Define oxidation and reduction according to the oxidation number.

Explanation

Oxidation involves increase in O.N while reduction involves decrease in O.N
`Sn^(2+)+2Hg^(2+)rarrSn^(4+)+Hg_(2)^(2+)`
here `Sn^(2+)` gets oxidised to `Sn^(4+)` because its O.N increase from +2 while `Hg^(2+)` gets reduced to `Hg_(2)^(2+)` because its O.N decrease from + 2 to +1