I'm trying to perform a potentiometric titration to determine the redox potential of a $\ce{Fe^3+|Fe^2+}$ couple. To do this, I am using an $\ce{Ag|AgCl}$ electrode connected to an inert $\ce{Pt}$ electrode, which is submerged in a solution containing $\ce{Fe^3+}$ ions. I then titrate it with $\ce{Co^2+}$, which reduces it to $\ce{Fe^2+}$. The reference and inert electrode are connected with a standard voltmeter.

What I don't understand is why the inert electrode measures the redox potential of the $\ce{Fe^3+|Fe^2+}$ couple. As per my understanding, there is an electron flow from the reference electrode into the inert electrode, which would get accepted into the Fe solution, but now, the transfer of electrons is only between the $\ce{Co^2+}$ and $\ce{Fe^3+}$ ions, so what happens to that electron? Can someone explain exactly how the redox potential of the $\ce{Fe^3+|Fe^2+}$ couple gets measured, with reference to electron transfer between the reference and inert electrodes?

  • $\begingroup$ Potential measurement ideally requires no current flow. You just refer the potential at which the Pt electrode is versus a fixed one, that of the Ag/ AgCl one. Look for electrode of the third kind. $\endgroup$ – Alchimista Mar 1 '19 at 8:10
  • $\begingroup$ That’s what I’m confused about - how is the Pt electrode able to measure the potential of the Fe(III)/Fe(II) couple? The electrons are being transferred between Cobalt and Iron, so how does the Pt electrode pick up on the potential of that reaction? $\endgroup$ – malhi3 Mar 1 '19 at 11:32
  • $\begingroup$ If you have a clear picture of a Pt electrode immersed in a, e.g., Fe+++/Fe++ solution, then you see how it sense the potential of your current system. $\endgroup$ – Alchimista Mar 1 '19 at 11:40
  • $\begingroup$ The point is that you your are ideally working without current. This part is what a voltmeter takes care. You can also build three electrodes cells, in which current flows to the extra added working electrode, and the reference ekectride is there just to serve as reference indeed. $\endgroup$ – Alchimista Mar 1 '19 at 11:55

It is a very good question. Your question has nothing to do with current being drawn or not or a reference electrode etc. Imagine a single Pt electrode dipped in an Fe(II) solution and forget about the presence of cobalt. Imagine that a metal is a source (sea) of electrons.

The moment you dip the electrode, instantaneous exchange of electrons takes place between the metal and the reducible or oxidizable ions. Electrons can be released by Fe(II) to form Fe(III) and similarly electrons can be gained by Fe(III) to form Fe(II). This exchange rate is dependent on the concentration ratios of Fe(II) and Fe(III).

As a result of this exchange, the Pt metal develops a charge. It is very surprising that you only need to remove a very very very small number of electrons to develop a measurable surface charge. Hence, there is no perceptible change in the concentration of Fe ions. If platinum electrode develops a charge, the interface (solution electrode boundary) develops an opposite charge until equilibrium is achieved.

Now given the fact that Pt electrode is charged i.e., if we connect Pt to another electrode, we can sense the potential difference between the Pt electrode and another electrode, which can be any other electrode. Now that you wish to have a stable. For that purpose you need an electrode whose potential is fixed i.e. a reference electrode.


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