[hilaryofoz]

My grade 12 students are doing a selective precipitation lab. The original mixture (blue) contains lead(II) nitrate and copper(II) nitrate. Both of these groups then added potassium iodide (to make lead(II) iodide).

I know sometimes copper(II) can change valences to copper(I) and produce a solid with iodide, but I have no idea why the right solution turned brown (and the other didn't). It looks like iodine was produced!

Any ideas on what happened here?
http://haydukchemistry.weebly.com/uploads/2/0/9/7/20974360/1051997_orig.jpg

[Borek]

No idea why it happened in one flask and not in another (were they using exactly the same reagents, in exactly the same amounts?), but why the iodine was produced is rather obvious - try to balance the redox equation describing the CuI precipitation from the Cu²⁺ solution.

[hilaryofoz]

Ohhhhhh - thanks! Maybe one group used a little more than the other. That would make sense.

[AWK]

Rather one group had sample without copper or added also thiosulfate to this sample.

[hilaryofoz]

No, they had the same solutions. All from the same containers!

[Borek]

Never underestimate your students.

Never.

[AWK]

For this mixture redox reaction goes first with iodine (brown solution) and CuI (almost white precipitate) production. Then start precipitation  of a canary yellow PbI₂. The easiest method of removing iodine (discoloring) is the addition of Na₂S₂O₃.