## Electrochemical potentials from first principles (part 1)

In my experience with reading introductory chemistry textbooks and tutoring students in chemistry, I’ve found that most existing approaches to explaining electrochemistry skimp on conceptual detail. This is understandable, in light of the fact that many students are already having trouble simply performing the calculations. If you’re on top of the material, though, you might have come to wonder about the first principles involved.

Let’s say we have the galvanic cell: Zn(s) | ZnSO4(aq) || CuSO4 | Cu(s). (This denotes a zinc electrode in contact with a zinc sulfate solution, separated by a salt bridge or porous disc from a copper sulfate solution in contact with a copper electrode.) We know experimentally that if we connect the two electrodes with a wire, electrons will flow from zinc to copper, or, equivalently, conventional current will flow from copper to zinc. We interpret this in terms of the electric potential being higher in the copper half-cell than in the zinc half-cell. Recall that the absolute electric potential at a point in space is the change in potential energy that would result if a unit positive test charge were brought from infinity to that point. If the electric potential is higher in the copper half-cell than in the zinc half-cell, the potential energy of the system will decrease when conventional current flows from the copper half-cell to the zinc half-cell. It’s the same as how, when an apple falls off a tree, it travels downward because the gravitational potential is lower closer to the ground. Note however that electrons flow from low potential to high potential, since they are negatively charged rather than positively charged.

Because we have connected the two electrodes with a wire, the two electrodes and the wire together constitute a conductor. Therefore the electric potential is some constant value, $V_e$, throughout the electrodes and the wire. If current is flowing, therefore, it must be due to a potential difference between the solutions in which the electrodes are immersed. Let’s call the potential inside the copper sulfate solution $V_+$, and the potential inside the zinc sulfate solution $V_-$. Then, the potential difference between the copper sulfate solution and the copper electrode is $V_e - V_+$, the potential difference between the zinc sulfate solution and the zinc electrode is $V_e - V_-$, and the potential difference between the copper sulfate solution and the zinc sulfate solution is $V_- - V_+$.

The values $V_e - V_+$ and $V_e - V_-$ from the previous paragraph are important properties of the half-cells. What is their significance? Well, $V_e - V_+$ is the change in potential energy that would occur if a unit positive test charge were moved from the copper sulfate solution to the copper electrode. If this is positive, then the change in potential energy for moving an electron (which is negatively charged) from the solution to the electrode is negative, and negative is favourable. We see that $V_e - V_+$ should be greater the more favourable it is to move an electron from the solution to the electrode.

It is not impossible to measure these values—the potential differences associated with half-reactions alone. They’re just potential differences between the electrode and the solution in contact with the electrode. In principle, it should be possible to measure this, if only you could measure the tiny current that would flow along a tiny wire of very precisely known resistance, connected to the electrode, at the instant it came into contact with the solution, before the charge buildup inhibited further electron flow. Presumably you can see why this is not practicable. I imagine practicable methods do exist, but there isn’t a method that’s both relatively practical and accurate.

That’s why, in practice, we measure cell potentials. The cell potential is the potential difference between the two solutions, which we can easily measure by connecting a voltmeter across the two electrodes. If the cathode solution is at potential $V_+$, the anode solution is at potential $V_-$, and the electrode+wire assembly is at potential $V_e$, then the cell voltage is $V_+ - V_-$, which we can rewrite as $(V_+ - V_e) - (V_- - V_e)$.

The first term, $V_+ - V_e$, is called the reduction potential at the cathode. It is called this because, when the solution is reduced, conventional current flows from the solution to the electrode and wire, and the reduction potential gives the change in free energy for a unit positive test charge undergoing this movement. Note the sign convention, however. If conventional current is flowing from the solution to the wire, then the change in free energy is actually given by $q(V_e - V_+) = -q(V_+ - V_e)$. The second term, $V_- - V_e$ is called the reduction potential at the anode. These reduction potentials are (up to sign) the half-cell potential differences that I said we can’t really measure.

