# Calculating oxidation number

In the previous section we learned of instances in which oxidation and reduction occur. From this, we know that oxidation occurs when the oxidation number increases and reduction occurs when the oxidation number decreases or reduces. We tend to calculate the oxidation state in the instances when we cannot directly tell if a substance has been oxidised or reduced, or if we are asked to show evidence of oxidation or reduction.
Oxidation number is determined via certain rules:

-Oxidation number for an unreacted elements is zero, example oxidation number of Mg =0 and H2(g) =0

-Oxidation number of an ion is its charge

-Oxidation number of oxygen is -2 unless in peroxides or compounds containing fluorine

-Oxidation number of hydrogen is -1 unless in metal hydrides in which case it is +

-The sum of the oxidation number in a neutral molecule is equal to zero, example MgO where Mg =+2 and O=-2 thus giving zero.

Examples

Find the oxidation number of Mn in KMnO4.

From the rules above, we see that oxidations number of potassium= 1 and the oxidation number of oxygen is =-2.

KMnO4

Since there is 4 oxygen atoms, then to calculate the oxidation state of Mn we must multiply the oxidation state of oxygen by 4.
KMnO4 is a neutral compound so the oxidation number ads up to 0.
1+ Mn + (-2) x 4=0

Mn-7=0

Mn=7, so the oxidation number of Mn in KMno4 is +7

SO42-

Calculate the oxidation state of S in the above compound.

Since the overall oxidation number of SO42 is -2, when calculating the oxidation of S we set the equation equal to -2, since SO42 is a polyatomic ion ( the oxidation state of an ion is its charge).

S + 4 × (-2) =-2

S= 6, The oxidation number of S in SO42- is +6

Exercise

Cr2O72– find Cr

SO2 find S

PbH4 find Pb

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