Oxidation States in Sugar Reactions

Oxidation States

A previous lecture discussed how you could determine the oxidation state of any atom in a molecule from the Lewis structure. Whenever the oxidation numbers of some atoms change from reactants to products, it is important that the electrons that are lost by any atoms exactly equals the number that are gained by other atoms.

Let's look at the oxidation states of the atoms in glucose. We get the oxidation states by (1) drawing the Lewis structure, (2) breaking all bonds and giving the electrons in the bond to the more electronegative element, and (3) comparing the number of electrons around each element to that element's number of valence electrons. Oxidation states are always given in Roman numerals to distinguish them from formal charges.

We can see that the oxidation state of all oxygen atoms is -II and the oxidation state of all hydrogen atoms is I. The oxidations states of the carbon atoms varies between I and -I with an average oxidation state of 0.

Now we'll look at the oxidations states for some other molecules.

Half Reactions for Oxygen + Glucose

First let's consider the combustion of sugar (or respiration). Glucose reacts with molecular oxygen to produce carbon dioxide and water.

The carbon atoms in glucose are oxidized. That is, they lose electron and go to a higher oxidation state.

The oxygen atoms in molecular oxygen are reduced. That is, they add electrons and go to a lower oxidation state.

If we want to represent only the oxidation process in an equation, we show only the reactant that becomes oxidized. We can use H2O or H+ to balance oxygen and hydrogen atoms on either side of the equation.

Each of the carbon atoms, on average, is oxidized by 4 electron for a total of 24 electrons. To balance the oxidation half reaction, we need to add 6 water molecules to add enough oxygen atoms to make all of the carbon dioxide molecules. There will also be 24 protons as products.

Now consider the reduction of molecular oxygen (0 oxidation state) to water (-II) oxidiation state. Each oxygen atom requires 2 electrons, for a total of 4. To balance the reaction, we can add 4 protons to the reactant side of the equation.

To get the net reaction, we multiply one of the equations by some factor so that the electrons produced equals the electrons used.

Half Reactions for Chlorate + Glucose

Chlorate anion is another oxidizing agent (see the gummy worm reaction on the index page) that can oxidize glucose and become reduced.

The oxidation of glucose to carbon dioxide is the same as above.

The Cl(V) in the chlorate ion is reduced to Cl(-I) in the chloride anion for a reduction of 6 electrons.

Multiplying the reduction half reaction by 4, to use 24 electrons, and combining the oxidation and reduction half reactions gives us the net reaction.

Half Reactions for Hydrogen + Glucose

In the reaction with hydrogen, hexane and water are produced. Let's look at the Lewis structure and oxidation states of the atoms in hexane.

In going from glucose to hexane, the carbon atoms are reduced by a total of 14 electrons.

Hydrogen is oxidized from oxidation state 0 in H2 to oxidation state I in water.

It is necessary to multiply the oxidation half reaction by 7 to have equal numbers of electrons in the two half reactions.

Professor Patricia Shapley, University of Illinois, 2012