Galvanic Cells and Free Energy

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 15:36


We see in the section on the Electromotive Force of Galvanic Cells that the emf of a galvanic cell can tell us whether the cell reaction is spontaneous. In other sections we show that the free-energy change ΔG of a chemical process also indicates whether that process is spontaneous. It is quite reasonable, then, to expect some relationship between ΔG and E, and indeed one exists.

In the section on Free Energy we stated that the free-energy change corresponds to the maximum quantity of useful workA mechanical process in which energy is transferred to or from an object, changing the state of motion of the object. which can be obtained when a chemical reactionA process in which one or more substances, the reactant or reactants, change into one or more different substances, the products; chemical change involves rearrangement, combination, or separation of atoms. Also called chemical change. occurs. In other words,


ΔG = –wmax


where the minus sign is necessary because the free energy decreases as the chemical system does useful work on its surroundings. If we are referring to a redox reaction, that work can be obtained in electrical form by means of an appropriate galvanic cell. It can be measured readily, because when a quantity of charge Q moves through a potential difference ΔV, the work done is given by


w = Q ΔV


Thus if one coulomb passes through a potential difference of one volt, the work done is


w = 1 C × 1 V = 1 A s × 1 J A–1 s–1 = 1 J


Now suppose we construct a Zn-Cu cell of the type described earlier:


Zn│Zn2+ (1 M)║ Cu2+(1 M)│Cu


and suppose we make the cell large enough that the concentrations of Cu2+ and Zn2+ will not change significantly even though 1 mol Zn is oxidized to 1 mol Zn2+ according to the cell reaction

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)      (1)


If this cell is discharged through a large enough resistance, the potential difference will have its maximum value, namely, the cell emf, E°; so if we know how much charge is transferred, we can calculate the electrical work done. For the oxidationThat part of a chemical reaction in which a reactant loses electrons; simultaneous reduction of a reactant must occur. of 1 mol Zn [that is, for the occurrence of 1 mol of reaction (1)], there must be 2 mol e transferred according to the half-equation


Zn(s) → Zn2+(aq) + 2e


Therefore the quantity of electrical charge transferred per mole of reaction is


Qm = 2 × F = 2 × 9.649 × 104 C mol–1 = 1.930 × 105 C mol–1


(The Faraday constantThe electric charge carried by one mole of electrons, 9.648 670 x 104 C mol-1; abbreviated F., F, is the quantity of charge per mole of electrons. It has the value 96,485 C/mol.) The maximum useful work per mole of reaction which the cell can perform while discharging is thus


wmax = QmE° = 2FE° = 2 × 9.649 × 104 C mol–1 × 1.10 V = 212 kJ mol–1


The standard molar free energy change for the cell reaction is thus


ΔGm° = –wmax = –2FE° = –212 kJ mol–1


A similar argument can be applied to any cell in which the reactants and products are all at their standard concentrations or pressures. If the standard emf of such a cell is E°, while ΔGm° is the standard molar free energy change for the cell reaction,these two quantities are related by the equation


ΔGm° = – zFE°      (2)


where z (a dimensionless number) corresponds to the number of moles of electrons transferred per mole of cell reaction.


A similar relationship holds even when reactants and products are not at standard concentrations and pressure:


ΔGm = – zFE


This connection between cell emf and free-energy change provides a means of measuring ΔGm, directly, rather than by determining ΔHm, and ΔSm, and then combining them.



EXAMPLE 1 The emf of the cell


Pt, H2(1 atmAbbreviation for atmosphere, a unit of pressure equal to 101.325 kPa or 760 mmHg.)│H+ (1 M)│O2(1 atm), Pt


is 1.229 V at 298.15 K. Calculate ΔGf° for liquidA state of matter in which the atomic-scale particles remain close together but are able to change their positions so that the matter takes the shape of its container water at this temperatureA physical property that indicates whether one object can transfer thermal energy to another object.


SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. The half-equations for the cell are


                      2H2(g) → 4H+(aq) + 4e

4e + 4H+(aq) + O2(g) → 2H2O(l)


so that the cell reaction is


2H2(g) + O2(g) → 2H2O(l)      1 atm, 298.15 K


Since there are 4 mol e transferred per mol cell reaction, z = 4 and


ΔGm = – zFE = – 4 × \frac{\text{9.649 }\times \text{ 10}^{\text{4}}\text{ C}}{\text{1 mol}} × 1.229 V = – 474.3 kJ mol–1


The reaction produces liquid water at standard pressure and the desired temperature, but 1 mol reaction produces 1 mol 2H2O(l), that is, 2 mol H2O. Therefore


ΔGf°[H2O(l), 298 K] = ½ΔGm = – 237.2 kJ mol–1