7.2 - Oxidation states

Oxidation states are used to see what has been oxidised and what has been reduced in a redox reaction.

~in an ionic compound, they tell us how many e- have been lost/gained
~in a molecule, they tell us about the distribution between elements of different electronegativity

-every element in its uncombined state = 0
-positive number = lost electrons (oxidation)
-negative number = gained electrons (reduction)
The more positive the number, the more it has been oxidised and the more negative the number, the more it has been reduced.

Rules:
1) uncombined = 0
2) some elements always have the same number, for others, sometimes it changes
3) sum of all o. numbers in a compound = 0 (they aren't charged)
4) sum of all o. states in a complex ion = the charge of the ion (NH4+ = +1)
5) the most electronegative element in a compound always has a - oxidation state

element

o. state

e.g.

+1 (in metal hydrides is -1)

HCl

Group 1

NaCl

Group 2

CaCl2

AlCl3

-2 (in peroxides and OF2 = +2)

Na2O

-1

NaF

-1 (except w/ F and O

working out oxidation states:
1. remember rules (e.g. H=+1)
2. using basic maths work out the values for the other elements

e.g. 1:
phosphorus pentachloride, PCL5
1. there are 5 Cl, each with a charge of -1
2. if Cl5=-5, then P has to equal +5

e.g. 2:
red copper oxide, Cu2O
1. oxygen=-2
2. Cu2 = +2
(3. divide Cu2 by 2 to get the value for EACH Cu element)
+2/2= +1, Cu=+1

nb: if an element's oxidation state has not changed, it is a spectator ion.

	Criado por Naomi Moylan-Torke mais de 8 anos atrás

AS - Level AS Chemistry (Physical Chemistry) Notas sobre Chapter 7 - Oxidation, reduction and redox reactions, criado por Naomi Moylan-Torke em 07-02-2016.