Electrochemistry, Free Energy, and Work

The most amazing thing about electrochemical cells is that by separating the oxidation and reduction reaction into different compartments, we now have the ability to directly measure the free energy of the reaction by measuring the electrical potential.

The potential or voltage or EMF or whatever you prefer to call it is a measure of the free energy of the reaction.  This is because the difference in free energy between the reactant and the products directly corresponds to the electrical potential of the cell.

When looking at chemical reactions, we often talked about the difference in free energy between the reactants and products.  This was either \(\Delta G\) at whatever conditions existed currently or if it was given at standard conditions, we had \(\Delta G^\circ \).

Previously, we discussed this in terms only of spontaneity.  However, \(\Delta G\) has another interpretation.  It corresponds to the maximum reversible work we can extract from this chemical reaction.  While previously you may have dealt with the concept of "work" in the context of pressure-volume work, the "work" we are interested in now is electrical work.

Electrical work is defined as charge times potential (\(w = q \cdot \mathcal{E}\) ).   This will give us an energy.  The total work would be the charge that is run through the cell (coulombs, C) times the potential (volts, V) of the cell.  Note that a coulomb·volt is equivalent to a joule (1 V·C = 1 J). For a reaction, we are typically interested in the work (or energy) per mole of reaction.  This is how we defined \(\Delta G\) with units of kJ mol^-1.  Looking at this, we get

\[{\rm work} = \Delta G = -n F \mathcal{E}\]

Here \(\Delta G\) is not for the standard conditions, but for the conditions now.  So the potential \(E\) is not the standard potential but the potential now.  \(n\) is the number of electrons per mole reaction as the equation is written.  \(F\) converts between mole of electrons and charge in coulombs.  Note: the energy you get is still per mole, but it is per mole reaction (not per mole of electrons).

With this relation, we can now convert directly between differences in free energy and differences in electrical potential!

Standard Potentials and K

Since we can relate free energy differences and electrical potentials at any conditions, then we can relate them at standard conditions.

\[\Delta G^{\circ} = -nFE^{\circ}\]

Since ΔG° is related to the equilibrium constant K, we can now relate K and E°.

\[\Delta G^{\circ} = -RT\ln K\]

\[E^\circ = {RT \over n F}\ln K\]

Assuming a temperature of 298.15K and converting from ln to log₁₀ we get

\[E^\circ = {0.05916 \over n}\log K \]

As such ΔG°, K, and E° are all different ways to state the same concept.

ΔG° < 0 matches K > 1, which matches E° > 0 and they all mean that the process is "spontaneous" as written and the products are favored.

ΔG° > 0 matches K < 1, which matches E° < 0 and they all mean that the process is "non-spontaneous" as written and the reactants are favored.

Standard Potentials and the Solubility Product, K_sp

Back in Unit 5 we learned that the solubility of a salt in a saturated solution can be characterized via the solubility product, K_sp. These reactions are really about the extent of reaction of dissociation and dissolution. Consider the solubility of silver chloride, AgCl:

AgCl(s)    ⇌    Ag⁺(aq)    +    Cl⁻(aq)

The overall reaction here is not a redox reaction. The silver is in the +1 state on both sides of the equation and the chlorine is in the -1 state (chloride) on both sides. However, this reaction can, in fact, be split into two half-reactions where one is a reduction and the other an oxidation. It is the metal cation in the salt that is reduced to the metal (cathode) and then the metal is oxidized back to the metal cation (anode). Because it is the same elemental metal that is oxidized and reduced, the net reaction appears to not be a redox reaction. Below are the two half-reactions for finding the K_sp for AgCl.

Notice that these two reactions combine to give the familiar solubility product reaction shown previously. The potentials were directly added to get the net reaction potential because the sign for the oxidation reaction has already been taken into account (-0.799 V is the negative or "flipped" version of what you find on a standard potential table). We also know that the equilibrium constant for any reaction can be calculated if either ΔG° or E° is known. Here, we now know E° for the reaction and use:

\[K = \exp\left({nFE^\circ \over RT}\right)\]

Remember that the "exp" function is the exponential function, e^x. Now we plug in the known standard values and calculate K which in this case is actually K_sp.

\[K_{\rm sp} = \exp\left({1(96485)(-.576) \over 8.314(298.15)}\right)\] \[K_{\rm sp} = \exp\left(-22.42\right)\] \[K_{\rm sp} = 1.8\times 10^{-10}\]

By manipulating the correct set of half-reactions, you can find the standard potential for numerous chemical equations for solubility. This is in fact the method used for the determination of many solubility products.

Nernst Equation

What if I'm not at standard conditions?

Imagine you have a voltaic cell at standard conditions to start.  Now, you let the chemistry start by wiring up the cell and allowing the electrons to flow from anode to cathode.  As the chemistry proceeds, the concentrations are changing and the potential is changing.  The reaction will "go" until the system reaches equilibrium.  At this point, the potential will be zero.  Clearly the potential must change as the concentrations are changing.  How can we determine the potential under any concentration conditions not just standard conditions?

