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"Valence and Molecular Structure," Lecture 3.

"Valence and Molecular Structure," Lecture 3. 1957.
Produced for the Institutes Program of the National Science Foundation. Robert and Jane Chapin, producers.

Lecture 3, Part 6. (6:42)

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Linus Pauling: Now, the assignment of oxidation numbers to the atoms in a molecule or complex ion, having been made, we may go ahead with the job of balancing the equation for a reaction.

Let us take, well I’ve been talking about hydrogen peroxide, I remember a nice oxidation-reduction reaction that involves hydrogen peroxide. If one has hydrogen peroxide in a beaker with some sulfuric acid, and then adds potassium permanganate, that is, adds the permanganate ion, a beautiful magenta color, the permanganate ion is destroyed, as shown by the loss of color, and oxygen is evolved. Hydrogen peroxide is oxidized by the permanganate ion in acid solution.

Of course, to write the equation for this reaction, we need to know what it is that is reacting and what the products of the reaction are. The way to find out is by experiment. Or, if somebody else has carried out the experiment already, by reading a textbook, or a paper in a journal, and if the statements made in the textbook and paper are reasonable, they can be accepted. Now, in this case, permanganate ion is reduced by the hydrogen peroxide to manganous ion, Mn++, and hydrogen peroxide is oxidized to oxygen. You can see bubbles of gas coming out of the reaction mixture and you can identify this gas easily as oxygen.

I like to balance oxidation-reduction reactions, I like to balance the equations for oxidation-reduction reactions, by writing electrode reactions. For example, let us take H2O2 and say that it forms hydrogen ions plus oxygen…which escapes. Now, let us assume that this reaction is taking place by itself, perhaps in an electrolytic cell where there are electrodes present that can provide electrons or take up electrons. Here, I have started to write the electrode reaction, but there is not conservation of electric charge. Hydrogen ions are formed so that I can write plus two electrons to conserve charge, and now everything is conserved. Hydrogen atoms are conserved because I wrote two in front of the H+, oxygen atoms are conserved, and the electric charge is conserved, which means electrons are conserved. The oxidation number of hydrogen is plus one, that of oxygen minus two. Hydrogen, again, is plus one here, so that there has been no change in valence in oxidation number of hydrogen. But, here, oxygen is minus one. This is a peroxide.

Oxygen has changed from minus one to zero, and if I look at the atoms that have changed their oxidation number, I see that two oxygen atoms have gone up from minus one to zero, which means that two electrons have been given off, and they are indicated here, so everything is rosy.

Now, the other reaction, with the permanganate ion, we can write as an electrode reaction too. This one is so simple that you don’t need to have any rules. This one is a little more complicated. Here, permanganate is reduced in acid solution to the manganous ion, bipositive manganese. In the permanganate ion, oxygen has oxidation number minus two, manganese plus seven, so that manganese has changed by five. This means that we need five electrons in order to carry out the reduction of manganese, the deelectronation, the electronation of manganese from plus seven to plus two. Two charges over on this side, plus seven, minus five gives us...oh, this is minus one, minus five and minus one is six over here. We need a total of eight plus charges. This is in acid solution so I write plus eight plus, with hydrogen ions. And now, to balance the atoms, I have plus four H2O.

These two electrode reactions can be combined with one another if we make the same number of electrons used up in this reaction as formed in this reaction. I think that I can multiply this reaction by five and this one by two without making any mistakes and write two MnO4- plus five H2O2 and here there would be five, ten electrons that balance the two electrons. Sixteen H+ and ten H+ leaves plus six H+ to form two Mn++ plus five O2 that comes off plus eight H2O.

This should complete the reaction. There should be conservation of charge, two minuses and six pluses and four pluses, and there are the four pluses, conservation of manganese atoms, conservation of oxygen atoms. Here are all together eight and ten is eighteen, there are eight and ten is eighteen. And conservation of hydrogen, ten and six, sixteen and a total of sixteen over there, so that we have succeeded in balancing this equation.


Associated: Linus Pauling, Robert Chapin, Jane Chapin, National Science Foundation
Clip ID: 1957v.2-06

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Creator: National Science Foundation
Associated: Linus Pauling, Robert Chapin, Jane Chapin

Date: 1957
Genre: video
ID: 1957v.2
Copyright: More Information

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