Rates of reaction - kinetics (A2)
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I am doing the iodine clock for my coursework, and I would like to examine the reaction mechanism. Can you help me with the structure of peroxodisulphate ions? I have also repeated this experiment with Cu2+ ions as a catalyst, but again I am stuck on the mechanism for this reaction. Can you help me here, what for example would the intermediate stage be?
Igloo writes ...
First, the structure for the peroxodisulphate ion:
As you can see, it effectively consists of two sulphate ions linked together via an O-O linkage: hence the name “peroxo …” (hydrogen peroxide is H-O-O-H). Incidentally, I have simply drawn a contributing displayed formula for the actual structure. In reality there will be delocalisation between each sulphur atom and the three outlying oxygen atoms, in an attempt to “spread out” the charge on one of the oxygen atoms.
The central O-O linkage is relatively weak (see Nuffield Data Book for bond energies on page 51) and is undoubtedly the one broken when the ion acts as an oxidizing agent forming two sulphate ions (SO42-) in the process: S2O82-(aq) +2I-(aq) --> 2SO42-(aq) + I2(aq)
As regards the involvement of Cu2+ ions as catalysts, you may get some ideas by studying the Students’ Nuffield Chemistry Book on page 468. Perhaps in this case the Cu2+ ions oxidize the I- ions to iodine, whilst themselves being simultaneously reduced to Cu+ ions, and then immediately these can be oxidized back to Cu2+ ions by the peroxodisulphate ions; this entire process would follow routes of lower activation energies than the one occurring in the absence of the Cu2+ ions. I am by no means certain that this represents the true course of events, but it’s a possibility.
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updated: 02 March 2004
