Question

In basic medium $\stackrel{+2}{N{a}_2}\stackrel{+4}{S_2}\stackrel{-6}{O_3}\to \stackrel{+2}{N{a}_2}\stackrel{+6}{S}\stackrel{-3}{O_4}$.

Write the balanced chemical equation.

Answer 1

$N{\stackrel{+2}{a}}_2{\stackrel{+4}{S}}_2{\stackrel{-6}{O}}_3⟶N{\stackrel{+2}{a}}_2\stackrel{+6}{S{O}_4^{-8}}$

oxidation no. of each atom.

$\stackrel{+1}{N{a}_2}\stackrel{+2}{S_2}{O}_3^{2-}⟶N{\stackrel{+1}{a}}_2\stackrel{+6}{S{O}_4^{-2}}$

An ionic from

Oxidation half

$S_{2}O-3^{2-}⟶S{O}_4^{2-}$

$S_{2}O-3^{2-}⟶2S{O}_4^{2-}$

$S_2{O}_3^{2-}+N_{2}O⟶2S{O}_4^{2-}+2H^{+}$

$S_2{O}_3^{2-}+N_{2}O⟶2S{O}_4^{2-}+2H^{+}+2\stackrel{-}{O}H$

$S_2{O}_3^{2-}+2o\overline{O}N⟶2S{O}_4^{2-}+N_2^O$

Final reaction

$2N{a}^{+}+S_2{O}_3^{2-}+2\overline{O}H⟶2N{a}^{+}+2S{O}_4^{2-}+N_{2}O$