Question

Balance the following equation by oxidation number method.
$NaI{O}_3+NaHS{O}_3\to N{a}_{2}S{O}_4+NaHS{O}_4+I_2+H_{2}O$.

Answer 1

Writing oxidation numbers of all the atoms.
$\stackrel{+1}{Na}\stackrel{+5}{I}\stackrel{-2}{O_3}+\stackrel{+1}{Na}\stackrel{+1}{H}\stackrel{+4}{S}\stackrel{-2}{O_3}\to \stackrel{+1}{N{a}_2}\stackrel{+6}{S}\stackrel{-2}{O_4}+\stackrel{+1}{Na}\stackrel{+1}{H}\stackrel{+6}{S}\stackrel{-2}{O_4}+\stackrel{0}{I_2}+\stackrel{+1}{H_2}\stackrel{-2}{O}$
The oxidation no. of I has decreased while that of $S$ has increased.
$Na\stackrel{+5}{I}{O}_3\to \stackrel{0}{I}$ .........(i)
$NaH\stackrel{+4}{S}{O}_3\to NaH\stackrel{+6}{S}{O}_4$ ..........(ii)
Decrease in Ox. no. of $I=5$ units per molecule $NaI{O}_3$
Increase in Ox. no. of $S=2$ units per molecule $NaHS{O}_3$
Eq. (i) is multiplied by $2$ and eq. (ii) is multiplied by $5$ as to make decrease and increase equal.
$2NaI{O}_3+5NaHS{O}_3\to I_2+3NaHS{O}_4+2N{a}_{2}S{O}_4$
To balance hydrogen and oxygen, one $H_{2}O$ molecule should be added on RHS. Hence, the balanced equation is
$2NaI{O}_3+5NaHS{O}_3\to I_2+3NaHS{O}_4+2N{a}_{2}S{O}_4+H_{2}O$.