept the electron and thus making platinum a medium for electron transfer in the first place, though. This all applies to the iron wire (or any other element that may be used for electron transfer) as well, but I would like to tackle the case of platinum overall.

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    $\begingroup$ Think of the Pt electrode as a capacitor plate. The surface gets charged and the charges have to cross the metal-liquid barrier. $\endgroup$ – MaxW Apr 7 at 17:56
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Reduction and oxidation are defined as gaining and donating electrons from the valence shell of an atom respectively.

Whereas in the case of electrons flowing in the electrodes and connecting wires the electrons are passing from the interatomic voids and not being released by the wires and the electrode.

(There will be some electrons which will interact with the conductor's atoms but what will happen is that they will get deflected. There will be a slim chance that a few electrons will replace an electron from the conductor's electron cloud but that doesn't change the situation as both the electrons are same in a classical sense. )

If your claim was true than the anode would have got some extra negative charge and the cathode extra positive which would have driven up the voltage higher than that expected from the cell alone.

In a similar way when we connect a wire it doesn't magically change the measured voltage.

Even when we perfom the same experiment with different inert electrodes and different kinds of wires the measured reading doesn't change much. (There are some changes in current flowing through the circuit due to the difference in resistances of different materials)

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    $\begingroup$ "Interatomic void" is a property of atoms I've not heard of. If the electrodes and wire are quickly pulled from the solutions during the electrolysis, what happens to the electrons still remaining in the electrodes and wire, since they did not complete their transfer? Are they stuck in this void? $\endgroup$ – Ian Apr 7 at 19:30
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    $\begingroup$ @Ian No electrons will be "left" behind when you pull out the wires. The total electrons in the wire remain constant as the same number of electrons enter and leave the wire. Interatomic void is nothing but the empty space between atoms. $\endgroup$ – Nisarg Bhavsar Apr 7 at 19:44
  • $\begingroup$ Why is it that none will be left behind? Is it because it's impossible to pull them out at exactly the same time, so the electrons will all be transferred either back into the anode half-cell or the cathode half-cell? $\endgroup$ – Ian Apr 7 at 20:01
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    $\begingroup$ @Ian You need to understand the condition just before pulling out the wires. The electrons are flowing in and out at the same rate at either ends and thus there are no "extra" electrons in the conductor and thus there will be no electrons left behind. For example when a pipe full of water is connected in a loop of pumps working at same rates and we pull the pipe, the pipe will have same water as before it was connected. I am comparing the wire with a filled water pipe as the conductor already has the free electrons which will cause the conduction. $\endgroup$ – Nisarg Bhavsar Apr 7 at 20:15
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The answer is no. If an electric current is flowing through a piece of metal (platinum or anything else) than the same number of electrons are getting in and getting out of the piece at either ends of the piece. No atoms are reduced or oxidized. The total number of electrons present in the metallic piece does not change. There is no oxidation and no reduction in the metallic piece.

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