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As an experiment for electrolysis and to show how a metal can be protected of corrosion, I am trying to anodize an aluminium plate, using sulfuric acid as electrolyte.

I know the aluminium plate is the cathode and, thus, connected to the positive terminal of the generator, while for the cathode I am using aluminium foil.

Supposing I have $H_{2}SO_{4}:80ml, 3M$. I have two questions:

$1.$ Knowing: $$\epsilon^{o}_{Al^{3+}/Al} = -1.67V\\ \epsilon^{o}_{O_{2}/H_{2}O} = 1.23V, pH = 0\\ \epsilon^{o}_{2H^{+}/H_{2}} = 0V, pH = 0\\ \epsilon^{o}_{2H_{2}O/H_{2}} = -0.83V, pH = 14$$ I want an explicit explanation using those potentials as to why the reaction at the cathode is $$2H^{+} + 2e^{-}\rightarrow H_{2}$$ and why the reactions at the anode are ( If i am not mistaken ) $$Al \rightarrow Al^{3+} + 3e^{-}\\ 2Al^{3+} + 3O^{2-}\rightarrow Al_{2}O_{3}\\ 2Al^{3+} + 3OH^{-} \rightarrow Al_{2}O_{3} + 3H^{+}$$ Especially the second and third reactions confuse me

$2.$ Knowing: $$\epsilon^{o}_{Cu^{2+}/Cu} = 0.34V\\ \epsilon^{o}_{Pb^{2+}/Pb} = -0.13V\\$$ Could I use a plomb or copper blade instead of the aluminium foil?

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closed as too broad by Mithoron, airhuff, pentavalentcarbon, Todd Minehardt, Pritt Balagopal Apr 20 '18 at 4:27

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Right now, people are flagging this as two broad: I think your two questions should be asked separately, as they seem relatively unrelated. If you edit this question to only contain one, and ask a new question, it may be better received. $\endgroup$ – pentavalentcarbon Apr 19 '18 at 20:09
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As the experiment is described, in the solution we have : $$H^{+}, H_{2}O, Al_{(s)}, SO_{4}^{2-}, OH^{-}$$ Taking that into account and seeing how the environment is acidic ( $pH<7$ ), we can see that the reaction of reduction reaction that can take place is $H^{+}/H_{2}$ Following the same kind of reasoning, reaction $(1)$ in the anode is caused by the odydation of the aluminium plate when placed in an acidic environment, $O^{-2}$ is present because of the oxygen in the atmosphere and $OH^{-}$ exists in the solution anyway(water autoionization).

As far as the second question is concerned, notice that the aluminium foil in the anode does not participate in any reaction. It is the $H^{+}$ that is reduced, while the aluminium foil serves as a conductor for the current. Knowing that, we could replace the aluminium foil with either $Cu_{(s)}$ or $Pb_{(s)}$, that may also be used as conductors ($Cu^{2+}, Pb^{2+}$ will not be present in the solution and we will have the same reduction reaction in the anode), no matter their standard electrode potentials.

Finally, it is interesting to point out that in this experiment the anode and cathode are in the same solution which is used at the electrolytic reactions (acidic environment ) and at the same time plays the role of the electrolytic bridge constantly providing $H^{+}$ and keeping the solution electrically neutral.

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