Chemistry What does this equation for an electrochemical cell mean?

[zenterix]

Homework Statement

Consider the electrochemical cell reaction given by

Relevant Equations

$$\mathrm{2Ce^{4+}(aq)+3I^-(aq)\rightarrow 2Ce^{3+}(aq)+I_3^-(aq),\ \ \ \ \ E^\circ_{cell}=+1.08V}$$

Item (a) of problem 1 of a chapter on "Galvanic Cells" asks us to calculate the Standard Gibbs free energy for this cell reaction.

I can do the calculation, but I am frustrated by the following.

The entire chapter focused on the electrochemical cell

$$\mathrm{Zn(s)|Zn^{2+}(aq)||Cu^{2+}(aq)|Cu(s)}$$

with explanations for what this means in terms of a physical setup. Namely, solid metals immersed in ionic solutions, connected by a conductor and a salt bridge.

The equation

$$\mathrm{2Ce^{4+}(aq)+3I^-(aq)\rightarrow 2Ce^{3+}(aq)+I_3^-(aq),\ \ \ \ \ E^\circ_{cell}=+1.08V}$$

doesn't have any solid metals in it.

It appears we have the redox half equations

$$\mathrm{2Ce^{4+}(aq)+2e^-\rightarrow 2Ce^{3+}(aq)}$$

$$\mathrm{3I^-(aq)\rightarrow I_3^-(aq)+2e^-}$$

Is this correct?

What does this cell look like in real physical terms?

 

[zenterix]

Here is my guess.

Not all electrode reactions include a conducting solid as a reactant or product.

A hydrogen electrode, for example, includes hydrogen gas and aqueous hydrogen cations.

There is, however, a chemically inert metallic conductor present, such as an unreactive metal or graphite or platinum (in the case of a traditional hydrogen electrode), which serves to carry electrons into or out of the electrode compartment.

So in the case of the cell reaction involve cerium and iodine, I suppose there would be something similar, ie some inert metal conductor present.

Perhaps we would have

$$\mathrm{Pt(s)|Ce^{3+}(aq),Ce^{4+}(aq)||I^-(aq),I_3^-(aq)|Pt(s)}$$

However, we don't have gases. Instead we seem to have two solutions, one at each inert metal, each containing two different ions.

The oxidized and reduced species are both in the same phase.

Not clear though how this works exactly.

Since $E^\circ_{\text{cell}}$ is positive we have electrons moving spontaneously from the electrode with the iodine ions to the electrode with the cerium ions.

In the case of the zinc/copper example, I understood that we had solid zinc being oxidized and going into solution and copper ions being reduced and becoming solid copper.

In the case of cerium and iodine, what happens exactly?

 

[Mayhem]

Pt act as an inert electrode (and possibly catalytic in some case) and simply connects the circuit.

 

[ketoenol]

In the case of cerium and iodine, what happens exactly?

Yes, you are correct, there would be inert electrodes of some sort

Physically, Cerium(IV) ions get close enough to the cathode that an electron can hop from the cathode to the cerium ion, reducing it to cerium(III). The cerium(III) ion then floats away in solution.

The iodine is likely slightly more complicated; each I^- ion will go to the inert electrode and give up its electron, but you're not going to get a free iodine atom floating off into solution. Probably you get a couple iodine atoms adsorbed to the electrode until they can meet up with each other to form the I₂ molecule, which will then react separately with an additional I^- ion to form the triiodide.

There's also water of solvation in all of these cases which can affect how the ions interact with the electrodes in subtle ways. In a real battery, various ligands or other additives might be added to either solution to make the interaction of the reactants with the electrode more efficient, or prevent various possible side reactions.

 

[Mayhem]

Yes, you are correct, there would be inert electrodes of some sort

Physically, Cerium(IV) ions get close enough to the cathode that an electron can hop from the cathode to the cerium ion, reducing it to cerium(III). The cerium(III) ion then floats away in solution.

The iodine is likely slightly more complicated; each I^- ion will go to the inert electrode and give up its electron, but you're not going to get a free iodine atom floating off into solution. Probably you get a couple iodine atoms adsorbed to the electrode until they can meet up with each other to form the I₂ molecule, which will then react separately with an additional I^- ion to form the triiodide.

There's also water of solvation in all of these cases which can affect how the ions interact with the electrodes in subtle ways. In a real battery, various ligands or other additives might be added to either solution to make the interaction of the reactants with the electrode more efficient, or prevent various possible side reactions.

To be add, the addition of ligands will likely drastically change the electrochemistry. The example in the original post is only (safely) applicable to simple Cerium salts.