So far we have cited the possibility of doing useful electrical work as the major reason for constructing galvanic cells. Now, we will define exactly what "useful electrical work" means and derive a relationship between work, free energy (G), and cell potential.

>From physics, we know that: Potential (E) = - Work (w) / Charge (q) therefore, w = -qE

Let us now define a quantity called a faraday (F) and let F equal the charge in coulombs per mole of electrons (96,485 C). Then q = nF and w= -nFE.

From thermodynamics we know that ΔG =
ΔU -TΔS +
Δ(PV) and U = heat + w.
Therefore, at constant T and P:
ΔG = w. Therefore:
ΔG = -nFE and at standard state:
ΔG^{o} = -nFE^{o}

Because the sign of ΔG predicts the direction of spontaneous reaction and G and E are directly related by the above equation, we can also use E to predict the direction of spontaneous reaction. As you know, if ΔG is greater than zero, the reaction is non-spontaneous but spontaneous in the reverse direction but if ΔG is less than zero, the reaction is spontaneous as written. ΔG and E have opposite signs from the above equation, therefore, spontaneous reactions will have positive cell potentials.

As alluded to in Galvanic Cells, Heading , E's are
intrinsic properties of reactions and therefore do not need to be multiplied by any factors when
computing the overall cell potential. However, when adding reduction potentials to calculate the
potential of a new reduction reaction there are additional mathematical complications. Those
complications arise because *potentials are not thermodynamic quantities.* According to Hess's
Law, only state functions (G, H, S) and not path functions (w, q, E) of a series of reactions may be
summed to generate a new value for the overall reaction. Because G is a state function and we have a
relationship between G and E (ΔG = -nFE) we can add the ΔG's of the reactions to compute the G of the
overall reaction which can then be converted into a value of E. For example, that you can compute the overall cell potential by adding the reduction and oxidation
potentials of the half-reactions.

Figure %: Proof of the Additivity of Oxidation and Reduction Potentials

Note that in the that the ΔG's were added together and then we solved for E^{o}
_{total} instead of simply adding the E^{o}'s. As that proof shows, if there are no "left-over" electr
ons in the overall balanced equation, then you can sum the potentials of the half-reactions.

However, if you are trying to add two reduction potentials to generate the
reduction potential of a new reaction, your balanced equation will have some
left-over electrons and you cannot simply add the two reduction potentials together. I will derive the formula for adding reduction or oxidation potentials together to generate a new half-reaction. I will use the reduction of Fe^{3+
} to Fe metal as my example. Be sure to understand the important conclusion that when summing reduction potentials, E^{o}
_{total} does not equal the sum of the individual
E^{o}'s.

Figure %: Derivation of the Rule for Adding Reduction Potentials

So far in our discussion of electrochemical cells, we have
only considered reactions at "standard state" which are, in reality,
impossible to achieve. The moment you hook up the wire connecting two
half-cells the reaction proceeds and changes the concentrations of
all reactants and products. Furthermore, if the reaction is exothermic
or endothermic, the reaction mixture will heat or cool making it
deviate from the standard temperature. Therefore, we need a way to
relate E^{o} at the standard conditions and E, the potential at any real condition. That
relationship, called the Nernst Equation, was first derived by Walther
Nernst and earned him the 1920 Nobel Prize in chemistry. The
is found below:

Figure %: Derivation of the Nernst Equation

Please note that the familiar form of the Nernst Equation is
only applicable when the reaction is carried out at 25^{o}C
(298^{o}K). At any other temperature you need to use the first
form of the Nernst Equation: E = E^{o} - (RT/nF) ln Q. One caveat in using the Nernst Equation: Q is the reaction
quotient, so you must have already balanced the redox reaction to be
able to place the correct power on each concentration term in Q. Make
sure you use consistent units for both R and T!

As you can tell by inspection of the Nernst equation the cell potential depends on concentration. In fact, the equation implies directly that you can construct a galvanic cell with half-cells of identical composition but differing concentrations--a concentration cell. As is intuitively obvious from our knowledge of osmotic pressure a concentration cell reacts in such a manner as to dilute the more concentrated half-cell and to concentrate the more dilute half-cell as shown in .

Figure %: Depiction of a Concentration Cell

As shown in the , the dilution of the cathode half-cell is
achieved by reducing Cu^{2+} to Cu metal and plating that metal onto the Cu
electrode. In the anode half-cell, the Cu anode is oxidized to Cu^{2+}
and, thus, dissolved into the solution, making the anode cell more concentrated.

Let's calculate the cell potential for the concentrations listed in (0.01M and 1M) at 25^{o}C:

The Nernst Equation tells us that:

E = E^{o} - (0.0592 / n) log Q

and we know that

E^{o} = 0 for any concentration cell so: