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Oxidation number of sulfur

The oxidation number is a bookkeeping of electrons. In a neutral compound the sum is zero. Elements in pure form have zero, hydrogen is usually +1, and oxygen is usually -2. You set up an equation for the unknown number and solve it. When the number increases, the substance is oxidised. When it decreases, the substance is reduced. In sulfuric acid sulfur is strongly oxidised, and the acid can act as an oxidising agent.

2⋅(+1)+S+4⋅(−2)=02\cdot(+1) + S + 4\cdot(-2) = 0the sum in H2SO4

Symbols

HHusually +1
OOusually -2
SSthe unknown

Example

In H2SO4: 2+S+4⋅(−2)=02 + S + 4 \cdot (-2) = 0, that is 2+S−8=02 + S - 8 = 0.

S=8−2=6S = 8 - 2 = 6, so the oxidation number is +6.

Exceptions: oxygen is -1 in peroxides, and hydrogen is -1 in metal hydrides.
Practise redox reactions for free →

← Neutralisation · Manganese in permanganate →

Part of Chemistry 1: Redox reactions.