|Chapter Four, Section Three|
I. Oxidation Reduction Reactions
In these reactions some species is oxidized (gives away electrons) and some other species is reduced (accepts electrons). A first-grade method of remembering which is which is LEO says GER. LEO stands for "loose electrons oxidized" and GER stands for "gains electrons reduced".
To recognize oxidation-reduction reactions, we have to look for changes in the oxidation numbers of the elements. Thus we have to be able to determine the oxidation numbers of the elements. So we have some rules to learn:
a.) in metal H compounds the oxidation number of H is -1. Example: NaH (oxidation number of Na is +1, oxidation number of H is -1. b.) in peroxides the oxidation number of O is -1. Example: H2O2(the oxidation number of H is +1 and the oxidation number of O is -1.
Using Rule 2 we know that the oxidation states of ions in compounds according to the columns of the periodic table are IA -> +1, IIA -> +2, IIIA -> +3, VIA -> -2, VIIA -> -1 (except for exceptions: Tl can have +3 and +1; Cl, Br, and I are usually -1 but can have a number of other OxN; S is often -2 but can have +6, +4, +2; Se is usually -2 but can be +6 or +4; Te is usually -2 but can be +6 or +4). In compounds and polyatomic ions we then use the OxN of the ones that we know to calculate the OxN of the ones we don't know.
So let's look at some examples:
H3PO4 is neutral and we know that the OxN of H is +1 and the OxN of O is -2 which gives the charges of 3*(+1) = +3 and 4*(-2) = -8 so that the total charge contributed by H and O is -5. Therefore the OxN of P has to be +5 in order for the compound to be neutral.
MnO4-1 has an overall charge of -1. We know the OxN of O is -2 which gives a total of 4*(-2) = -8, so Mn must have an OxN of +7.
S2O3-2 has an overall charge of -2. We know that the OxN of O is -2 for a total charge of 3*(-2) = -6 so the oxidation number of S must be +2.
HNO3 is neutral. The OxN of O is -2 which gives a total of -6, H is +1 which gives a total of +1, so the sum of H and O is -5 and thus the OxN of N must be +5.
Now we are ready to recognize some Ox-Red reactions (often called Redox reactions). Look at
ZnO + C CO + Zn and see which atom is oxidized and which is reduced.
Write the OxN below each atom to get:
Zn O + C C O + Zn +2 -2 0 +2 -2 0So you see that C looses two electrons in the reaction and Zn gains two electrons. Therefore C is oxidized and Zn is reduced. Whatever is oxidized is called the reducing agent and whatever is reduced is called the oxidizing agent. Except in rare instances, metals acting alone are reducing agents and so are oxidized. You see the reverse of the above reaction has Zn acting alone and you see that in that reverse reaction Zn is oxidized and hence is the reducing agent.
S + O2 S O2 0 0 +4 -2So S is oxidized (0 to +4) and oxygen is reduced (0 to -2).
3Zn S + 8 H N O3 Zn S O4 + 8 N O + 4H2O +2 -2 +1+5-2 +2 +6-2 +2-2 +1-2Here we see that N is reduced and S is oxidized.
Important examples of oxidation-reduction reactions are: combustion, respiration, rusting, bleaching, and batteries.
II. Heats of Reactions
As we said in a previous section, a chemical reaction may also give off (or absorb) heat and we can indicate that as:
R1 + R2 P1 + P2 + heat
R1 + R2 P1 + P2 - heat
A reaction that gives off heat is called an exothermic reaction and a reaction that absorbs heat is called an endothermic reaction (we wrote "-heat" above, we could have just as well put it on the reactant side of the equation). Heat is one form of energy, the reaction could also give off or absorb another form of energy such as light or electricity. The general term to account for any type of energy absorbing reaction is endergonic and for the giving off of any type of energy is exergonic.
A combustion reaction is an exothermic reaction such as
CH4 + O2CO2 + 2 H2 + heat
Such reactions are very important for they are the source of heat for our home.
After you have studied this material and practiced some problems, take quiz three. If you score at least 80 on the test then you are ready to continue to the next section.
Web Author: Dr. Leon L. Combs
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