Electrochemical potentials from first principles (part 2)

Now let’s revisit Hess’s law and the standard values \Delta H_f^\circ, \Delta G_f^\circ, S^\circ found in tables in the back of most textbooks.

In part 1, I pointed out, just as any textbook would, that we can compute the voltage of any full redox reaction by breaking it down into the reduction and oxidation half-reactions and adding their voltages. (You get the oxidation voltage by flipping the sign on the corresponding reduction potential.) This is true even if the half-reactions involve different numbers of electrons as written, because the fundamental meaning of voltage is energy per electron; when we combine the two half-reactions properly, the oxidation half-reaction must produce as many electrons as the reduction half-reaction consumes.

I said that if we have N half-reactions, then this lets us compute the voltages of all N^2 full reactions we can get by combining them pairwise, just by tabulating the N half-reaction potentials. This might sound like it’s “good enough”, but it isn’t. There are millions of known substances, but a practically unlimited number of half-reactions we could potentially compose from them. Tabulating all half-reaction voltages, in the hopes of being able to compute full-reaction voltages from them, is impossible.

But that’s okay, because half-reaction voltages aren’t all independent of each other, just like full-reaction voltages turned out not to be independent of each other (which we used in part 1 to justify our gauge fixing).

Here’s a simple example. Consider the following three half-reactions, under some uniform, specified set of conditions. (If they aren’t all occurring under the same set of conditions, you need to use Nernst’s equation to transform some or all of the voltages to a standard set of conditions; but more on that later.)

  1. Fe3+ + e → Fe2+
  2. Fe2+ + 2e → Fe
  3. Fe3+ + 3e → Fe

Denote their standard reduction potentials by E_1, E_2, E_3, respectively. Can we derive a relationship between the three? The answer is yes, and the reason is that these half-reactions all involve the same substances (loosely speaking). But here, caution must be exercised. Half-reaction 3 is the sum of half-reactions 1 and 2, but that does not mean its voltage is the sum of the voltages of half-reactions 1 and 2. The only time we can directly add is when the result is a full-reaction (with no dangling electrons) rather than a half-reaction. The bottom line is that E_3 \neq E_1 + E_2 in general, when adding half-reactions.

The correct method for solving this kind of problem is to rewrite the voltages as free energy changes and apply Hess’s law. Since reaction 3 is the sum of reaction 1 and reaction 2, it must also be true that \Delta G_3 = \Delta G_1 + \Delta G_2. In the first reaction one electron is transferred, so \Delta G_1 = -FE_1; likewise in reactions 2 and 3, \Delta G_2 = -2FE_2 and \Delta G_3 = -3FE_2, respectively. The result is -3FE_3 = -2FE_2 - FE_1. Cancelling a factor of -F gives our desired relationship: 3E_3 = E_1 + 2E_2. In general, n_3 E_3 = n_1 E_1 + n_2 E_2.

This raises a pressing question, though. We used the formula \Delta G = -nFE for half-reactions here, but only derived it for full reactions in part 1. In fact, what does \Delta G even mean for a half-reaction anyway? It’s the free energy of the right side minus the free energy of the left side… but what if one side or the other contains dangling electrons?

We can deal with this problem in two ways. One is to combine each half-reaction with the standard hydrogen electrode half-reaction, which I will halve, for the sake of mathematical convenience, that is, 1/2 H2 → H+ + e. Because this half-reaction is defined to have voltage zero, when we add it to any of our half-reactions, we can get a full-reaction whose voltage equals the voltage of the original half-reaction. Thus:

  1. Fe3+ + 1/2 H2 → Fe2+ + H+
  2. Fe2+ + H2 → Fe + 2H+
  3. Fe3+ + 3/2 H2 → Fe + 3H+

Now the voltages of these three full-reactions are as follows: E_a = E_1; E_b = E_2; E_c = E_3; again, this is true because E_a = E_1 - E_0^\circ and so on, where E_0^\circ = 0 is the standard reduction potential of our standard hydrogen electrode reference reaction. Now we can apply the formula \Delta G = -nFE rigorously. So what we get is \Delta G_c = \Delta G_a + \Delta G_b, or -3FE_c = -FE_a -2FE_b. But since each full reaction voltage equals the corresponding half-reaction voltage, we end up at the same conclusion: -3FE_3 = -FE_1 -2FE_2, and so on. So what we can do is just apply the formula \Delta G = -nFE to half-reactions willy-nilly because, as this example demonstrates, it’s shorthand that allows us to reach the same conclusion as we would have if, more rigorously, we had used full-reactions by combining with the reference cell.

The other way we can think about this is to carefully think about how we defined half-reaction potentials in the first place. I omitted a small detail in part 1, which is that the absolute half-reaction potentials actually depend on the nature of the wire—wires made of more electropositive metals don’t want electrons as much, for example. But that doesn’t matter because the number of electrons in the wire doesn’t ever change; all the wire does is ferry electrons from one half-cell to the other. Mathematically, the absolute potential inside the wire, V_e, always cancels out. Indeed, we might as well represent the transfer of electrons from anode to cathode by the sum of a different set of two processes: one in which an electron moves from the anode to infinity, and another in which an electron moves from infinity to the cathode. If we do this, then the half-reaction potentials truly depend only on the nature of the participating substances (and their states), and not at all on the wire; and furthermore we can now define \Delta G properly for a half-reaction, by declaring that whenever “e” appears on either the left or right side, its free energy is considered to be zero, since it now represents a standard reference point derived from our hypothetical half-processes—an infinitely dilute electron gas located infinitely far away from everything else.