What’s up with this sign convention? Well, unfortunately, it was decided that a spontaneous galvanic cell ought to have a positive cell voltage. The cathode solution is at a higher absolute potential than the anode solution, so that when conventional current flows from cathode to anode, the free energy change $q(V_- - V_+)$ is negative (spontaneous). The positive absolute value of the potential difference is $V_+ - V_-$, and it is this that we call the cell voltage. Now, we’ve decided to break down the process of conventional current flowing from cathode solution to anode solution into two steps. In the first step, conventional current flows from cathode solution to cathode/wire/anode assembly, and in the second, it flows from cathode/wire/anode assembly to anode solution. The overall process’s voltage is the sum of the voltages for the two steps. Since we want the sum to be $V_+ - V_-$, the first step must be given the sign convention $V_+ - V_e$, which is the negative of the actual potential difference $V_e - V_+$ associated with the flow of conventional current from cathode solution to wire. It’s thus set up so that if this flow is favourable (because the cathode solution doesn’t like positive charge very much), then the cathode reduction potential $V_+ - V_e$ is positive. The second step is like the first step but reversed, since the conventional current is flowing from wire/electrode to anode solution. We thus define the quantity $V_- - V_e$ analogously, which is the negative of $V_e - V_-$, the actual potential difference associated with the flow of conventional current from anode solution to wire, and then subtract this since the current is actually flowing in the other direction for our overall process. (If this sounds confusing, it’s because it is; I got the signs wrong several times while writing this post.) Overall, $V_- - V_+ = (V_e - V_+) - (V_e - V_-)$ if you’re using the absolute sign convention, or $V_+ - V_- = (V_+ - V_e) - (V_- - V_e)$ if you’re using the usual sign convention in electrochemistry. If you’re wondering why we don’t just define the oxidation potential at the anode as $V_e - V_-$ so we can add it instead of subtracting it, it’s because you can swap out the cathode for a different half-cell, so that the original anode becomes the cathode and the new cell becomes the anode, if you should happen to pick the new half-cell such that its absolute potential is less than that of the original anode (since, now, conventional current will want to flow from the original anode to the new half-cell, thus making the former the new cathode and the latter the new anode). Since any pairing of half-cells is valid, and any half-cell can be either the cathode cell or the anode cell depending on its partner, we arbitrarily choose to define only reduction potentials, treating each half-cell as though it is a cathode, and flipping sign whenever that half-cell turns out to be an anode. You could make a table of oxidation potentials but each value would just be the negative of the corresponding reduction potential.

Recall that electric potential is defined based on the potential energy of a unit positive test charge. This gives us, immediately, the interpretation of cell potential. The cell potential $E = V_+ - V_-$ is the negative of the potential difference $V_- - V_+$. Thus, when conventional current flows from cathode to anode, the change in free energy is $\Delta G = -qE$. Let’s say we write the cell reaction in such a way that, every time it occurs once, $n$ electrons are transferred. Then, when a mole of this reaction occurs, the change in free energy is $\Delta G = -nN_0 eE$, where $e$ is the elementary charge and $N_0$ is Avogadro’s constant. We define Faraday’s constant $F = N_0 e$, so $\Delta G = -nFE$, an equation which will be familiar to you.

So, I said we can’t really measure the half-reaction reduction potentials $V_+ - V_e$ and $V_- - V_e$. But suppose we really did know those half-reaction reduction potentials. Then we could combine any two half-reactions and instantly calculate the cell potential. Indeed, we could calculate $\Delta G$ for any redox reaction by treating it as the sum of two half-reactions—reduction, which would occur at a cathode, and oxidation, which would occur at an anode; this computation would be valid even if we didn’t set up the reaction as a galvanic cell, because Hess’s law tells us that the $\Delta G$ for a transformation from reactants to products doesn’t depend on how the transformation was accomplished. Needless to say, then, knowing all the half-reaction reduction potentials would be really useful. Let’s say we know of $N$ different half-reactions. Then we could form $N^2$ combinations just by knowing their potentials ($N$ choices for the cathode and $N$ choices for the anode). We could tabulate $V_s - V_e$ values (where “s” stands for “solution”) for each half-cell, and then, once we put together two half-cells to make a cell, we’d figure out which one is the cathode, and which one is the anode, and take $V_s - V_e$ for the cathode and subtract $V_s - V_e$ for the anode. (From this procedure it is obvious how to figure out which one is the cathode and which one is the anode; you have to pick the half-cell with the greater reduction potential as the cathode, so that the difference will be positive and the cell will be spontaneous. Basically, this means the cathode is the half-cell that prefers being reduced the most, because, since it is the cathode, its solution is going to be reduced.) In this way we’d always end up with our desired $V_+ - V_-$ value. This saves us from having to measure the cell potentials of all $N^2$ reactions individually.

