Units of Free Energy and Electrochemical Cell

Units of free energy and electrochemical cell potentials "The devil is in the details." Units of DG for chemical transformations There is a subtlety about the relation between DG and K, namely how DG changes when the amounts of reactants and products change. Gibbs free energy G= H -T S is an extensive quantity, that is, it depends on how much of the system we have, since H and S are both extensive (T is intensive). This means if we double the reactants and products, 2aAV2bB then DG must also double, but the expression DG = R T lnHQ ê KL found in textbooks does not seem to contain the amounts of reactants and products! In fact it does contain these, because the stoichiometric coefficients appear as exponents in Q and K. Thereby, doubling the moles of reactants and products will change Q and K into their square, which will double DG. Here is how. DGHdoubledL = R T lnHQ2 ê K 2 L = R T lnHQ ê KL2 = 2 R T lnHQ ê KL = 2 DGHoriginalL This result evidently means that the units of DG are energy, not energy per mole, but with the understanding that it is the change in Gibbs free energy per mole of reaction as written. The question arises, why is this not taken account by the explicit units in the equation for DG? The answer is that the appropriate unit has been omitted. Here is how to see what is going on. The connection between DG, Q, and K comes from equalities like a R T lnH1 atm ê pA L = R T lnH1 atm ê pA La 2 Units of free energy and electrochemical cell potentials Now, we know that exponents must be dimensionless, so such an equality means that the exponent a is the numerical value of the stoichiometric coefficient, without any units. The units of the stoichiometric coefficient remain behind, multiplying the gas constant R. Strictly, therefore, we should make this explicit by writing a R T lnH1 atm ê pA L = mol R T lnH1 atm ê pA Laêmol which then leads to the explicitly correct result that DG = mol R T lnHQ ê KL. The problem is, this is never done in textbooks. The bottom line is that when we see the relation DG = R T lnHQ ê KL, we just need to understand that (1) exponents are pure numbers, and (2) the units are energy for the reaction, not energy per mole. Units of electrochemical cell potentials Proper treatment of mol units in the relation between DG, Q, and K carries over into the Nernst equation. Here is how to see this. The Nernst equation is based on the alternative expressions for free energy change, DG = mol R T lnHQ ê KL and DG = -n - , Setting these equal, and using the special cases of these expressions that DG° = - mol R T lnHKL and DG° = -n - ,° the result is mol R T , = ,° - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ lnHQL nThis form of the Nernst equation differs from the form in standard textbooks by the explicit mol unit in the numerator of the last term. That this unit must be present can be seen by considering the units in the factor multiplying lnHQL. mol R T ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ n- mol Joule ê HKelvin molL Kelvin = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Hmol of electronsL HCoulomb ê mol of electronsL Joule = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Coulomb = Volt Units of free energy and electrochemical cell potentials That is, by having properly accounted for mol units in the expression for free energy, we achieve proper units in each term of the Nernst equation, and so in particular the Q-dependent term of the Nernst equation is an explicitly intensive quantity. 3