Half-reactions

In this topic and the next few topics we're going to explore how concepts like "electrode potential", "redox potential", and "standard hydrogen electrode" look in the $V_i$ world. These are all built around electron transfer, aka "redox".

The general idea of redox chemistry is considering half-reactions of the following generic form:

$$\mathrm{Ox} + z\mathrm{e}^- \rightleftharpoons \mathrm{Red}$$

I.e. we consider $z$ electrons transferred onto an oxidized species $\mathrm{Ox}$ to become a reduced species $\mathrm{Red}$ (note the charges satisfy $z_{\mathrm{Ox}} - z = z_{\mathrm{Red}}$).

A half-reaction can't happen on its own since the solvent generally doesn't contain a population of free electrons: for a half-reaction to move forward, the electron has to be taken from an electrode or another half-reaction, and likewise for the reaction to move backward it needs to find somewhere to give away its electron.

In terms of electrochemical potentials, this defines an electrochemical potential of electrons by equilibrium:

$$\bar\mu_{\mathrm{Ox}} + z \bar\mu_{\mathrm{e}^-} = \bar\mu_{\mathrm{Red}},$$

which in our $V_i$ terms becomes

$$V_{\mathrm{e}^-} = \frac{\bar\mu_{\mathrm{Ox}} - \bar\mu_{\mathrm{Red}}}{zF}.$$

(we're describing $\mathrm{Ox}$ and $\mathrm{Red}$ with electrochemical potentials for now, instead of converting them to $V_i$, because one of them might be an uncharged species)

Previously (in the Equilibrium topic), we associated this $V_{\mathrm{e}^-}$ with an electrode. That is, we talked about what happens when such a half-reaction is exchanging its electrons with an electrode, and so at equilibrium this reaction $V_{\mathrm{e}^-}$ equalizes with the $V_{\mathrm{e}^-}$ in the electrode. But now we'll take a different point of view: The central premise of redox chemistry is that yes, it is useful to talk about the reaction $V_{\mathrm{e}^-}$ even when there is no equilibrated electrode nearby.

Implied $V_{\mathrm{e}^-}$ of a reaction

In the absence of an equilibrated electrode, it is indeed useful to calculate this $V_{\mathrm{e}^-}$ as an "implied" voltage, representing a real thermodynamic availability of electrons even though no electrons roam freely in solution. For clarity, we write these implied values as $V_{\mathrm{e}^-}(\mathrm{Ox}/\mathrm{Red})$ since they are specific to a certain reaction. This implied $V_{\mathrm{e}^-}$ value is useful in the following ways:

1. In-solution coupling between different half-reactions

   $$V_{\mathrm{e}^-}(\mathrm{\mathrm{Ox1}/\mathrm{Red1}}) \rightleftharpoons V_{\mathrm{e}^-}(\mathrm{\mathrm{Ox2}/\mathrm{Red2}})$$

   • A solution may have multiple available half-reactions.
   • All half-reactions are thermodynamically driven to equalize their $V_{\mathrm{e}^-}$ values by directly exchanging electrons with each other ('electron transfer reactions'), even in the absence of an electrode. But, electron transfer can be kinetically slow, in which case it is valuable to distinguish multiple $V_{\mathrm{e}^-}$. (for example, natural ground water is notoriously redox disequilibrated^[1])
   • Where there is no electrode, electron transfer between two species can be described in terms of a mismatch in their half-reactions' $V_{\mathrm{e}^-}$ values.

2. Coupling between half-reactions and an electrode

   $$V_{\mathrm{e}^-}(\mathrm{metal}) \rightleftharpoons V_{\mathrm{e}^-}(\mathrm{Ox}/\mathrm{Red})$$

   • A simple electrode should couple to one half-reaction, but equalization of $V_{\mathrm{e}^-}$ only occurs when the electrode has no current. At driven electrodes (with nonzero current) we can describe overpotentials at the interface as a step between $V_{\mathrm{e}^-}$ in the electrode and the equilibrium $V_{\mathrm{e}^-}$ implied by its ideal reaction.
   • Real electrodes may couple to more than one half-reaction, giving 'mixed potentials'.^[2]

On our diagrams we will depict these implied levels as dashed lines, inside the solution.

The idea that a solution can have an implied electronic $V_{\mathrm{e}^-}$ (or $\bar\mu_{\mathrm{e}^-}$) is not at all new. This is often called a 'redox Fermi level'^[3] and what's attractive about it is that $\bar\mu_{\mathrm{e}^-}$ can be directly plotted on a traditional electronic energy band diagram. However, I've found that past visualizations can be a bit confusing,^[4] so I hope to present these diagrams in a fresh light.

