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: ΔGo = -nFEo
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.
Note that in the that the ΔG's were added together and then we solved for Eo total instead of simply adding the Eo'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 Fe3+ to Fe metal as my example. Be sure to understand the important conclusion that when summing reduction potentials, Eo total does not equal the sum of the individual Eo's.
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 Eo 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:
Please note that the familiar form of the Nernst Equation is only applicable when the reaction is carried out at 25oC (298oK). At any other temperature you need to use the first form of the Nernst Equation: E = Eo - (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 .
As shown in the , the dilution of the cathode half-cell is achieved by reducing Cu2+ to Cu metal and plating that metal onto the Cu electrode. In the anode half-cell, the Cu anode is oxidized to Cu2+ 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 25oC:
The Nernst Equation tells us that:
E = Eo - (0.0592 / n) log Q
and we know that
Eo = 0 for any concentration cell so: