Definition+4

So far, we have gone through 3 definitions of oxidation. But this last definition, is actually the most useful and popular.

The oxidation number/state of an atom in a compound is equal to its theoretical charge if all the atoms were joined together by electrovalent bonds.

Thus, ** an increase in oxidation number means the substance has been oxidised. **

Before we can apply this definition, there are 4 basic rules to follow:

S  Cl 2 || 0 0 0 || the same as the charge on the ion ||  K +  Al 3+  S 2- || +1 +3 -2 || in the formula of a compound add up to zero. ||  CaCO 3 [Ca 2+ ; CO 3 2- ] || +2 – 2 = 0 || a polyatomic ion is equal to the charge on the ion. ||  SO 4 2- || +6 + 4(-2) = -2 || Remember, when writing the oxidation state, the charge comes first than the charge value. for example, the charge of an oxide ion is 2-, but its oxidation state is -2.
 * **Rule ** || **Example ** || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 10pt; text-align: center;">** Oxidation **** State ** ||
 * The oxidation state of a free element is zero || Cu
 * The oxidation state of a simple ion is
 * The oxidation states of the atoms present
 * The total of the oxidation states of the atoms in

You can read more about the rules of oxidation states. click on this link. []

media type="youtube" key="EHe8-AFMsMA" height="344" width="425"

<span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif;">After you have viewed the video, let's apply the concept of oxidation numbers to chemical reactions.

<span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif;">__Example 1__ <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif;">Fe <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif; vertical-align: super;">2+ <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif;"> + H <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif; vertical-align: sub;">2 <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif;">O <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif; vertical-align: sub;">2 <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif;"> → Fe <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif; vertical-align: super;">3+ <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif;"> + H<span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif; vertical-align: sub;">2 <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif;">O <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif;">In this reaction, has Fe<span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif; vertical-align: super;">2+ been oxidised or reduced? <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif;">the oxidation number of Fe<span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif; vertical-align: super;">2+ has been increased from +2 in Fe<span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif; vertical-align: super;">2+ to +3 in Fe<span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif; vertical-align: super;">3+. <span style="color: #ff00ff; font-family: 'Trebuchet MS',Helvetica,sans-serif;">Thus, Fe<span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif; vertical-align: super;">2+ has been oxidised. The oxidising agent is H<span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif; vertical-align: sub;">2 <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif;">O <span style="color: #008000; font-family: 'Trebuchet MS',Helvetica,sans-serif; vertical-align: sub;">2.

<span style="color: #ff4e00; font-family: 'Trebuchet MS',Helvetica,sans-serif;">__Example 2__ 3SO <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">2 + Cr <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">2 O <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">7 <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: super;">2- <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">+2H <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: super;">+ <span style="color: #ff4e00; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> ® 3SO <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">4 <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: super;">2- <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">+ 2Cr <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: super;">3+ <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;"> + H <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">2 O  <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">In this reaction, has dichromate(VI) been oxidised or reduced? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">the oxidation number of Cr in Cr<span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">2 O <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">7 <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: super;">2- has been decreased from +6 in Cr<span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">2  O <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">7 <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: super;">2- to +3 in Cr<span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: super;">3+. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Thus, Cr<span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">2 O <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">7 <span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: super;">2- has been reduced. The reducing agent is SO<span style="color: #ff4e00; font-family: 'Trebuchet MS'; font-size: 11pt; line-height: 200%; vertical-align: sub;">2.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">After understand all the four definitions of oxidation, we will move on to learn more about the oxidising and reducing agents.