I've watched a YouTube video "Soda Can Battery - Power From Trash - DIY Project" by Source Port Services. I understand that the anode can be oxidised, but what is the reduction reaction at the cathode? They are using a salt water solution $\ce{Na+}$ $\ce{Cl-}$ and a copper cathode.

$$ \begin{array}{cc} \hline \text{Half-reaction} & E^\circ/\pu{V} \\ \hline \ce{Cu^2+ + 2 e− <=> Cu(s)} & 0.337 \\ \ce{Na+ + e- <=> Na(s)} & -2.71 \\ \ce{Al^3+ + 3 e− <=> Al(s)} & -1.662 \\ \hline \end{array} $$

Thus we can see that the aluminium is oxidised, but there are no $\ce{Cu^2+}$ ions to be reduced also $\ce{Na+}$ reduction would lead to a negative $E_\mathrm{cell}$ standard and a positive Gibbs energy. Is it an air cathode where $\ce{O2}$ is being reduced?

  • $\begingroup$ I took a liberty to convert a list of reduction potentials for elements into a table. Standard potentials determined for half-reactions, not elements. $\endgroup$ – andselisk Nov 1 '19 at 6:06

Thus we can see that the Al is oxidised but there are no Cu2+ ions to be reduced also Na+ reduction would lead to a negative Ecell standard and a positive Gibbs. Is it an air cathode where O2 is being reduced?

Well, there is no free lunch in this world. This is not a conventional cell in which each electrode is sitting in its own ions and one can easily calculate their Nernst potential. What we are probably seeing here is the very concept of contact potential. When two dissimilar metals are brought into contact via some electrolyte, a potential difference appears between the two. Recall the first experiment of Volta. He made a long series of -silver[wet cardboard]zinc-silver[wet cardboard]zinc-silver[wet cardboard]zinc- cells, where the cardboard was wetted with salt-water. He was able to generate large potential differences this way by making very long series of Ag-Zn couple. Is this set-up good for drawing current-probably not.

These experiments were done in the 17th century by Volta. Atoms and ions did not exist then. These are contact potentials and you do not need a redox system to begin the experiment (metal dipping in its salt). Let us say you start with Al, Al will start to dissolve in a corrosive medium (salt is very corrosive) to form Al ions. You have a valid point, there are no copper ions to reduce? What else do we have in the system? Water, it has some ionized $\ce{H+}$, they start to reduce, one should see hydrogen bubbles at the copper electrode. Certainly you can draw small currents to power a LED from such batteries but not for long.

I do have some doubt about the design of the experiment. All soda cans are lined with a very very thin polymer layer. So the contact of Al is not that good with the copper coil. If one scrubs the can very well, only then that polymer layer would come off. The claim of seven LEDs is a stretch.

  • $\begingroup$ But this setup is being used to power 7 LEDs and a potato battery can be used to power a motor but it too has no ion to reduce. I am still wondering what is the reduction reaction? $\endgroup$ – ChemEng Nov 1 '19 at 19:59
  • $\begingroup$ @ChemEng, I edited the answer to clarify your query. $\endgroup$ – M. Farooq Nov 1 '19 at 20:52
  • $\begingroup$ I actually made a mistake he said 9 LEDs and it lasted for a week plus in the vid $\endgroup$ – ChemEng Nov 1 '19 at 21:35
  • $\begingroup$ M. Farooq pointed to the cathodic reduction: H2O + e- --> OH- + H. ; then 2H. --> H2. In a dry cell, the H2 bubbles (or H.) would be oxidized to H2O by MnO2 to prevent pressure build-up. In these soda can cells, if the sand is not saturated, some air might get to the Cu wire and be reduced to water: 4H. + O2 --> 2H2O. $\endgroup$ – James Gaidis Nov 2 '19 at 14:32

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