The balanced ionic equation for oxidation of ferrous oxalate by MnO₄^– in acidic medium is as follows :

3MnO₄^- + 5FeC₂O₄ + 24H⁺ $$ \to $$

3Mn²⁺ + 10CO₂ + 12H₂O + 5Fe³⁺

Thus, 5 moles of FeC₂O₄ require 3 moles of MnO₄^-.

So, 1 mole of FeC₂O₄ requires

= $${3 \over 5}$$ = 0.6 moles of MnO₄^-

H₂O + $${\mathop {Br}\limits^0}$$₂ $$ \to $$ HO$$\mathop {Br}\limits^{ + 1} $$ + H$$\mathop {Br}\limits^{ - 1} $$

In the above reaction the oxidation number of Br₂ increases from zero (in Br₂ ) to +1 (in HOBr) and decreases from zero (in Br₂ ) to –1 (in HBr). Thus Br₂ is oxidised as well as reduced and hence it is a redox reaction.

SO₃^2– : oxidation state of ‘S’ is +4

S₂O₄^2– : oxidation state of ‘S’ is +3.

S₂O₆^2– : oxidation state of ‘S’ is +5.

So, the order is S₂O$${_4^{2 - }}$$ < SO$${_3^{2 - }}$$ < S₂O$${_6^{2 - }}$$.