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They gave us an unbalanced reaction of permanganate and iodide in acidic medium, and ask to balance the reaction using redox methods and the write the cell configuration notation for the resultant cells. The last part is where I get into trouble.

Unbalanced reaction:

$$\ce{MnO4- (ac) + H+ (ac) + I- (ac) <=> Mn^2+ (ac) + I2 (s)}$$

I balanced the two half reactions and got:

\begin{align} 2 \times&& \big(\ce{MnO4- (ac) + 8 H+ (ac) + 5 e- &<=> Mn^2+ (ac) + 4 H2O (l)}\big)&& \text{(red./cat.)}\\ + 5 \times&& \big(\ce{2 I- (ac) &<=> I2 (s) + 2 e-}\big)&& \text{(ox./an.)}\\\hline &&\ce{2MnO4- (ac)+ 16 H+ (ac) + 10 I- (ac) &<=> 2Mn^2+ (ac) + 8 H2O (l) + 5 I2 (s)} \end{align}

Now, for writing the half cell notation for the cells, we are to use platinum as electrode. The question is, should I put it in both the cathode and anode? Because I heard people saying that since the anode already has a solid substance (solid iodine) it's not necessary to add platinum to the anode. I digress because I think platinum is necessary for the electrons to flow from anode to cathode.

I would write

$$\text{anode} || \text{cathode}$$ $$\ce{I- (ac) | I2 (s), Pt (s) || MnO4- (ac), Mn^2+ (ac) | H2O (l) | Pt (s)}$$

I'm not sure what the correct cell configuration notation is, would anyone clarify this for me please?

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Because I heard people saying that since the anode already has a solid substance (solid iodine) it's not necessary to add platinum to the anode. I digress because I think platinum is necessary for the electrons to flow from anode to cathode.

Your reasoning is correct. Iodine will be oxidized in the anode, as iodide will enter the electrolyte and electrons will flow to the cathode. The way to accomplish this, is to have an electronic conductor for the electrons to reach the cathode. If this is absent, it is impossible for the reduction of permanganate to take place, since the charge transfer happens at the solid/electrolyte interface. If you can't ensure one reduction and one oxidation, then the cell cannot operate, because one cannot occur without the other taking place at the same time.

This phase can be $\ce{Pt}$ or any other solid phase. The requirement is that it should be an electronic conductor, and do not interfere with the reactions (it is “inert”). It would be pretty embarrassing to use a solid phase, for which its reduction has more tendency to occur than the one you seek (in this case the reduction of permangante).

I will make some corrections to your cell notation:

  • Every time you identify a potential difference between two phases, you must use a bar | or a slash /. Writing “$\ce{I2(s),Pt(s)}$” is not correct. Commas are used for separating different species in the same phase, e.g., ions in an electrolyte, like you did with permanganate and manganese ions.
  • Although it seems that $\ce{l}$ and $\ce{aq}$ are different phases, both are at the same liquid phase, or the same potential. They are all together in the electrolyte at the cathode.
  • The double bar $||$ represents a potential difference which is negligible to the other potential differences in the cell. I imagine that you have put those in order to separate both electrolyte phases from the cathode and the anode. If you have that information in the problem, then put both bars.

With this in mind, the cell configuration is the following:

$$\ce{Pt|I2|I^-||MnO4^2-,Mn^2+,H^+|Pt}$$

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  • $\begingroup$ You still need an electrode in the solution hence Pt since it might be nonreactive. $\endgroup$
    – jimchmst
    Nov 20, 2023 at 21:27
  • $\begingroup$ I looked it up in Lange's Compilation of potentials. the potentials for I2[s] and I3- ion both require platinum electrodes. The potentials are the same. $\endgroup$
    – jimchmst
    Mar 9 at 18:42
  • $\begingroup$ @jimchmst You are totally right, and $\ce{Pt}$ (or another inert electrdoe) is needed. Huge mistake from my part, thanks! $\endgroup$ Mar 9 at 23:50

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