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How does redox couple develop its potential if electrode isn't the part of the redox system like platinum in SHE or Cl2/Cl- in water solution for example? Usually it is discussed how electrode develops its potential when it isn't inert like Cu/Cu2+ system when copper electrode itself is important for developing potential. What about inert electrodes?

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    $\begingroup$ What is mysterious in providing an electron from the electrode to reduce a molecule/ion, or accepting it when oxidizing them? $\endgroup$ – Poutnik Apr 14 at 13:17
  • $\begingroup$ Who said it was? So, electrode acts as electron donor or acceptor we can say for equilibrium to develop? $\endgroup$ – Dario Mirić Apr 14 at 13:28
  • $\begingroup$ @Dario Then why don't you ask similar questions for the connecting wires? $\endgroup$ – Nisarg Bhavsar Apr 14 at 18:04
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So, let's take a solution of copper sulfate in a beaker and insert an active electrode: a copper wire. OK so far? We connect this wire to a voltmeter.

Then lets connect another wire (we are quite well off, even rich, so we will use a shiny platinum wire) to the other connection of the voltmeter. Platinum is nice, inert, always shiny. How shall we connect to the solution? Just stick the inert wire into the solution! The wire is inert chemically in that it does not replace copper from copper sulfate or hydrogen from water, but it is not an insulator - it does do something: it carries electrons when they are pushed or dragged.

If we had inserted an inert electrode (Pt) into the solution first so we had two inert Pt electrodes, well, nothing would happen. Nothing that we could measure. Those little ions and molecules are hopping all over the place, but to no net effect.

So the copper atoms near the solution can donate electrons which then float thru the voltmeter to the platinum which can accept those electrons, handing them off to water, forming OH$^-$ and H$^.$ (two of which will form H$_2$). This occurs more easily than the other way around. (It's even a bit more complicated: copper doesn't do this so easily, but if O$_2$ is available, it will consume the hydrogen atoms produced at the Pt wire and drive the equilibrium toward copper dissolution. I just didn't want to start off the discussion with an active metal like zinc, and zinc wires, etc., because it's more difficult to paint a picture with unfamiliar objects. I probably could have used aluminum wires...)

The philosophical question then is: Is the inert electrode really inert? At the copper electrode, a copper ion just plops down on the metal (yeah, some activity), while at the platinum electrode, electrons come in, go out into H$_2$O, split it, form OH$^-$ and H$^.$, help that form H$_2$, then bubbles - it's a crazy mess compared to the copper electrode.

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  • The kinetic rate constant of oxidation reactions ( electrons taken away by an electrode ) climbs exponentially with increasing electrode potential.
  • The kinetic rate constant of reduction reactions ( electrons provided by an electrode ) climbs exponentially with decreasing electrode potential.

The electrode potential converges to the value where both reaction rates match each other. In case of multiple redox systems, then the potential converges to the value when the total oxidation rate equals to the total reduction rate.

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