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March 2 2012 6 02 /03 /March /2012 14:51

A comproportionation reaction sounds extremely complicated. In reality it is quite simple.Two differently charged atoms orr ions of the same element react to form two equally charged ions. Three common examples will be discussed.


First: Copper. When an acidic copper(II) chloride solution is reacted with copper metal, the solution darkens. This is the result of the formation of copper(I) ions. The reaction that is occurring is: Cu(0) + Cu(2+) --> 2 Cu(+) This dark solution is a complex of copper(I) ions with additional copper(II) ions. When the solution is made neutral, the copper(I) chloride precipitates out. These pictures shows the difference in color between the bluish green copper(II) chloro complex and the brownish green copper(I)-copper(II) chloro complex.



Second: Iron. When an acidic iron(III) chloride solution is placed in contact with iron, the iron reduces the iron(III) to iron(II). This reaction is occurring: Fe(0) + 2 Fe(3+) --> 3 Fe(2+) Yellow iron(III) chloride is easily obtained by dissolving rust in hydrochloric acid or by simple purchase. This reacts with iron metal to form a yellow-green iron(II) chloride solution. The test for iron(III) is a brown precipitate when sodium bicarbonate is added. Iron(II) forms from a dark green to a white precipitate. Since my solution was very acidified, all of the oxygen had been displaced from solution by the baking soda - acid reaction, and the precipitate was quite white. Top left: Original solution at experiment start. Top right: Solution after 30 minutes. Bottom left: solution after 9 hours. Bottom right: Iron(II) carbonate. The brown spot is a control, a dab of iron(III) chloride that escaped the iron metal. It forms the brown iron(III) precipitate when reacted with baking soda.


I did this experiment just for my readers.


Third: Tin. When a tin(II) chloride solution is desired, it of course must be pure. However, tin(II) chloride oxidizes to tin(IV) chloride and tin(IV) oxychloride in air. To prevent this oxidative contamination of the solution, tin metal is added. It reacts with any tin(IV) formed, converting it back into tin(II). This reaction occurs: Sn(0) + Sn(4+) -> 2 Sn(2+) All of these solutions are colorless, so pictures are pointless. The formation of a precipitate of tin(II) or (IV) oxychloride may occur, but this precipitate is out of solution and does not act as a contaminant of concern. It is only the tin(IV) in solution that is a problem.

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Published by LanthanumK - in Experiments
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