The very fact that cell potentials obey these relations, however—that they are given by the difference between the cathode reduction potential and the anode reduction potential, and that these half-potentials are independent of the half-cell’s partner—implies that cell potentials of different reactions are not all independent. And indeed you can see this just by examining overall cell reactions. For example, consider the following three full reactions, which are all spontaneous under standard conditions:

1. Cu2+ + Zn → Cu + Zn2+
2. Cu2+ + H2 → Cu + 2H+
3. Zn + 2H+ → Zn2+ + H2

Call the free energy changes $\Delta G_1^\circ = -2FE_1^\circ, \Delta G_2^\circ = -2FE_2^\circ, \Delta G_3^\circ = -2FE_3^\circ$, respectively. (Notice how two electrons are transferred in each reaction.) As written, reaction 1 is the sum of reaction 2 and reaction 3, so $\Delta G_1^\circ = \Delta G_2^\circ + \Delta G_3^\circ$, which implies $E_1^\circ = E_2^\circ + E_3^\circ$. Thus, the three cell potentials are not independent; once you know any two of the three, you also know the third. This is because the three cells only involve between them three different half-cell reactions:

1. Cu2+ + 2e → Cu
2. Zn2+ + 2e → Zn
3. 2H+ + 2e → H2

Reaction 1 has cathode (reduction) half-reaction (a) and anode (oxidation) half-reaction (b), so $E_1^\circ = E_a^\circ - E_b^\circ$. Reaction 2 has cathode half-reaction (a) and anode half-reaction (c), so $E_2^\circ = E_a^\circ - E_c^\circ$. And reaction 3 has cathode half-reaction (c) and anode half-reaction (b), so $E_3^\circ = E_c^\circ - E_b^\circ$. Clearly, $E_1^\circ = E_a^\circ - E_b^\circ = (E_a^\circ - E_c^\circ) + (E_c^\circ - E_b^\circ) = E_2^\circ + E_3^\circ$. So even though we might not know $E_a^\circ, E_b^\circ, E_c^\circ$ directly, we can derive relations between the full reaction potentials obtained from them.

The key is to, by convention, choose one half-reaction as a reference point. Because consider what happens when we know a bunch of cell potentials, all for cells having one half-reaction in common, for example, the following set (note: not all the following reactions are spontaneous, since in some of them the metal should be at the anode instead of the cathode):

• Mg2+ + H2 → Mg + 2H+
• Ca2+ + H2 → Ca + 2H+
• Fe2+ + H2 → Fe + 2H+
• Ni2+ + H2 → Ni + 2H+
• Cu2+ + H2 → Cu + 2H+
• Zn2+ + H2 → Zn + 2H+
• Sn2+ + H2 → Sn + 2H+
• Hg2+ + H2 → Hg + 2H+
• Pb2+ + H2 → Pb + 2H+

Because all these cells share a common half-reaction, we can take any two of the non-common half-reactions above and combine them, and immediately deduce the cell voltage. For example, E(Fe2+ + H2 → Fe + 2H+) – E(Zn2+ + H2 → Zn + 2H+) = E(Fe2+ + Zn → Fe + Zn2+)

Now we’re getting somewhere! We might not be able to measure the half-cell reaction potentials, Mg2+ + 2e → Mg, and so on, directly, but even still, we’ve found a set of $N$ reactions to study, such that once we’ve measured all their potentials, we’ll be able to assemble them to reconstruct the potentials of $N^2$ different reactions. The key was to use the a common half-reaction for all of them. This means that each of the reactions above is the difference of some reduction potential $E_i$ and a fixed reduction potential $E_0$, and when we subtract, the $E_0$ values cancel out. In fact, since the $E_0$ value can’t be measured directly, and since it always cancels out when we compute a measurable value (a full cell potential), why don’t we just set it to zero? This would make the cell potential for each of the above full reactions numerically equal to the other half-reaction potential, i.e., E(Mg2+ + 2e → Mg) = E(Mg2+ + H2 → Mg + 2H+), and so on.

And thus the standard reduction potential is born. The standard reduction potential for any given half-reaction is the cell potential when a cathode half-cell with that half-reaction under standard conditions is combined with the standard hydrogen electrode anode half-cell, in which the process H2 → 2H+ + 2e takes place. Note that the standard reduction potential might be negative for a half-reaction, if the left side is not as easily reduced as H+. In this case the theoretical galvanic cell used to define the half-reaction potential for this half-reaction also has a negative potential and is thus not spontaneous, but this is a mere technicality.

The setting of the standard hydrogen electrode reduction potential to zero is a simple example of gauge fixing. We had a degree of freedom in computing the standard reduction potentials, since, if we increase each of them by a fixed amount $k$, all their differences (and it’s only the differences, full cell potentials, that we measure in the lab) stay the same. We fixed the gauge by destroying the degree of freedom by making an arbitrary choice. Had we decided to set the standard hydrogen electrode reduction potential to +0.5291 V, for example, nothing observable would change. Even if we could measure the actual half-reaction reduction potential for the standard hydrogen electrode (and Wikipedia gives the value $4.44 \pm 0.02$ volts), knowing the actual value doesn’t give us any additional useful information for computing full cell voltages; it just makes our choice of gauge slightly less arbitrary.