Question

Balance the following equation by oxidation number method.
$I_2+NaOH\to NaI{O}_3+NaI+H_{2}O$.

Answer 1

Writing oxidation numbers of all the atoms,
$\stackrel{0}{I_2}+\stackrel{+1}{Na}\stackrel{-2}{O}\stackrel{+1}{H}\to \stackrel{+1}{Na}\stackrel{+5}{I}\stackrel{-2}{O_3}+\stackrel{+1}{Na}\stackrel{-1}{I}+\stackrel{+1}{H_2}\stackrel{-2}{O}$
The Ox. no. of iodine has increased as well as decreased
$\stackrel{0}{I}\to Na\stackrel{+5}{I}{O}_3$ .......(i)
$\stackrel{0}{I}\to Na\stackrel{-1}{I}$ .......(ii)
Increase in Ox. no. of $I=5$ units per I atom
Decrease in Ox. no. of $I=1$ unit per I atom
Eq. (ii) should be multiplied by $5$ as to make increase and decrease equal.
$3I_2\to NaI{O}_3+5NaI$
To balance Na, $6$ molecules of NaOH should be added on LHS.
$3I_2+6NaOH\to NaI{O}_3+5NaI$
To balance hydrogen and oxygen, $3H_{2}O$ should be added on RHS. Hence, the balanced equation is
$3I_2+6NaOH\to NaI{O}_3+5NaI+3H_{2}O$.