Indeed, in a certain sense, it’s rather unfortunate that we had to use convoluted reasoning simply because we’re dealing with electrochemical reactions, in which electrons are transferred, rather than “ordinary” chemical reactions. A lot of chemical reactions can’t be broken down in a straightforward manner into two halves the way electrochemical reactions can, and yet, we manage to compute their \Delta H and \Delta G values, rather than just looking up each reaction separately in a table. We accomplish this using the standard enthalpies and standard free energies of formation in conjunction with Hess’s law. In theory, as long as we know these values for all individual substances under our reaction conditions, we can predict the \Delta H and \Delta G values for all chemical reactions under that set of conditions. The number of possible reactions that can be obtained from a given set of substances is vastly greater than the number of substances, so tabulating the standard electrode potentials for “all” half-reactions we’re interested in is far less practical than tabulating the standard enthalpies and free energies of formation for all substances we’re interested in.

In essence, basic electrochemistry looks harder than “normal” chemistry because many students are confused by what to do with the electrons, and because it seems like we use voltages to describe electrochemical reactions whereas we use free energies (which are perhaps simpler to understand) for non-electrochemical reactions. Thermochemically, however, the electrons ought to be treated as first-class citizens, that simply have their \Delta H_f and \Delta G_f set to zero under all conditions. If we think of half-reactions in this way, we can treat them in the same way as other reactions; we should be able to compute the potential of any half-reaction if we know just the \Delta G_f^\circ of all participants.

To flesh out this idea, let’s consider the following reaction: 1/2 Cl2 + e → Cl. This looks like a standard formation reaction for chloride ion. Certainly we cannot form an ion only from neutral elements, so we define the standard formation reaction for an anion as the half-reaction with one mole of the anion on the right side, and the constituent elements in their standard states together with an appropriate number of electrons on the left side; for a cation we define it the same way except that electrons go on the right side. We can measure this reaction’s voltage in the laboratory, and thus obtain the free energy change of this reaction. We will refer to this as the standard free energy of formation \Delta G_f^\circ for chloride ion, just like with neutral substances. This standard formation reaction trivially satisfies Hess’s law, because its \Delta G is equal to \Delta G_f^\circ of the right side minus the sum of all \Delta G_f^\circ values on the left; but the former describes the same reaction, and the latter is zero since we defined it to be zero for the elements in their standard states plus the dangling electron e.

Then, when considering a more complex half-reaction, such as: MnO4 + 8H+ + 5e → Mn2+ + 4H2O, we can compute its free energy change by using the tabulated free energies of formation of all participants, and, again, noting that this is zero for the dangling electrons 5e.

Recall that we fixed a gauge by selecting the standard hydrogen electrode as the reference point for voltage. This gauge freedom exists in the free energy formulation as well. For, under this choice of gauge, the free energy of formation of H+ is exactly zero; but for any other choice of gauge, it is a nonzero number, and the free energies of formation of all other ions are shifted also; but once electrons are balanced, we still get the same value for the free energy change of a full-reaction as we would under any other choice of gauge.

That being said, the equation \Delta G = \Delta G^\circ + RT \log Q applies to half-reactions. When we treat the electron as a reactant or product, the expression for Q should include a factor of “[e]”. This is always one, so you can just cross it out, because the electron gas is always in its standard state. Dividing through by -nF gives the Nernst equation: E = E^\circ - \frac{RT}{nF} \log Q.

Okay, if you agree with everything I’ve said, then I’ve achieved my “agenda” this whole time, which is equal rights for the electron. Okay, I’m joking, but also half-serious. Some textbooks pretend that half-reactions are not “real” and that they are only used as a bookkeeping mechanism, i.e., so you can balance redox reactions by the half-reaction method and compute their voltages by adding the half-reaction voltages. But I feel that they stretch this a bit when they tabulate free energies of formation for ions, such as sulfate, in the back. If you don’t accept the reality of a half-reaction, then how can you make sense of \Delta G_f^\circ values for ions?

This leads logically to the approach I have taken: accept that half-reactions are just like other reactions; accept that they have real \Delta G values; see that this allows us to consider the standard free energies of formation of neutral molecules and ions in the same way; and accord the electron status as a separate “substance” in half-reactions, whose \Delta G_f^\circ is zero. I think this is the least confusing way to regard the entirety of basic electrochemistry.


About Brian

Hi! I'm Brian Bi. As of November 2014 I live in Sunnyvale, California, USA and I'm a software engineer at Google. Besides code, I also like math, physics, chemistry, and some other miscellaneous things.
This entry was posted in Uncategorized. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s