According to my book, metals want to "lose" electrons to achieve noble gas configuration. But at the same time, copper ions in a galvanic element are going to gain electrons from a zinc electrode. Why is it so? Is it because one of them HAS to gain/lose electrons, and zinc wins out in terms of achieving noble gas configuration?
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3 Answers
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Typically, when one uses terms such as "likes to lose" or "likes to gain", we should think of that phrase as "likes to lose relative to...". In your case, when we think about electron configurations, it might be more helpful to think "metals like to lose rather than gain electrons to achieve a noble-gas-like configuration". When comparing two metals, both of which tend to lose electrons, we need to look at different concepts to answer the question.
We can explain the transfer of electrons by looking at the the metal activity series such as the one shown here. The way one reads this table is that a Galvanic cell created with two metals will have the metal higher on the list as its anode (where oxidation occurs) and the other metal will be the cathode. Thus, we would expect Zn to be oxidized to Zn(II) and Cu(II) to be reduced to Cu. These tables are based on standard reduction potentials:
$$\ce{Cu^{2+} + 2 e^- -> Cu(s)}\ \tag{E = 0.34 V}$$ $$\ce{Zn^{2+} + 2 e^- -> Zn(s)}\ \tag{E = -0.76 V}$$
By flipping the 2nd reaction (thus changing the sign of E) one obtains a cell potential of about 1.1 V; a positive number meaning that this reaction will happen spontaneously:
$$\ce{Cu^{2+} + Zn(s) -> Cu(s) + Zn^{2+}}\tag{E = 1.1V}$$
The standard reduction potentials tell us that the electrons would rather be on copper than on zinc.
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Reactive systems seek lowest free energy equilibrium. In aqueous solution, Zn + Cu(II) is higher in energy than Zn(II) + Cu(O) re the electromotive series.
This can be manipulated. In hard Lewis base water, soft Lewis acid Cu(I) is unstable to disproportionation to give hard Lewis acid Cu(II) and Cu(0). In soft Lewis base acetonitrile, Cu(II) and Cu(0) are unstable to conproportionation to Cu(I). That last is quite a reaction. Anhydrous CuCl2 and copper powder (slight excess) are refluxed with stirring in MeCN. The solution turns a terrible opaque inhomogeneous khaki color, then flashes colorless and transparent.
It would be interesting to try it with Fe(II) and Fe(O).
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Metals want to lose electrons because the process requires very less energy. Some metals require more energy than others, e.g. consider Cu and Zn. Zn will require lesser amount of energy to lose electrons than Cu. This is easily seen by the greater Oxidation potential values of Zinc.
When a metal electrode is dipped in an electrolyte containing its own ion, then it may lose or gain electrons depending on whether it has a negative or a positive oxidation potential, again involving energy $(W=qV)$. Thus, Zn having a negative potential loses electrons and Cu having a positive potential gains electrons. Thus, a connected Zn and Cu system will allow passage of electrons from Zn to Cu.
