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Is it possible to create a galvanic cell without a half cell?

In the book I was reading, Illustrated Guide to Home Chemistry Experiments, the following lab (Laboratory 16.4) was suggested:

add 2 M HCl and add copper metal to one end of the beaker, with magnesium at the other end. Then, add patch cables to connect the metal plate to an LED to have physical evidence of success.

However, I don't understand how useful work could be generated without half cells.

To my understanding, we add HCl to react with the magnesium and copper, to produce ions ($\ce{Cu+}$ and $\ce{Mg^2+}$). Then, they can spontaneously produce work due to their eV differences.

What I am missing is how the work can be useful, as opposed to just generating heat in the solution.

I would believe that, rather than traveling though the patch cables, electrons can travel directly through the solution and produce heat.

Thanks for reading and I really appreciate any help!

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  • $\begingroup$ HCl won't react with Cu. $\endgroup$ – Mithoron Apr 5 '15 at 22:13
  • $\begingroup$ Please consider giving the green checkmark (or upvote or both) to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this! $\endgroup$ – Ed V Jul 15 '20 at 0:17
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This differs from the lemon galvanic cell (aka, 'battery') in several important ways: 1) Mg is used as the anode, rather than Zn, 2) there is no salt bridge, unlike in the the lemon cell, where the lemon serves as a crude 'salt bridge', and 3) the electrotyle is 2 M HCl rather than citric acid (and other stuff) in lemon juice. An open circuit voltage will be produced, easily measured with a digital multimeter in voltage input mode. A short circuit current will be produced, easily measured with a digital multimeter in current input mode. It may be possible to light a red LED, since they have low threshold voltage and low current requirements. I suspect it would light the red LED, though feebly.

But regardless of whether or not there is an external connection of the Mg and Cu electrodes, there will be a rapid redox reaction at the Mg electrode: $$\ce{Mg(s) + 2 H^+ <=> Mg^{2+} + H_2 (g)}$$

This unwanted competing redox reaction is the same as the net reaction obtained from the two relevant half cell reactions and it justs wastes some of the available energy as heat in the cell's vessel. So some electrical power is available by connecting an electrical load (e.g., red LED or resistor) between the two electrodes. But the parasitic oxidation of the Mg anode will continue until the electrode is consumed or the HCl is almost entirely consumed. (Magnesium will very slowly react with boiling water.)

Notes: 1. Clean copper, i.e., without surface compounds or so-called 'patina', does not react with 2 M HCl at room temperature. So the 2 M HCl does not contain any $\ce{Cu^+}$ or $\ce{Cu^{2+}}$ ions. 2. The reaction of Zn with 2 M HCl is considerably slower than is the reaction of Mg with 2 M HCl.

For more information, and at a much deeper level, see K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", Journal of Chemical Education, 95 (2018) 1801-1810.

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Electrons cannot move through the solution. Only ions can. Therefore the circuit will still complete properly and work will be done.

Similar idea to the lemon battery. Two dissimilar metals will have different potentials and things will work out.

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