Chapter 14 Homework Problem 14.13(b)

[Isabel Strouse 1H]

Problem 13 from the chapter 14 homework asks us to balance the half reactions of redox reactions and then set up a cell diagram for the reactions.

The skeleton reaction for 13b.) is Ce⁴⁺ + I^- --> I₂ + Ce³⁺

I had no trouble finding the half-reactions and balancing the reaction, and my balanced result was:
2I^- (aq) + 2Ce₄₊ (aq) --> I₂ (s) + 2Ce³⁺ (aq)

My question is that, in the solutions manual, rather than the cell diagram being simply
I₂ (s) | I^- (aq) || Ce⁴⁺ (aq), Ce³⁺ (aq) | Pt (s), the diagram is

Pt (s) | I^- (aq) | I₂ (s) || Ce⁴⁺ (aq), Ce³⁺ (aq) | Pt (s).

Why is there a platinum anode when there is solid I₂ available to be the electrode? I thought platinum/graphite were only needed when a solid wasn't present (like how it does when it acts as the cathode for the reduction half of the reaction in this problem)? Why is the I₂ (s) now in the middle rather than the outside, I thought galvanic cell diagrams always went anode | ions of oxidation || ions of reduction | cathode...

The solutions manual says that "an inert electrode such as Pt is necessary when both oxidized and reduced species are in the same solution," but this doesn't help me much. If anyone can explain this reasoning/why the cell diagram is set up this way I would greatly appreciate it! Thank you!! :)

[Cindy Chen_2I]

Though I₂ is a solid, it is not a conductor. That's why we need Pt here as an electrode. And since Pt is an inert electrode, it will not be involved in the reaction. My TA said he thinks the order of I₂ and I^- doesn't matter but he was not entirely sure.

Hope that helps :)