# Copper brightness change

In an electrolysis situation I have identified a slight brightness change on a copper electrode. What might have happened?

I'm dealing with different experiments and this reaction occurs in some of them. In this experiment I am doing an electrolysis using a water solution with $$\ce{NaCl}$$, a Lechanché and two copper bars. The brightness change happens on the anode.

I also get bubbles and a color change in the solution of the cathode, I'm guessing it is due to the reduction of water.

• – user7951
Jun 10 '20 at 19:42

If an aqueous solution of $$\ce{NaCl}$$ is electrolyzed between copper electrodes, the most important reaction taking place on the anode is the dissolution of copper, according to : $$\ce{Cu -> Cu^{2+} + 2e^-}$$ If the copper plate was well polished and shiny like a mirror, the corrosion reaction will first occur on the smallest crystals, which are inserted between bigger regular crystals. As a consequence, the surface will be irregularly attacked, and its aspect will loose its shiny appearance.

Of course, at the cathode, some Hydrogen gas is produced.

My take on the possible chemistry starting with the electrolysis of salt water, which can produce some chlorine. The latter can further react with water as follows:

$$\ce{Cl2 + H2O <=> H+ + Cl- + HOCl}$$

The created Hypochlorous acid may react with cuprous (here present as a $$\ce{Cu2O}$$ on the copper metal electrode). A fenton-type reaction based on $$\ce{HOCl}$$ could then proceed as follows for $$\mathrm{pH > 5}$$:

$$\ce{Cu(I) (s) + HOCl (aq) -> Cu(II) (aq) + ^.OH (aq) + Cl- (aq) }$$

The dissolving of the $$\ce{Cu2O}$$ results in a de facto cleaning of the copper metal surface, which per a reference to quote:

Copper is a beautiful burnished-gold color when it is clean and well maintained, but like all metals, copper can become discolored when exposed to air and water.

For background on the fenton-type reaction, see "Fenton chemistry in biology and medicine" by Josef Prousek, to quote reaction (15) on page 2330, to quote:

For $$\ce{Fe(II)}$$ and $$\ce{Cu(I)}$$, this situation can be generally depicted as follows [20,39]

$$\ce{ Fe(II)/Cu(I) + HOX -> Fe(III)/Cu(II) + HO^. + X- }$$ (15)

where $$\ce{X = Cl, ONO}$$, and $$\ce{SCN}$$.