|Electrochemistry, which is the study of the interaction of electricity and chemical reactions, is a central theme in the story of the discovery of cisplatin. In the redox chemistry module, we saw how electrons are transferred from one species to another in a chemical reaction, and we also learned some of the formalisms for describing these types of reactions. Now we will take these concepts a step further and discuss how various chemical species vary in their ability to "pull" electronsthat is, to be reduced. The implications of the different pulling powers of various species are far reaching. As a result of this phenomenon, we are able to power batteries, produce aluminum, extract metals from their salts, and protect metals through electroplating. We can also gain a more complete understanding of the serendipitous discovery of cisplatin.|
Pt0(s) ® Pt2+(aq) + 2 e-Chemists can determine whether this oxidation reaction is spontaneous and if so, how much energy is released. Likewise, they can determine whether the reaction is nonspontaneous and how much energy is required to make the reaction proceed. Knowing the amount of energy released or required for various oxidation and reduction reactions gives us an idea of the ability these reactions (called cell reactions) have to push or pull electrons through a circuit; the measure of this ability is called the cell potential. The cell potential is a useful value when it can be compared to other cell potentials obtained under the same standard set of conditionsthat is, when all participating gases exist at a pressure of 1 atm, and all ions are present in a concentration of 1 mol L-1. A cell potential under standard conditions is called a standard cell potential, or a standard electrode potential, E°. Standard cell potentials are reported only for reduction reactions; for this reason, they are also called standard reduction potentials. Standard electrode potentials are all measured in reference to the hydrogen electrodes standard potential, which is arbitrarily set at zero. If another species oxidizes a molecule of H2 to H+ ions, the other species is itself reduced and has a positive standard reduction potential. On the other hand, if H+ ions oxidize another species, that species has a negative standard reduction potential. Most general chemistry textbooks have tables of standard cell potential values. When we look up the standard cell potential for the reduction of Pt2+ to Pt0 (the opposite of the oxidation reaction written above), we see that this reaction has a positive value for E°, meaning that Pt2+ oxidizes H2 to H+ under standard conditions.
Pt2+(aq) + 2 e- ® Pt0(s) E° = +1.20 VAs stated above, we are really more interested in the reverse reaction, the oxidation of Pt0 to Pt2+. The standard cell potential for the oxidation reaction has the same value but the opposite sign as that for the reduction reaction:
Pt0(s) ® Pt2+(aq) + 2 e- E° = -1.20 VNow lets take a look at what these cell potential values mean. We will see that the cell potential is related both to the reaction free energy, which tells us whether or not the reaction is spontaneous, and to the equilibrium constant for the reaction, which tells us the extent of the reaction at equilibrium. The standard cell potential is related to the standard reaction free energy through the following equation: DG° = -nFE° Here, n is the number of moles of electrons participating in the reaction. F is the Faraday constant, which is the magnitude of charge per mole of electrons, and is equal to 9.6485 x 104 C mol-1. We can now calculate the standard free energy for the oxidation reaction written above: DG° = -nFE° = -2 x (9.6485 x 104 C mol-1) x (-1.20 V)
= +231,564 C.V mol-1
= +231,564 J mol-1
= +232 kJ mol-1When the reaction free energy is positive (and the cell potential is negative), the reaction is nonspontaneous in the direction written. Therefore, the oxidation of Pt0 to Pt2+ is nonspontaneous; energy must be supplied in order for the reaction to proceed. The energy required for the oxidation of Pt0 (from the platinum electrodes) can be supplied in the form of an electric current; such a current was applied to the continuous culture chamber containing E. coli bacteria. The standard cell potential is also related to the equilibrium constant, K, for the redox reaction: DG° = -RTlnK Here, R is the gas constant, 8.31451 J K-1 mol-1, and T is the temperature. Given this expression and the relationship above between reaction free energy and cell potential, we can relate the cell potential directly to the equilibrium constant: DG° = -RTlnK = -nFE°
lnK = nFE°/RT
K = e(nFE°/RT)Therefore, at 25° C (298.15 K), the equilibrium constant for the oxidation of Pt0 to Pt2+ is calculated to be the following:
K = e(nFE°/RT) = e-93.41
= 2.70 x 10-41As this calculation shows, when Pt0 is oxidized to Pt2+, the reaction strongly favors the reactantsunless energy is supplied to the reaction. Now that we have seen how Pt2+ can be generated under the reaction conditions described in the electric fields module, we can understand the control experiment in which the potassium iodide-starch test was used as a way to detect the presence of an oxidizing agent. We saw in the control experiments and redox chemistry modules that in this test, an oxidizing agent (here, Pt2+) reacts with iodide ions to form the reduced, neutral metal and elemental iodine:
Pt2+(aq) + 2 I-(aq) ® Pt0(s) + I2(s)One way to see the electron transfer more clearly is to break this redox reaction into two half-reactions:
2 I-(aq)® I2(s) + 2 e-
Pt2+(aq) + 2 e- ® Pt0(s)The two iodide ions are each losing one electronin a net two-electron oxidation. As we learned in the redox chemistry module, I- is the reducing agent. Likewise, Pt2+ is gaining two electrons and is reduced, so Pt2+ is the oxidizing agent. Using half-reactions is a convenient way to keep track what is going on in a redox reaction. However, it is important to realize that the electrons are not ever actually free in solution, as the half-reactions suggest; rather, they are in transit between the reducing agent and the oxidizing agent. Now, if we add these two half-reactions together, we see that we obtain the original balanced redox equation. Rules for balancing simple redox equations are given in the redox chemistry module.
2 I-(aq) ® I2(s) + 2 e-
Pt2+((aq) + 2 e- ® Pt0(s)
Pt2+(aq) + 2 I-(aq) ® Pt0(s) + I2(s)
2 I-(aq) ® I2(s) + 2 e- E° = -0.54 V
Pt2+(aq) + 2 e- ® Pt0(s) E° = +1.20 V
Pt2+(aq) + 2 I-(aq) ® Pt0(s) + I2(s) E° = + 0.66 V
(1) Atkins, P. W., Jones, L. L. Chemistry: Molecules, Matter, and Change, 3rd ed. W. H. Freeman and Company: New York, 1997, Chapter 17.