Cisplatin
3. Redox Chemistry

The transfer of electrons from one species to another is called oxidation-reduction chemistry, or redox chemistry. Oxidation refers to a loss of electrons, whereas reduction refers to a gain of electrons.
When discussing redox chemistry, it is useful to assign oxidation numbers to elements undergoing a transfer of electrons. The use of oxidation numbers is a bookkeeping method that chemists use to keep track of electrons. The oxidation number is defined so that oxidation corresponds to an increase in oxidation number, and reduction corresponds to a decrease in oxidation number. Another way to think about redox chemistry and oxidation numbers is that in redox chemistry we exaggerate the ionic character of bonds; in other words, redox chemistry is based on a representation in which the atoms are pictured as ions. Assigning oxidation numbers requires following three rules:
1. The oxidation number of an element in its elemental form is 0. (For example, the oxidation number of Na is 0, and the oxidation number of Cl2 is also 0.) 2. The oxidation number of an element when it exists as a monatomic ion is the same as its charge. (For example, the oxidation number of Na+ is +1, and the oxidation number of Cl- is -1) 3. The sum of the oxidation numbers of all the atoms in a species is equal to its total charge. (For example, the sum of the oxidation numbers in SO42- is -2.)
In addition, there are a few elements that have specific values for their oxidation numbers:
The oxidation number of hydrogen is +1 when bonded to a nonmetal and -1 when bonded to a metal. The reason for this is that hydrogen is less electronegative than nonmetals. When we treat a hydrogen-nonmetal bond as an ionic bond, hydrogen donates its electron to the nonmetal, resulting in a +1 oxidation number for H. For example, in a molecule of water, two hydrogen atoms are bonded to an oxygen atom (a nonmetal that is more electronegative than hydrogen). H has an oxidation number of +1, and O has an oxidation number of -2. (To restate rule #3 above: the sum of the oxidation numbers in a neutral molecule is 0.) The oxidation number of hydrogen atoms in metal-hydrogen bonds can also be explained by the electronegativity differences between H and M. Hydrogen is more electronegative than metals. Therefore, in an ionic treatment of a metal-hydrogen bond, the metal atom donates an electron to the hydrogen atom, resulting in a -1 oxidation number for H. For example, in sodium hydride (NaH), sodium has an oxidation state of +1, whereas hydrogen has an oxidation state of -1. The oxidation number of oxygen is almost always -2. This can also be explained on the basis of electronegativity. Oxygen is the second most electronegative element—after fluorine. Therefore, in an ionic treatment of a bond containing oxygen and another element—with the exception of fluorine—oxygen will accept two electrons and take on a -2 oxidation number. For example, in sulfate ion (SO42-), each oxygen has an oxidation number of -2, whereas sulfur has an oxidation number of +6. (Again, see rule #3 above: total charge on ion = -2 = 4 x [oxid. no. of O] + [oxid. no. of S].) The oxidation number of fluorine is always -1. The oxidation of the other halogens is -1 unless the halogen is bonded to oxygen or another halogen higher in the group. Again, these values can be explained based on electronegativity differences. Fluorine is the most electronegative element of all. Using an ionic model, fluorine always accepts an electron from an atom it is bonded to, giving fluorine both a -1 charge and a -1 oxidation number. Other halogen atoms will also have a -1 oxidation state unless they are bonded to a more electronegative atom, such as oxygen or another halogen higher in the group, such as fluorine. For example, in the molecule FI, fluorine has a -1 oxidation number, whereas iodine has a +1 oxidation state.
The species that causes oxidation is called the oxidizing agent. It does this by accepting electron(s) from the species being oxidized. In other words, the oxidizing agent is reduced. According to our bookkeeping method of calculating oxidation numbers, the oxidizing agent contains an atom that undergoes a decrease in oxidation number. Likewise, the species that causes reduction is called the reducing agent. The reducing agent donates electron(s) to the species being reduced. Therefore, the reducing agent is itself oxidized and undergoes an increase in oxidation number. For example, in the following redox reaction—which occurs during the potassium iodide-starch test described in the control experiments module—a metal with an oxidation state of +2 reacts with iodide ion. Let us say that the metal is platinum, and let’s write the oxidation numbers below each reactant and product:

Pt2+ + 2 I- Pt0 + I2

+2       2 x -1      0     0

In this case, Pt2+ (also called Pt(II)) reacts with iodide ion to form Pt0 (or Pt(0)) and iodine. Using our oxidation number bookkeeping method, we see that the oxidation number of platinum decreases by 2; therefore, platinum undergoes a two electron reduction. Pt(II) acts as the oxidizing agent by oxidizing iodide ion to iodine. Two iodide ions each undergo a one electron oxidation to give one molecule of I2, resulting in a two electron oxidation overall. Likewise, the two iodide ions act as the reducing agents because they reduce Pt(II) to Pt(0). Just as with all other chemical reactions, redox equations must balance. There are two rules to keep in mind when balancing redox equations:
1. Atoms must balance. (In the example above, there is one Pt atom and two I atoms on each side of the reaction; therefore the atoms balance.)

2. Total charge on each side must balance. (In the example above, the charge on the left is equal to [+2 + 2(-1)], or 0 overall. Likewise, the charge on the right is equal to [0 + 0], or 0 overall. Therefore, the equation is also balanced with respect to charge.)

    (1) Atkins, P. W., Jones, L. L. Chemistry: Molecules, Matter, and Change; 3rd ed. W. H. Freeman and Company: New York, 1997, Chapter 3: Sections 3.12 to 3.15.


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