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Oxidation States of Copper
A demonstration of stabilization of Cu(I) by formation of the ammonia complex.
John C. Wheeler 5/12/97
Copper(I) ion is very unstable in aqueous solution where it, spontaneously disproportionates to Copper(II) and Cu(0). It can be stabilized by appropriate ligands, one of which is NH3.
Materials
- 15M NH3 (aq) - (a dropper bottle is convenient)
- 6 M H2SO4(aq) - (again, a dropper bottle is convenient)
- 0.2M CuSO4 (aq) - (a beaker with enough in it for the class to see)
- Na2S2O4 (solid) - (sodium dithionite) (sodium hydrosulfite)
- Large Beaker
- Stirplate and Stirbar
Procedure
- Start with 0.2M CuSO4(aq) solution (a clear light blue solution) in a beaker with a stirring bar on a stirrer. Gradually add 15M NH3. At first, a white precipitate of Cu(OH)2 will form, while simultaneously the solution becomes a dark blue. Continue adding 15M NH3 until the precipitate redissolves and the solution is a deep clear blue, due to the Cu(II)(NH3)5 complex ion.
- Now add, gradually, sodium dithionate Na2S2O4. This reduces Cu(II) to Cu(I), producing the Cu(I)(NH3)2 complex ion, which is colorless (filled d shell). As the endpoint is neared, the solution will become paler, then suddenly colorless. If not too much excess dithionite is added, the solution will become blue at the air-water meniscus, then blue throughout due to air oxidation of the Cu(I) back to Cu(II). Addition of a bit more dithionite turns it colorless again.
- While the solution is colorless, but with no great excess of dithionite, add H2SO4 to make the solution acidic. Immediately redish copper metal will precipitate out and the solution will become an opaque redish brown. After stirring for a moment, turn off the stirrer and let the precipitate of Cu(0) settle to see the blue (actually slightly green, for reasons I don't know) color of the Cu(II) ion that has resulted from 2Cu(I) -> Cu(0) + Cu(II)
