Chem Question, help greatly appreciated?

Question

Use this reaction for the following questions:
2Al(s)+ 6H2O(g) --> 2Al(OH)3+ 3H2

a. determine the oxidation numbers of each element in both the reactants and the products of this oxidation reduction reaction of aluminum and water vapor.

b. which element is oxidized in this reaction?

c. which element is reduced in this reaction?

Answer 1

Aluminum as a reactant in this case will have an oxidation state of zero, because any element in its standard state has an oxidation state of zero. In water, each hydrogen has an oxidation state of +1 and the oxygen has -2. Oxygen is generally -2 unless it is in a peroxide. For the products, hydrogen has a oxidation state of zero, because it is in its standard state. Each hydrogen is once again +1 and each oxygen is -2. From this you can determine that the aluminum has an oxidation state of +3 because each compound has to be electrically neutral.

Oxidation means a loss of electrons, so in this case the element oxidized is aluminum. It goes from 0 to +3, a loss of three electrons.

Reduction is a gain of electrons, so in this case the element reduced is hydrogen. It goes from -1 to zero, a gain of an electron.

Hope this helps! Sorry if I made any mistakes.