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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

1. 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.

2. 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.

3. 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)