oxidation+numbers

Oxidation Numbers
The first step in identifying a reduction/oxidation (redox) reaction is to determine whether or not electrons have been transferred from one element to another.

When this happens, the oxidation state of the elements involved in the redox process will change.

Recall that oxidation is the loss of electrons, and that reduction is the gain of electrons. Both of these processes must occur in a redox reaction.



__How to determine oxidation numbers__

First, elements in their natural states are going to have an oxidation number of zero.

Here are some examples:
 * all metals, such as Cu (s), Fe (s), Na (s)
 * the seven diatomics: H 2, N 2 , O 2 , F 2 , Cl 2 , Br 2 , and I 2
 * elements such as: P 4, S 8

Next, any elements involved in compounds will have oxidation numbers of greater than or less than zero.

Some elements will always have the same oxidation numbers. We will call these //"anchor" elements//.

Here are the **anchor elements** to remember:
 * Group IA elements are always +1
 * Group IIA elements are always +2
 * Group IIIA elements are always +3
 * fluorine is always -1
 * oxygen is always a -2, except when it is part of the peroxide ion (as a -1)

Finally, after identifying anchor elements, use their oxidation states to calculate those of any remaining elements. If the compound is neutral, the total of all oxidation numbers should be zero; if it is ionic, the oxidation numbers should have a sum equal to the overall charge of the ion.

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To find an element's oxidation number in a compound, use the following method. The example will be calcium carbonate.

1. separate each element in the compound using a chart. For example: CaCO 3 would be separated into: The top row will be the individual oxidation number, and the bottom row will be the total oxidation number.
 * = Ca ||= C ||= O 3 ||
 * = Ca ||= C ||= O 3 ||

2. Identify the "anchor" ions. Anchor ions are those that will always have the same oxidation state in any compound.

For CaCO 3, these would be Ca and O. Ca is always +2 and O is almost always -2.
 * = +2 ||=  || -2 ||
 * = Ca ||= C ||= O 3 ||

Because there are three O's, the total charge from O is 3 x -2 = -6.
 * = +2 ||=  || -2 ||
 * = Ca ||= C ||= O 3 ||
 * = +2 ||=  ||= -6 ||

3. Now the only unknown is C. Since the total of all oxidation numbers should be zero, do some math to determine carbon's charge.

+2 + x + -6 = 0

x = +4


 * = +2 ||= +4 || -2 ||
 * = Ca ||= C ||= O 3 ||
 * = +2 ||= +4 ||= -6 ||