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?

  • $\begingroup$ @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. $\endgroup$ – Ben Norris Dec 12 '12 at 16:24
  • $\begingroup$ @BenNorris Yes, however, I don't get why the water is turning the colour of copper oxidization $\endgroup$ – fr00ty_l00ps Dec 12 '12 at 16:48
  • $\begingroup$ 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. $\endgroup$ – Ben Norris Dec 12 '12 at 16:57

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.


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