I emphasize though that not all solutions have a meaningful redox Fermi level, and some disequilibrated solutions have multiple redox Fermi levels.

Nernst equation

We are interested in the case where the reactants have activities $a_{\mathrm{Ox}}$ and $a_{\mathrm{Red}}$. We can then formally break this down using standard states of the reactants (noting that $\bar\mu^\circ_i$ are also 'floating' for the charged reactants):

$$\bar\mu_{\mathrm{Ox}} = \bar\mu^\circ_{\mathrm{Ox}} + RT\ln(a_{\mathrm{Ox}})$$

$$\bar\mu_{\mathrm{Red}} = \bar\mu^\circ_{\mathrm{Red}} + RT\ln(a_{\mathrm{Red}}).$$

Note that not all reactants need to be solutes, so activity (and standard state) can be defined in various ways.

Substituting these in, we arrive at the following equation which I call the "floating Nernst equation":

$$V_{\mathrm{e}^-}(\mathrm{Ox}/\mathrm{Red}) = V^\circ_{\mathrm{e}^-}(\mathrm{Ox}/\mathrm{Red}) + \frac{RT}{zF} \ln\bigg(\frac{a_{\mathrm{Ox}}}{a_{\mathrm{Red}}}\bigg) ,$$

where we define the standard redox level for the $\mathrm{Ox}/\mathrm{Red}$ reaction:

$$V^\circ_{\mathrm{e}^-}(\mathrm{Ox}/\mathrm{Red}) = \frac{\bar\mu^\circ_{\mathrm{Ox}} - \bar\mu^\circ_{\mathrm{Red}}}{zF} .$$

These levels float alongside our $V^\circ_i$ ionic standard states. Note that we can substitute $z_i F V^\circ_i$ for $\bar\mu^\circ_i$ for the ion reactants, to get formulae involving $V^\circ_i$ (see below for a more general formula).

The above equation for $V_{\mathrm{e}^-}$ looks extremely similar to the regular Nernst equation, though it involves $V_{\mathrm{e}^-}$ instead of $E$. We'll get back to what this traditional electrochemical $E$ actually means in the next topic, but we don't need it for now.

It's helpful to visualize an example of where these redox levels lie in relation to the ionic levels we've been discussing previously, in this case with the ferric/ferrous redox couple. Note that in order to read out $V_{\mathrm{e}^-}(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+})$ potentiometrically without influence from other reactions, a glassy carbon electrode could be used, since a platinum electrode may also pick up hydrogen or oxygen redox couples and drift toward a mixed potential.^[5]

General form

In general we might consider a half-reaction containing more than one species on each side, coming in more than one number:

$$a\mathrm{A}^{z_{\mathrm{A}}} + b\mathrm{B} + z\mathrm{e}^- \rightleftharpoons c\mathrm{C}^{z_{\mathrm{C}}} + d\mathrm{D}$$

where $A$, $C$ are generic charged species (ions), and $B$, $D$ are generic neutral species ($z_{\mathrm{B}} = z_{\mathrm{D}} = 0$). The Nernst equation is then (writing just "rxn" for short instead of "$\mathrm{A},\mathrm{B}/\mathrm{C},\mathrm{D}$"):

$$V_{\mathrm{e}^-}(\mathrm{rxn}) = V^\circ_{\mathrm{e}^-}(\mathrm{rxn}) + \frac{RT}{zF} \ln\bigg(\frac{(a_{\mathrm{A}})^a (a_{\mathrm{B}})^b }{(a_{\mathrm{C}})^c (a_{\mathrm{D}})^d }\bigg) ,$$

and for the standard redox level we can use $V^\circ_i$ for the charged species:

$$V^\circ_{\mathrm{e}^-}(\mathrm{rxn}) = \frac{az_{\mathrm{A}}}{z}V^\circ_{\mathrm{A}} - \frac{cz_{\mathrm{C}}}{z}V^\circ_{\mathrm{C}} + \frac{b\mu^\circ_{\mathrm{B}} - d\mu^\circ_{\mathrm{D}}}{zF} .$$

This provides a general recipe and it's easy to see how to extend it to more ionic or more neutral reactants. (Note that the weighting of $V^\circ_i$ levels is always balanced to a total of 1, since the original reaction is charge-balanced: $az_{\mathrm{A}} - z = cz_{\mathrm{C}}$ in this case.)

