Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So during school today, I am running an electrolysis experiment with water and sodium bicarbonate copper leads on both the anode and cathode. I am noticing the usual $\ce{NaO}$ on the anode, and $\ce{H2C2}$ on the cathode. However, on the anode, the copper is obviously oxidizing, and changing the hue of the water to a bright green-blue. How can this be calculated into the chemical equation?

Also, is this equation correct?

$$\ce{H2O + NaC2 + energy -> NaO + H2C2}$$

(with the order of anode then cathode)

Yes, I am aware that this is the best possible situation, but in general, is this what is supposed to be happening?

share|improve this question
@CodeAmdiral - you have written what looks like a valid chemical equation, but it appears unrelated to sodium bicarbonate and copper. You have sodium carbide on the left and sodium oxide or maybe peroxide on the right. – Ben Norris Dec 12 '12 at 16:24
@BenNorris Yes, however, I don't get why the water is turning the colour of copper oxidization – fr00ty_l00ps Dec 12 '12 at 16:48
Probably because copper is being oxidized. A good answer, which I may get around to posting, would compare the standard potentials of the various possible oxidation half reactions to show that, for example, copper is easier to oxidize than water. – Ben Norris Dec 12 '12 at 16:57
up vote 5 down vote accepted

Sodium compounds:

$\ce{Na2C2}$ is sodium carbide. It reacts violently with water. $\ce{Na2CO3}$ is sodium carbonate. $\ce{NaO}$ is not a valid formula for any of the oxides of sodium: $\ce{Na2O},\ \ce{Na2O2},$ and $\ce{NaO2}$ are all real compounds, and they all react with water to produce $\ce{NaOH}$, which is probably what is formed.


If indeed you had $\ce{Na2C2}$, then you would form $\ce{H2C2}$ and $\ce{NaOH}$ when it was placed in water. However, you would not need electrolysis to do so. This is an acid-base reaction that is spontaneous under normal circumstances. $\ce{H2C2}$ cannot be the brown sludge on your cathode. It is a gas.

$$\ce{Na2C2 + 2H2O -> 2NaOH + H2C2 ^}$$

What actually happened?

In an electrolytic cell, the process that occurs (at least under standard conditions) is the one with the least negative standard potential $E^o_{cell}$.

The possible oxidations are either the oxidation of copper or the oxidation of water. Sodium carbonate is the supporting electrolyte.

  1. $\ce{2Cu + 2OH- -> Cu2O + H2O +2e-}\ \ \ E^o=+0.36 \ \text{V}$
  2. $\ce{Cu -> Cu^{2+} + 2e-} \ \ \ E^o=-0.34 \ \text{V}$
  3. $\ce{4OH- ->O2 + 2H2O +4e-}\ \ \ E^o=-0.40 \ \text{V}$
  4. $\ce{2H2O -> O2 + 4H+ +4e-} \ \ \ E^o=-1.23\ \text{V}$

Since the oxidation of copper to $\ce{Cu^{2+}}$ has the least negative $E^o$, it is the most likely oxidation. $\ce{Cu^{2+}}$ also explains the color, since aqueous solutions of $\ce{Cu^{2+}}$ are blue. The first equation is also likely, since it has a positive $E^o$, however it does not explain the color.

The possible reductions are the reduction of water/hydroxide and the reduction of $\ce{Cu^{2+}}$, since we know that it is being formed.

  1. $\ce{2H2O + 2e- -> H2 + 2OH-} \ \ \ E^o=-0.83 \ \text{V}$
  2. $\ce{Cu^{2+} + e- -> Cu+} \ \ \ E^o=+0.16 \ \text{V}$

The reduction of $\ce{Cu^{2+}}$ to $\ce{Cu^{+}}$ is most likely, but not initially. Initially, the is no $\ce{Cu^{2+}}$, so the only reduction that can happen at the cathode is the first one, which produces hydrogen gas. The redox then might be: $$\ce{Cu + 2H2O -> H2 + Cu^{2+} + 2OH-} \ \ \ E^o =-1.19 \ \text{V}$$

Once the $\ce{Cu^{2+}}$ diffuses over to the cathode, a different reduction starts to compete, with a different net redox reaction: $$\ce{2Cu + H2O -> H2 + Cu2O} \ \ \ E^o=-0.47 \ \text{V}$$

This redox reaction still produces hydrogen, but it has a much less negative $E^o_{cell}$. The other peculiarity is that $\ce{Cu2O}$ is produced at the cathode after $\ce{Cu^{2+}}$ has diffused through the cell. $\ce{Cu2O}$ is your brown sludge, since like rust, it is both brown and completely insoluble in water.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.