EasyChemistryHigh School (11–12)The p-Block Elements › Groups 15 and 16: Nitrogen and Oxygen FamiliesConceptualA-LevelIB
A wet red flower is put in a jar of sulphur dioxide and loses its colour…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A wet red flower is put in a jar of sulphur dioxide and loses its colour. After a day in the open air, some of the colour comes back. A second flower is put in a jar of chlorine and also loses its colour, but the colour does not come back. Explain the difference. Name the type of change in each case, and say what the dichromate test would show for the sulphur dioxide.
Need a reminder? The key ideaShow
Ozone gives one mole of iodine for each mole of ozone with potassium iodide. Roasting pyrites follows 4FeS2 + 11O2 giving 2Fe2O3 + 8SO2, so each FeS2 gives 2 SO2. Permanganate and dichromate oxidise sulphur dioxide: 5 SO2 for each 2 and 3 SO2 for each . Work in moles: the amount of a solute is the volume in litres times the concentration.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Sulphur dioxide. The wet gas bleaches by reduction. It adds hydrogen to the dye, or removes oxygen from it, and the product is colourless.
Step 2. Why the colour returns. Oxygen from the air slowly oxidises the colourless product back to the dye, so the bleaching is temporary.
Step 3. Chlorine. Wet chlorine bleaches by oxidation. The hypochlorous acid that forms destroys the coloured part of the dye for good, so the colour does not return.
Step 4. The dichromate test. Sulphur dioxide turns orange acidified dichromate green, because the chromium(VI) is reduced to chromium(III).
Answer: reduction for sulphur dioxide (temporary) and oxidation for chlorine (permanent).
Watch out. Both gases need moisture to bleach. The dry gases do little.
Discussion (0)
Log in to join the discussion.
Loading…