Plating couples

The simplest case is a metal plating couple, where the reduced species is the pure metal itself. The front page quietly relied on this one:

$$\mathrm{Li}^+ + \mathrm{e}^- \rightleftharpoons \mathrm{Li(s)},$$

for which the implied level needs no Nernst machinery at all:

$$V_{\mathrm{e}^-}(\mathrm{Li}^+/\mathrm{Li}) = \frac{\bar\mu_{\mathrm{Li}^+} - \mu_{\mathrm{Li}}}{F} = V_{\mathrm{Li}^+} - \frac{\mu_{\mathrm{Li}}}{F}.$$

Under our convention (and at the reference temperature and pressure) $\mu_{\mathrm{Li}} = 0$, so $V_{\mathrm{e}^-}(\mathrm{Li}^+/\mathrm{Li}) = V_{\mathrm{Li}^+}$: the couple's implied electron level plots exactly on top of the ion's own level, at every concentration, since $\bar\mu_{\mathrm{Li}^+}$ carries the whole activity dependence. This cashes the promise made in the front page's footnote: the line we drew there as $V_{\mathrm{Li}^+}$ doubles as the electrolyte's redox Fermi level, in the manner of Gerischer. The same holds for every metal plating couple $\mathrm{M}^{n+}/\mathrm{M}$ (compare the $\mathrm{Fe}^{2+}/\mathrm{Fe(s)}$ row in the table below), and it is the same accident that put $V_{\mathrm{e}^-} = V_{\mathrm{Zn}^{2+}}$ at the zinc electrode: the elemental $\mu = 0$ convention wearing another hat.

Keep in mind what kind of coincidence this is. The two lines are different objects, $V_{\mathrm{Li}^+}$ being the level of ions really present in the electrolyte while $V_{\mathrm{e}^-}(\mathrm{Li}^+/\mathrm{Li})$ is an implied electron level. They separate the moment $\mu_{\mathrm{Li}} \neq 0$: pick a different convention and the implied level shifts away; or, more physically, put a temperature gradient across the cell and $\mu_{\mathrm{Li}}(T)$ varies from place to place, peeling the implied level off of $V_{\mathrm{Li}^+}$ by a real, position-dependent amount.

Standard redox levels in water

Also known as the standard electrode potential, the standard reduction potential $E^\circ$ is the reduction potential for a reaction that involves species in their standard states. In particular this means that dissolved ions are at a hypothetical ideally-dilute concentration of $c^\circ = 1~\mathrm{mol/L}$; practically this means that these standard reduction potentials are best extrapolated from dilute solutions.

It is also assumed that the temperature is 25 °C and the pressure is 1 bar. (Actually, 1 atm is commonly used, which tweaks $\mu_{\mathrm{H}_2} / 2F$ by a sub-millivolt correction ($\approx 0.2~\mathrm{mV}$), but we'll ignore that.)

The consequence of the standard-ideal-concentration (or unit activity) condition is that all $V_i$ for dissolved ions are replaced by $V^\circ_i$.

As with our ionic standard states, we can tabulate all the relative positions of the $V^\circ_{\mathrm{e}^-}(\mathrm{Ox}/\mathrm{Red})$ ladder, by defining one half-reaction (usually $\mathrm{H}^+/\mathrm{H_2}$) as a reference level.^[6] We'll call this $E^\circ$, because this is in fact the standard electrode potential (we'll discuss the meaning of "electrode potential" more in the next topic):

$$E^\circ = V^\circ_{\mathrm{e}^-}(\mathrm{Ox}/\mathrm{Red}) - V^\circ_{\mathrm{e}^-}(\mathrm{H}^+/\mathrm{H_2})$$

Wikipedia's standard electrode potential data page is a fantastic resource to find more of these.

It's worth visualizing the $V^\circ_{\mathrm{e}^-}(\mathrm{Ox}/\mathrm{Red})$ levels alongside the ionic levels $V^\circ_i$. We plot the standard redox levels as dashed (representing that they are 'implied' levels), and as thin lines (representing that they are only standard states):

As can be seen, when we change the electrical offset, both 'ladders' move in lockstep. However, the ionic standard states are sensitive to our choice of zero of neutral elements' chemical potentials, whereas the standard redox levels are totally immune to that (since they really are electronic in nature).

Takeaways

And so we've arrived at a description of redox half-reactions in terms of a virtual or implied value $V_{\mathrm{e}^-}(\mathrm{Ox}/\mathrm{Red})$, and in relating this to reactant activities, we've identified a new ladder of "standard redox levels" $V^\circ_{\mathrm{e}^-}(\mathrm{Ox}/\mathrm{Red})$.

In the next topic, we'll talk about how we can explain traditional electrochemical variables in terms of these redox levels.

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