We can use our free energy relations.  Before, we could figure out if a reaction was at equilibrium by comparing the ratios of the concentrations "now" to what they would be at equilibrium.  This was a comparison between the mass action expression now Q and what it is at equilibrium K.   Thinking back to free energy and equilibrium we can remember that

\[\Delta G = \Delta G^{\circ} +RT\ln Q\]

If we put in our relationships for ΔG and E we get

\[E = E^{\circ} - {RT \over nF}\ln Q\]

This equation is called the Nernst Equation. The last term (\(RT/nF \ln Q\))can be converted to a more "user friendly" form by calculating for the constants R, F, and T (assuming 298.15K) and then converting from natural log (ln) to log base 10 (log). That means using the mathematical relation: ln x = (ln 10)(log x) = (2.303)(log x)

\[E = E^{\circ} - {(8.314)(298.15) \over n (96485)}(2.303)\log Q\]

Which gives our "nicer" version of the Nernst Equation:

\[E = E^{\circ} - {0.05916 \over n}\log Q\]

Why do we need or even want to switch to "log" instead of "ln"? There is no other reason that for our human preferences of thinking better and faster in multiples of 10 as opposed to multiples of e which is 2.7182818... (not exactly the multiple most of us humans think of).

Calculating a potential

The Nernst equation can be used to calculate the potential of an electrochemical cell under any conditions.  Let's look at an example of this using the very simple cell

\[\rm{ Zn(s) \;\bigl| \;Zn^{2+}(aq) \;\bigl|\bigl| \;Cu^{2+}(aq) \;\bigl| \;Cu(s) }\]

The balanced equation for this cell is

\[\rm{Zn(s)\; + \;Cu^{2+}(aq)\; \rightarrow\; Zn^{2+}(aq)\; + \;Cu(s)}\]

For this reaction, we can look up the standard half reduction potentials and calculate the standard potential E°.  In this case, you find that E° = 1.100 V.

Now the question is, what is the potential under other conditions.  For this let's look at an example where we [Zn²⁺] = 0.1 M and [Cu²⁺] = 0.001 M.

To find the potential, we need to use the Nernst equation

\[\mathcal{E} = \mathcal{E}^{\circ} - {0.05916 \over n}\log Q\]

Here n=2 since the equation as written has 2 electrons per mole reaction.  Now we need Q.

\[Q = {\rm [Zn^{2+}] \over [Cu^{2+}]} = {0.1 \over 0.001} = 100\]

Plugging this into the Nerst Equation with the standard potential, we get

\[\eqalign{ \mathcal{E} &= +1.100 - {0.05916 \over 2}\log 100 \cr &= +1.100 - 0.05916 \cr &= +1.041\;{\rm V} \cr }\]

Since the conditions are closer to equilibrium (more Zn(II) ion than Cu(II) ion) than the standard conditions the potential is lower.

You should be able to manipulate the Nernst equation in any way.  Given a potential and standard potential, you should be able to determine ratios of concentrations or an unknown concentration.   Given concentration and standard potential, you should be able to find the potential, etc...

Concentration Cells

It turns out (and is quite easy to confirm via calculations) that you can generate real potential voltage just from the concentration difference of any chemical species that is part of a half reaction. Let's have an example with the Zn/Zn²⁺ half reaction.

Step 1: Build a cell with identical half reactions. For this example we are using the Zn/Zn²⁺ half reaction. Here is the resulting cell notation for this set up:

\[\rm{ Zn(s) \;\bigl|\; Zn^{2+}(aq) \;\bigl|\bigr| \;Zn^{2+}(aq)\; \bigr|\; Zn(s) }\]

The rather obvious standard cell potential is 0.00 V because this is the same reaction and same exact conditions in the anode compartment and in the cathode compartment. The standard potential for the Zn/Zn²⁺ couple is -0.76 V. The "calculation" of the standard cell potential is shown here:

\[E^\circ_{\rm cell} = E^\circ_{\rm cathode}-E^\circ_{\rm anode} = -0.76 - (-0.76) = 0.00\;{\rm V}\]

Realize that 0.00 V is always the standard cell potential when the same half reaction is used for both reduction and oxidation. Even the overall reaction looks like nothing is happening:

\[\rm Zn(s) + Zn^{2+} \longrightarrow Zn(s) + Zn^{2+}\]

Nothing does happen if and only if the two cells are identical. Notice that Q for this reaction is equal to 1.

\[Q = {\rm [Zn^{2+}]_{anode}\over [Zn^{2+}]_{cathode}} = {1.0\over 1.0} = 1 \]

Step 2: Now lets change the conditions so that the two half cells are not identical anymore. How? By changing the concentrations of the aqueous species in the cells so that they are different. This will make each of the cells have a different non-standard potential and this results in a net potential to drive electrons from one cell to the other.

New concentrations: Let's now have the anode compartment have a Zn²⁺ concentration of only 0.001 M and we will leave the cathode compartment Zn²⁺ at standard conditions of 1 M. Let's calculate the new cell potential, starting with the new value of Q:

\[Q = {\rm [Zn^{2+}]_{anode}\over [Zn^{2+}]_{cathode}} = {0.001\over 1.0} = 0.001 \]

Step 3: Now the cell potential can be calculated with the Nernst equation:

\[\eqalign{ E &= 0 - {0.05916 \over 2}\log 0.001 \cr &= 0 - 0.02958(-3) \cr &= +0.08874\;{\rm V} \cr }\]

This is 89 mV of potential created by simply having a concentration difference. This is important because gaining a potential through concentration difference is a key concept in developing a membrane potential in biology. A cell has a mechanism in place to help keep a difference in concentration of K⁺ inside and outside the cell membrane. Just a small difference in concentration develops a voltage that can be utilized by the cell. A basic knowledge of electrochemistry allows one to easily calculate the value of this membrane potential in the same exact way that you calculate the potential of any concentration cell.

© 2013 mccord/vandenbout/labrake

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