While in lab, I accidentally used two paper clips while trying to electrolyze potassium iodide and all I saw was was phenolphthalein turning pink at the cathode and no brown at the anode.

I asked my instructor to allow me to repeat the experiment with pencil lead, which produced brown and pink, aluminum foil, which only produced pink, silicon which produced brown and pink, and copper, which produced brown and pink.

Then I used one paperclip at the cathode and a piece of pencil lead at the anode and I found that the anode turned pink and the cathode turned brown. Then I flipped the pencil lead and paper clip (anode to cathode and vice versa), and only the cathode (pencil lead) turned pink and there was no brown at the anode (paper clip).

My original hypothesis was that the increased resistance in the silicon and graphite lowered the current since the battery provided a constant $\pu{9V}$, but being able to get brown and pink with copper invalidated that hypothesis. This has been bugging me all day, and my instructor says she does not know why this is happening.

  • 1
    $\begingroup$ Obviously, more active metallic anodes oxidize themselves instead of oxidizing iodide, hence no brown. $\endgroup$ – Ivan Neretin Mar 1 '17 at 6:45

To distill down your observations: you always saw pink at the cathode, but only some materials produced a brown color when used as the anode. This means that the anodic materials that didn't produce brown didn't oxidize the iodide to iodine.

So, the question is what was happening at the anode when the iodide was not being oxidized. The answer is simply that the anode material itself was easier to oxidize than the iodide in those cases, so the anode itself was oxidized and the iodide was not, so there was no brown color produced.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.