In terms of electronegativity, from what I understand electronegativity increases going across the period, so surely this should mean that zinc less readily loses its outer shell electrons than copper? Zinc has a greater nuclear charge but the outer shell electrons are in the same shell, so should the outer shell electrons in zinc not experience a stronger attraction to the nucleus? I have a feeling that this has something to do with which sub shells the electrons are removed from in copper and zinc...
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$\begingroup$ By reactivity, do you mean standard reduction potential? $\endgroup$– Brinn BelyeaFeb 16, 2015 at 16:44
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2$\begingroup$ I am learning about half cells, so it is in this context. Zinc loses its valence electrons more readily than copper, and also when zinc is placed into copper sulfate solution it will displace the copper, so is more reactive... $\endgroup$– MeepFeb 16, 2015 at 16:46
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2$\begingroup$ For an extremely thorough discussion of the findamentals, see this paper: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95(10 (2018) 1801-1810. The Zn and Cu Daniell cell is addressed at great length (way too long for an answer here) and the metal cohesive energy is the source of more than 75% of the energy in the Daniell cell (p. 1805.). $\endgroup$– Ed VJul 24, 2019 at 15:17
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$\begingroup$ In a metal, Zn isn't a standalone atom, it is a part of a weird large lattice. Unlike a standalone atom, Zn has no valence electrons of its own. Instead it shares the delocalized electron cloud with all the other Zn atoms in the Zn crystalline structure. There is no particular reason why Zn inside a lattice would be attached stronger to its delocalized electron cloud than Cu. $\endgroup$– akhmedNov 16, 2022 at 23:11
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3 Answers
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You have to think about the whole process. When a metal loses electrons to make a metal ion the following happens:
- The metallic bonds holding the metal atoms together are broken.
- The metal atom loses the electrons.
- The resulting metal ion is hydrated.
In your analysis you are only focusing on step 2. The enthalpy and entropy of the entire process factor into the reduction potential.
answered Feb 16, 2015 at 16:54
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1$\begingroup$ So why do we need to ionize two electrons? Do we need two electrons from sodium or silver? Is two electrons enough for aluminum or chromium? $\endgroup$ Dec 7, 2017 at 17:20
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1$\begingroup$ I edited the answer to make it more general for any number pf electrons lost. $\endgroup$ Dec 8, 2017 at 0:22
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Zinc has last sub orbital as $\mathrm{4s^2}$ which it gives willingly to become electronically stable. Also, zinc's pull on its last electronic shells is much less and is hence more willing to combine. Copper has a last sub-orbital as $\mathrm{4s^1}$ which decreases its capacity to form bonds like zinc. Ideally, copper should form more bonds but due to lack of electrons in the last shell, it cannot form many bonds.
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Zinc's relatively higher reactivity than copper is not because its highest occupied atomic orbital is $\ce{4s^2}$ while copper's is $\ce{4s^1}$; according to that magnesium should be more reactive than sodium as the highest occupied atomic orbital is $\ce{3s^2}$ while for sodium it is $\ce{3s^1}$. Therefore, this is not the case.
Now, the question that arises is: what is the real reason zinc is more reactive than copper?
The answer is that zinc is able to lose its outer electron more readily than copper. This phenomena occurs because copper metal is able to delocalize its outer electrons more readily than zinc. The metallic bond of copper is therefore stronger than zinc, so more energy is required to break this bond in order for it to react. Since more energy is required for copper to react, we can say that zinc is more reactive than copper.
