32
3 Life Itself (A): How Do We Get Energy to Live by?
use Roman numerals to indicate the oxidation state of an atom in a chemical compound,
and Arabic numerals for the electric charge of a chemical entity. A convention used
in many textbooks is to use Arabic numerals to express both the oxidation state of
an atom and the electric charge of a chemical species; and this is confusing].
Another convention (rule 2) is that hydrogen atom in all compounds (except
when it binds to a metallic element) has +I oxidation state. So the oxidation state of
carbon in methane, CH
4
, must be −IV. How about the oxidation state of the carbon
in methanol, CH
3
OH? There are four hydrogen atoms, which contribute +4, and the
oxygen carries −II; therefore, the carbon must carry “−II.” How about the oxidation
state of carbons in ethanol C
2
H
5
OH? [the answer is “−II”]. The carbon atoms in
acetic acid, CH
3
COOH, the ingredient of vinegar, should be “0, zero” in oxidation
state [try it]. Carbon dioxide CO
2
has its carbon in +IV oxidation state. As you see
here, carbon atoms in compounds can take different oxidation states, from −IV to
+IV. “−IV” is the most reduced state and “+IV” is the most oxidized state of carbon.
So, for example, in the process of reaction:
, the carbon
atom can be regarded to have lost two electrons (try to see it) and hence to have been
oxidized. Addition of oxygen to a compound is usually regarded as “oxidation”
reaction. The chemical agent that oxidizes others is defined as “oxidizing agent.” Of
course, oxygen (O
2
), the oxidizing agent itself, is reduced, starting from “zero” and
becoming “−II”; so the other chemical that reacts with oxygen can be said to be a
“reducing agent.”
[Note: The oxidation state does not necessarily represent the actual electric
charge, though it does so in certain cases including that in ionic compounds. It is a
convenient way to keep track where and how electrons move in oxidation–reduction
reactions].
Let us look at nitrogen cases, using rules 1 and 2. N in ammonia NH
3
is −III, the
most reduced. N in N
2
is obviously zero. N in nitrate, NO
3
−
? Try it. [“+V” is the right
answer]. So the reaction to form ammonia:
is a reduction as far
as nitrogen is concerned, and the reducing agent here is hydrogen which is formally
oxidized. To produce nitrate from ammonia you have to oxidize NH
3
; that is,
The oxidation state of N in NH
3
is −III and that in
HNO
3
is +V. Therefore, eight electrons have moved from the nitrogen atom to
oxygen atoms in this process.
Your car can rust; it is the oxidation of the iron of your car by oxygen in the air
to form iron oxide. It can be expressed formally as
. In this
process, iron changes its oxidation state from “zero” to +III. Thus, iron loses elec-
trons (be oxidized) to oxygen. Iron can be also in +II oxidation state; that is,
is also possible. As a matter of fact, iron can change very readily
its oxidation state between +II and +III. And this change is very widely used in
biological systems (see Chap. 6).
[Note: Rule 3 (of determining oxidation states) Halogen, F, Cl, Br, or I takes −I
oxidation states in compounds bound with other elements; for example, C is C
−II
,
and H
+I
and Cl
−I
in CH
3
Cl. Exception is when they bound with each others; in this
case, the lighter element carries −I oxidation state. For example, in FCl F is F
−I
and
Cl
+I
, whereas Cl
−I
and I
+III
in ICl
3
].