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The $E^θ(\ce{M^{2+}}/\ce{M})$ value for copper is positive ($\pu{+0.34 V}$). What is possibly the reason for this? (Hint: consider its high $Δ_aH_V$ and low $Δ_{hyd}H_V$)

We know that we have to first break the metallic lattice to turn the metal into gas and hence there should be some sort of relation between the two.

My confusion is that when we split the lattice, aren't we technically subliming the substance as well?

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tl; dr: In the case of a metal, sublimation and atomization can be regarded as synonymous when atomization refers to the process of converting the solid into a dilute atomic gas. In particular, the end points (states) of the process should be identical.

There are usually conventions on what you call "enthalpy of X". However the official source of chemical nomenclature - IUPAC - has not bothered to post a definition for enthalpy of atomization (or atomization enthalpy) in its Gold Book. There are entries for atomize:

To subdivide a liquid into very small particles; methods include: impact with a jet of gas, use of a spinning disk generator, vibrating orifice generator, etc.

and for atomization (in analytical flame spectroscopy):

The conversion of volatilized analyte into free atoms

These definitions do not refer to the same process, which leads to some ambiguity. Are you speaking of a transition from a condensed into the gas phase that preserves molecular structure, or of a chemical process that converts a sample in some phase into an atomic gas?

The wikipedia meanwhile has an entry for enthalpy of atomization which provides the following definition:

The enthalpy of atomization (also atomisation in British English) is the enthalpy change that accompanies the total separation of all atoms in a chemical substance (either a chemical element or a chemical compound)

There is an entry in the IUPAC gold book for bond energy (theoretical):

The energy required to break a given type of bond between atoms in certain valence states. An averaged bond energy is commonly derived by dissecting the heat of atomization of a molecule into contributions of individual bonds. For molecules with localized bonds, the heats of atomization (formation) are usually well approximated by the sum of pertinent averaged bond energies.

This reference to "heat of atomization" (which is not otherwise defined in the Gold Book) is consistent with that in the wikipedia.

If you apply this last definition then yes, the sublimation enthalpy and the atomization enthalpy of a solid metal should be identical, since both refer to the process:

$$\ce{M(s)->M(g)}$$

(assuming identical P,T end points). Note that sublimation enthalpy means standard sublimation enthalpy when referring to the process at standard pressure (1 bar) and a fixed temperature. Similarly for atomization enthalpy.

As regards the question:

The $E^θ(\ce{M^{2+}}/\ce{M})$ value for copper is positive ($\pu{+0.34 V}$). What is possibly the reason for this? (Hint: consider its high $Δ_aH_V$ and low $Δ_{hyd}H_V$)

Different uses of the word "atomization" render the meaning of "enthalpy of atomization" ambiguous. This is why I open my answer with an exposition on the definition of "atomization". Your problem refers to an ideal solution, but I assume here atomization refers to sublimation since reference to hydration would not seem to make sense otherwise. The question asks to break down the reaction into steps including atomization (sublimation), ionization and solvation. Ignoring the ionization step, a larger atomization enthalpy and smaller hydration enthalpy is consistent with the metal being more stable than the solubilized ion, ie the reduced form is favored (relative to standard hydrogen electrode).

A libre text and the accompanying diagram explain some of this

enter image description here

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    $\begingroup$ I got the big picture of this but I still have a question of atomisation energy of copper, it'd be great if you could check it out chemistry.stackexchange.com/questions/123802/… $\endgroup$ – Buraian Nov 15 '19 at 19:22
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    $\begingroup$ Sublimation is not exactly the same as atomisation, as many metals form diatomic molecules, especially this with lower boiling point. Rather an exception is mercury, which does not, what has close relation to its low melting point $\endgroup$ – Poutnik Oct 11 at 8:27
  • $\begingroup$ @Poutnik That seems an important caveat. Feel free to add an answer, else I'll edit mine. $\endgroup$ – Buck Thorn Oct 11 at 9:08
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    $\begingroup$ You can add it to your answer,it was just minor comment. See e.g. dirubidium $\endgroup$ – Poutnik Oct 11 at 9:10

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