You might be expecting that since hydrogen peroxide is an oxidizing agent, the copper will be oxidized to black copper oxide.
Let's try it. Here's an American penny, which is mostly copper on the outside. I've cleaned the oxide coating off by washing it in 0.1 M HNO3.
After a half hour bath in 30% H2O2:
Nothing much happened. A few bubbles formed on the penny. However, dropping a few drops of 0.1 M HCl into the peroxide made the reaction go much faster. The solution above the penny aquired a faint bluegreen tinge.
Why? Look at a table of standard oxidation potentials and pick out reactions that involve metallic copper and hydrogen peroxide. You'll find these likely half reactions:
|Cu(s) = Cu2+(aq) + 2e-
||E° = -0.34 V
|2 H2O(l) = H2O2(aq) + 2 H+(aq) + 2e-
||E° = -1.78 V
The oxidation potential of the first reaction is highest, so it goes
as written. The fact that the copper(II) ion is blue and copper chloride complexes are green in solution is consistent with our observations.
The second reaction will run in reverse. It doesn't explain the bubbles we saw, though. The bubbles are probably from the decomposition of the peroxide catalyzed by the copper(II) ions, a separate reaction: H2O2 = H2O + (1/2) O2.
Adding the two half reactions together gives
+ H2O2(aq) + 2 H+(aq)
= Cu2+(aq) + 2 H2O(l)
and the overall cell voltage under standard conditions will be (+1.78 - .34) or
+1.44 V. We can estimate the equilibrium constant for this reaction. Since
K = 10nE/0.0592
at 25°C, K = 102(1.44 V)/(0.0592 V) = 1048.6, which is very, very large. This indicates that hydrogen peroxide can oxidize copper metal on the surface of the penny by displacing it as copper(II) ions, and it's consistent with what we actually see.
Author: Fred Senese email@example.com