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<h1><!--mstheme--><font color="#000099"><img src="../../media/redox/redoxtitle.gif" alt="redoxtitle.gif" width="124" height="39"><!--mstheme--></font></h1>
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<!--mstheme--></font><!--msthemelist--><table border="0" cellpadding="0" cellspacing="0" width="100%">
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme--><a href="#oxnum">Oxidation Numbers</a>
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme--><a href="#redox">Redox Reactions</a>
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<p>We are now going to look at some reactions that involve
a transfer of electrons.</p>
<br>
<P>The oxidation number is the charge an atom in a substance would have if
the pairs of electrons in each bond belonged to the more electronegative
atom. Now this means that in a compound made up of monatomic ions, like
NaCl, in which the bonding pairs do belong to the more electronegative
atom, the oxidation number equals the ionic charge. The sodium ion has an
oxidation number of +1 while the chlorine ion has an oxidation number of
-1.</p>
<br>
<p>Here are some rules that are helpful in finding the oxidation number of
a specific atom.
<br>
<!--mstheme--></font><!--msthemelist--><table border="0" cellpadding="0" cellspacing="0" width="100%">
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme-->The oxidation number of an atom in an elementary substance is 0.
This means that the oxidation number of an O atom in O<sub>2</sub> is
0.
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme-->The oxidation number of a Group IA atom in any compund is +1;
The oxidation number of a Group IIA atom in any compund is +2.
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme-->The oxidation number of fluorine is -1 in all of its compounds.
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme-->The oxidation number of chlorine, bromine, and iodine is -1 in any
compound containing only two elements.
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme-->The usual oxidation number of oxygen in a compound is -2. The
major exceptions of this rule are peroxides, like
H<sub>2</sub>O<sub>2</sub>,
which have an oxidation number of -1.
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme-->The oxidation number of hydrogen in most compounds is +1.
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme-->The sum of the oxidation numbers in a compound is always zero. For
something that is an ion consisting of two atoms (a polyatomic ion), the
oxidation numbers add up to the charge on the ion.
<!--msthemelist--></table><!--mstheme-->
<br>
<p>You might be heartily confused at this point as to what oxidation
numbers are and how to find them. Don't worry, you can usually find the
oxidation
numbers of elements in ionic compounds by looking at the individual ions
in
the compound.</p>
<br>
<br>
<p>Example:
<p>In H<sub>2</sub>SO<sub>4</sub>, H = +1 and O = -2. Since there are 4
oxygen atoms, the total charge from oxygen is -8. Therefore the charge of
S must be +6, since the net charge of the molecule is zero.
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<a href="#oxnum">Oxidation Numbers</a> |
<a href="#redox">Redox Reactions</a> |
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<img src="../../media/redox/oxredreact.jpg" alt="Oxidation-Reduction
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<p>When we take an iron metal nail and dip it into a blue solution of
copper (II) sulfate, the nail becomes covered with a reddish substance.
This
reddish susbtance is metallic copper. The equation for this is:</p>
<br>
<br>
<center>
Fe(s) + CuSO<sub>4</sub>(aq) -> FeSO<sub>4</sub>(aq) + Cu(s)
</center>
<br>
<br>
<p>The net ionic equation, that is, the equation in which we have
cancelled
out the spectator ions (in this case the spectator ion is
SO<sub>4</sub>).</p>
<br>
<br>
<center>Fe<sup>0</sup>(s) + Cu<sup>2+</sup>(aq) -> Fe<sup>2+</sup>(aq) +
Cu<sup>0</sup>(s)
<center>
<br>
<br>
<p>What we have done here is to show you the oxidation numbers above the
elements in the equation. This shows that the iron metal loses electrons
to form iron(II) ions while copper(II) ions gain electrons to create
copper metal. The Cu<sup>2+</sup> on the reactancts side of the equation
has a 2+ because SO<sub>4</sub> has a charge of 2-. Any reaction like this
that involves a transfer of electrons is an oxidation-reduction
reaction.</p>
<br>
<br>
<p>We can write the equation above in the form of two half-reactions. A
half-reaction is one of two parts of an oxidation-reduction reaction, one
of which involves a loss of electrons, and the other a gain of electrons.
The half reactions for the iron nail reaction would be:</p>
<br>
<br>
<center>
1. Fe(s) -> Fe<sup>+2</sup>(aq) + 2e<sup>-</sup>
<br>
2. Cu<sup>+2</sup>(aq) + 2e<sup>-</sup> -> Cu(s)
<br>
<br>
<p>The first equation shows us that electrons are lost by Fe,
(2e<sup>-</sup>). The second equation shows electrons being gained by
Cu<sup>2+</sup>. You know, we can also express this reaction in terms of
what happened to the oxidation number. The oxidation number of iron
increases after the reaction, while the oxidation number of copper
decreases.</p>
<br>
<br>
<p>Basically, an oxidation-reduction reaction (redox reaction) is a
reaction in which
electrons are transferred between species or in which atoms change
oxidation numbers. Oxidation is the portion of the redox reaction in
which there is a loss of electrons by a species or an increase in the
oxidation number of an atom. Reduction is the part of a reaction in which
there is a
gain of electrons by a species or a decrease in the oxidation number of an
atom. Something that is oxidized loses electrons or contains an atom that
increases in oxidation number. Something that is reduced gains electrons
or has
an atom that decreases in oxidation number. An oxidizing agent is a
something that oxidizes something, meaning that the oxidation agent is
itself reduced. A reducing agent is a something that reduces something
else, and is itself oxidized. Notice the example below.</p>
<br>
<br>
<img src="../../media/redox/equa1.gif" alt="equation 1" width="455" height="170"></a>
<p>Fe is oxidized and Fe is the reducing agent.
<p>Cu<sup>+2</sup> is reduced and Cu<sup>+2</sup> is the oxidizing agent.
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