Question

plz answer as soonas possible

Answer 1

Oxidation state of 'S' in S₂O₃²⁻ :

2x + 3(-2) = -2

2x = -2 + 6

2x = 4

x = +2

Oxidation state of 'S' in S₄O₆²⁻ :

4x + 6(-2) = -2

4x - 12 = -2

4x = 10

x = +2.5

So,

In the above reaction, the oxidation state of sulphur changes from +2 to +2.5.

Thus the change in oxidation state = 2.5 - 2 = 0.5

→ But the change occurs in two sulphur atoms, hence the total change in oxidation state is 2 x 0.5 = 1

Thus, n-factor of Na₂S₂O₃ = 1

So, Equivalent weight = molecular weight / 1 = Molecular weight

Hope it helps!