Summary Redox I

Topic 3: Redox I
1. know what is meant by the term ‘oxidation number’
Oxidation number: tells you how many electrons an atom has accepted or donated to form an
ion/compound.

2. be able to calculate the oxidation number of elements in compounds and
ions (The use of oxidation numbers in peroxides and metal hydrides is expected.)
Oxidation Rules:
 Elements have an oxidation number of 0
o Ag  0
o O2 0
 Single ions have the same oxidation number as their charge
o Na+  +1
o Mg2+  +2
 In molecular ions, the sum of each atom’s oxidation numbers equals the overall charge
o SO42-  O: -2, S: +6
Oxygen almost always has an oxidation number of -2 and there are 4 oxygens so
there is a -8 charge from the oxygens. Therefore, to create an overall -2 charge
the S must be +6.
 For a neutral compound the sum of the oxidation numbers is 0
o MgCl2  Cl: -1, Mg: +2
 Hydrogen always has an oxidation number of +1, except in metal hydrides where it’s -1
o HCl  Cl: -1, H: +1
o CaH2  Ca: +2, H: -1
 Oxygen has an oxidation number of -2, except in peroxides where it’s -1
o H2O  O: -2, H: +1
o H2O2  H: +1, O: -1
o F2O  F: -1, O: +2

Example: What is the oxidation state of Cr in Na2CrO4?
Na: +1, O: -2. 2(+1) + 4(-2) = -6  Cr: +6
Example: What is the oxidation state of Cr in Cr(H2O)63+?
H: +1, O: -2. 6(+2-2) = 0  Cr: +3

3. understand oxidation and reduction in terms of electron transfer and
changes in oxidation number, applied to reactions of s- and p-block elements
4. understand oxidation and reduction in terms of electron loss or electron gain
Oxidation: a loss of electrons i.e. oxidation number increases
Zn(s)  Zn2+(aq) + 2e-
0 +2



Reduction: a gain of electrons i.e. oxidation number decreases
Cu2+(aq) + 2e-  Cu(s)
+2 0



Redox: a reaction in which both reduction and oxidation takes place
CuSO4 (aq) + Zn(s)  Cu(s) + ZnSO4 (aq)

Copper is reduced, and zinc is oxidised.