2
$\begingroup$

What is the full chemical equation when you use the following reactants to etch copper:

  • vinegar / acetic acid (5%) $\ce{CH3COOH (aq)}$
  • hydrogen peroxide (3%) $\ce{H2O2 (aq)}$
  • salt $\ce{NaCl (s)}$
  • copper $\ce{Cu (s)}$

Yes there are previous questions asking the same thing (one and two). However, the answers to both give only parts of the reaction, not the full equation.

Looking elsewhere on the web, I've come across this Instructables article that says copper(II) chloride $\ce{CuCl2}$ is formed when the copper ions bond with chloride ions from the salt. And therefore the solution is greenish in color instead of the blue it would be if copper(II) acetate $\ce{Cu(CH3COO)2}$ was formed as answered in question two. (When doing this at home, I also got a distinctly greenish color when salt was used, and a distinctly blue color when it was omitted).

Table salt, or $\ce{NaCl}$, brings chloride ions $\ce{Cl-}$ to which $\ce{Cu^2+}$ ions will bond to form cupric chloride or $\ce{CuCl2}$ instead of being left in solution (those ions would endlessly come back and forth from $\ce{Cu}$ to $\ce{Cu^2+)}$. You can see this during the reaction (if you leave the reaction with no salt it will turn blue which is the color of copper(II) acetate and stop, whereas if you put salt it will turn green, the color of cupric chloride, and carry on).

Another article on the web says both copper(II) acetate and copper(II) chloride are produced, but no equation is provided:

Equal parts vinegar and peroxide worked fine, dump in plenty of salt (to the point of having undissolved salt in the bottom of jar) since it is consumed in the reaction. You should see a vigorous fizzing at the copper surface. Add more salt as the reaction slows.

The solution will turn green as the copper is converted to copper acetate and/or copper (II) chloride

Can someone please give the complete chemical equation of the reaction that occurs when acetic acid, hydrogen peroxide, and salt are used to etch copper? I can't seem to find it anywhere on the web.


Context:

I want to dispose the products of the etching process safely. If copper(II) acetate is the product as suggested in answer two, then following the recommended approach for disposal is safe:

You can bypass the problem by collecting the etch solution, and submerge aluminum foil.

$$\ce{3Cu(CH3COO)2 (aq) + 2Al (s) -> 2Al(CH3COO)3 (aq) + 3Cu (s)}$$

React with the aluminum until the bluish color disappears, then collect the solid copper + remaining aluminum to throw in the trash. You can now flush the solution down the drain.

However, if the product is instead copper(II) chloride, the approach is not safe as the $\ce{AlCl3}$ formed is considered a hazardous waste and shouldn't be flushed down the drain.

$$\ce{2Al (s) + 3CuCl2 (aq) -> 3Cu (s) + 2AlCl3 (aq)}$$

$\endgroup$
  • 3
    $\begingroup$ Possible duplicate of What reaction occurs when etching copper to create circuit boards? $\endgroup$ – Mithoron Sep 25 '18 at 23:21
  • 1
    $\begingroup$ @Mithoron I quote that specific question (as one) in my question! And I even go on for a bit explaining why I do not find the given answer adequate! $\endgroup$ – Jet Blue Sep 25 '18 at 23:22
  • $\begingroup$ It's exactly the same question and is answered therefore it's a duplicate. That's it. $\endgroup$ – Mithoron Sep 25 '18 at 23:31
  • $\begingroup$ If you aren't satisfied you can comment under old post or put a bounty on it. Asking again isn't proper solution on this site. $\endgroup$ – Mithoron Sep 25 '18 at 23:32
  • 2
    $\begingroup$ @Mithoron the question was asked three years ago. The OP hasn't been on the site since. Even if I placed a bounty on it, there's no way for me to specify why I find the current answer inadequate. $\endgroup$ – Jet Blue Sep 25 '18 at 23:47
1
+100
$\begingroup$

The acetic Acid is used to dissolve the copper, however it cannot do so directly as the reaction of copper with protons is not spontaneous.

$$\require{cancel} \ce{Cu_{(s)} + 2H+_{(aq)} \cancel{->} Cu^2+_{(aq)} + H2_{(g)}}$$ The hydrogen peroxide is used to oxidize the copper since copper does not spontaneously react with acids directly. $$\ce{\frac12 H2\color{red}{O2} +e- -> OH-}\\ \ce{Cu_{(s)} -> Cu^2+_{(aq)} + 2e-}$$ However, hydrogen peroxide is unstable in acid so the acetic acid is used because it reacts with hydrogen peroxide to form a peroxyacid which is more stable.

$$\ce{CH3C(O)OH + H2\color{red}{O2} -> CH3C(O)\color{red}{OO}H + H2O}$$

This peroxyacid can react directly with copper to dissolve it.

$$\ce{CH3C(O)\color{red}{OO}H_{(aq)} + Cu_{(s)} -> Cu[CH3C(O)O]OH_{(aq)}} \\ \ce{Cu[CH3C(O)O](OH)_{(aq)} + CH3C(O)OH_{(aq)} -> Cu[CH3C(O)O]2_{(aq)} +H2O}$$

As far as the sodium chloride goes, I'm not sure what that is for. My best guess is that it is catalytic in dissolving the copper similar to how salt corrodes steel

$$\ce{CH3C(O)\color{red}{OO}H_{(aq)} + Cu_{(s)} ->[\text{cat.} Cl-_{(aq)}] Cu[CH3C(O)O]OH_{(aq)}}$$

As far as the final species goes, you probably have a complex from the excess chloride

$$\ce{Cu[AcO]2 (aq) + 2NaCl (aq) -> [Cu(AcO)2Cl2]^2- (aq) + 2Na+ (aq)}$$ which is in dynamic equilibrium $$\ce{[Cu(AcO)2Cl2]^{2-} (aq) <=>[Cl-][AcO-] [Cu(AcO)Cl3]^{2-} (aq) <=>[Cl-][AcO-] [CuCl4]^{2-} (aq)}$$

The color comes from the excess chloride that acts as a ligand to the copper From Wikipedia:

enter image description here
Aqueous solutions of copper(II) chloride. Greenish when high in [$\ce{Cl−}$], more blue when lower in [$\ce{Cl−}$].

$\endgroup$
  • $\begingroup$ Thanks for the response! Is the "chloride complex" (not sure if that's the right term) responsible for the green color that appears when salt is added (instead of the blue it would otherwise be)? Does it have a common name? $\endgroup$ – Jet Blue Sep 30 '18 at 1:00
  • $\begingroup$ @JetBlue see edits. Probably not the answer you want , but yes it is likely copper chloride. $\endgroup$ – A.K. Sep 30 '18 at 1:17
  • $\begingroup$ Ah I see. So the CuXCl compound should be disposed off as if it were CuCl2? $\endgroup$ – Jet Blue Sep 30 '18 at 2:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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