oxidation no. in S4O62-

Let n1 be the averageoxidation number of S in S2O3²⁻.
The sum of oxidation numbers = charge on the species
2n1 + 3(-2) = -2
The average oxidation numberof S in S2O3²⁻, n1 = +2
(Note that it is the average oxidation number, as the oxidation numbers of the 2S atoms in S2O3²⁻ are NOT the same.)

Let n2 be the averageoxidation number of S in S4O6²⁻.
The sum of oxidation numbers = charge on the species
4n2 + 6(-2) = -2
The average oxidation numberof S in S4O6²⁻, n2 = +2.5
(Note that it is the average oxidation number, as the oxidation numbers of the 4S atoms in S4O6²⁻ are NOT the same.)

There are 2 S and 3 O atoms in an S2O3²⁻ ion, but 4 S and 6 O atomsin an S4O6²⁻ ion. Hence, 2 S2O3­²⁻ ions react to form 1 S4O6²⁻ ion. The charge-unbalancedequation should be :
2S2O­3²⁻ → S4O6²⁻

Total oxidation numbers of S on the left hand side = (+2) x 2 x 2 = +8
Total oxidation numbers of S on the right hand side = (+2.5) x 4 = +10

The total oxidation number of S on the right hand side is +2 higher than thaton the left hand side. Hence, 2e⁻ should be added to theright hand side. The balanced half equation should be :
2S2O­3²⁻ → S4O6²⁻ + 2e⁻