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Cisplatin 4. Electrochemistry

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    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" electrons—that 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.

    In the modules on control experiments and the role of platinum electrodes (as well as the discovery of cisplatin), we saw that platinum-containing electrolysis products were responsible for causing both elongation of bacterial cells and regression of murine tumors. We now discuss the origin of these electrolysis products. Electrolysis occurs when an electric current from an external source is used to bring about a nonspontaneous chemical reaction. The researchers found that when they applied an electric current through platinum electrodes in a continuous culture chamber containing E. coli bacteria, a reaction occurred in which the supposedly inert platinum metal (Pt0) of the electrodes was oxidized to form both platinum(II) (Pt2+) and platinum(IV) (Pt4+). Complexes containing oxidized forms of platinum—particularly Pt2+—were found to be the causative agents in bacterial elongation. Let us now examine the oxidation reaction in which Pt0 is converted into Pt2+:

    \[Pt^0_{(s)} → Pt^{2+}_{(aq)} + 2 e^- \nonumber \]

    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 conditions—that 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, . 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 electrode’s 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 , meaning that Pt2+ oxidizes H2 to H+ under standard conditions.

    \[Pt^{2+}_{(aq)} + 2 e^- → Pt^0_{(s)}\;\;\;\; E^o = +1.20\; V \nonumber \]

    As 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:

    \[Pt^0_{(s)} → Pt^{2+}_{(aq)} + 2 e^-\;\;\;\; E^o = -1.20\; V \nonumber \]

    Now let’s 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:

    \[ΔG° = -nFE^o \nonumber \]

    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:

    \[ΔG^o = -nFE^o = -2 \times (9.6485 \times 10^4\, C\, mol^{-1}) \times (-1.20\; V) \nonumber \]

    \[= +231,564\; C \cdot V\; mol^{-1} \nonumber \]

    \[= +231,564 \;J \;mol^{-1} \nonumber \]

    \[= +232 \;kJ \;mol^{-1} \nonumber \]

    When 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:

    \[ΔG^o = -RT\ln K \nonumber \]

    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:

    \[ΔG^o = -RT\ ln K = -nFE^o \nonumber \]

    \[\ln K = \dfrac{nFE^o}{RT} \nonumber \]

    \[K = e^{(\frac{nFE^o}{RT})} \nonumber \]

    Therefore, at 25° C (298.15 K), the equilibrium constant for the oxidation of Pt0 to Pt2+ is calculated to be the following:

    \[K = e^{(\frac{nFE^o}{RT})} = e^{-93.41} \nonumber \]

    \[= 2.70 \times 10^{-41} \nonumber \]

    As this calculation shows, when Pt0 is oxidized to Pt2+, the reaction strongly favors the reactants—unless 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:

    \[Pt^{2+}_{(aq)} + 2 I^-_{(aq)} → Pt^0_{(s)} + I^2_{(s)} \nonumber \]

    One way to see the electron transfer more clearly is to break this redox reaction into two half-reactions:

    \[2 I^-_{(aq)} → I^2_{(s) }+ 2 e^- \nonumber \]

    \[Pt^{2+}_{(aq)} + 2 e^- → Pt^0_{(s)} \nonumber \]

    The two iodide ions are each losing one electron—in 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)

    Just as we added together the two half-reactions to obtain a balanced redox equation, we can add together the two standard cell potentials to obtain the net potential.

    2 I-(aq) I2(s) + 2 e- = -0.54 V

    Pt2+(aq) + 2 e- Pt0(s) = +1.20 V
    Pt2+(aq) + 2 I-(aq)
    Pt0(s) + I2(s) = + 0.66 V

    We see that the net potential for the redox reaction occurring in the potassium iodide-starch test is a positive number; this will lead to a negative value for the reaction free energy, meaning that the reaction is spontaneous as written. This means that Pt2+ is a strong enough oxidizing agent to effect the oxidation of iodide ions, explaining why Pt2+gives a positive result in the potassium iodide-starch test. In summary, we have seen how Pt2+ is generated when an electric current is applied to the platinum electrodes in the continuous culture chamber. We have also seen how the presence of Pt2+ gives a positive result when subjected to the potassium iodide-starch test. We now have the pieces of the puzzle to understand how platinum-containing complexes such as cisplatin are generated under the reaction conditions. We will do this in the transition metal chemistry module.


    1. Atkins, P. W., Jones, L. L. Chemistry: Molecules, Matter, and Change, 3rd ed. W. H. Freeman and Company: New York, 1997, Chapter 17.

    This page titled Cisplatin 4. Electrochemistry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by ChemCases.

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