Balancing the following equation
$\mathrm{Cr}(\mathrm{OH}{)}_3+{\mathrm{IO}}_3^{-}\stackrel{\mathrm{OH}}{\to }{\mathrm{I}}^{-}+{\mathrm{CrO}}_4^{2-}$
(or)
Balance the above equation basic medium.

Equation

Oxidation half reaction $\mathrm{Cr}(\mathrm{OH}{)}_3\to {\mathrm{CrO}}_4^{2-}$
1) Balance the atoms other than H and O
$\mathrm{Cr}(\mathrm{OH}{)}_3\to {\mathrm{CrO}}_4^{2-}$
2) Balance of oxygen atoms
$\mathrm{Cr}(\mathrm{OH}{)}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CrO}}_4^{2-}$
3) Balance of Hydrogen atoms
$\mathrm{Cr}(\mathrm{OH}{)}_3+{\mathrm{H}}_2\mathrm{O}+5{\mathrm{OH}}^{-}\to {\mathrm{CrO}}_4^{2-}+5{\mathrm{H}}_2\mathrm{O}$
4) Balance of charges
$\mathrm{Cr}(\mathrm{OH}{)}_3+5{\mathrm{OH}}^{-}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CrO}}_4^{2-}+5{\mathrm{H}}_2\mathrm{O}+3{\mathrm{e}}^{-}\dots (1)$
Reduction half reaction  ${\mathrm{IO}}_3^{-}\to {\mathrm{I}}^{-}$
1) Balance the atoms other than H and O
${\mathrm{IO}}_3^{-}\to {\mathrm{I}}^{-}$
2) Balance of oxygen atoms
${\mathrm{IO}}_3^{-}\to {\mathrm{I}}^{-}+3{\mathrm{H}}_2\mathrm{O}$
3) Balance of Hydrogen atoms
${\mathrm{IO}}_3^{-}+6{\mathrm{H}}_2\mathrm{O}\to {\mathrm{I}}^{-}+3{\mathrm{H}}_2\mathrm{O}+6{\mathrm{OH}}^{-}$
4) Balance of charges
${\mathrm{IO}}_3^{-}+6{\mathrm{H}}_2\mathrm{O}+6{\mathrm{e}}^{-}\to {\mathrm{I}}^{-}+6{\mathrm{OH}}^{-}+3{\mathrm{H}}_2\mathrm{O}\dots \left(2\right)$
$(1\times 2)$
$2\mathrm{Cr}(\mathrm{OH}{)}_3+10{\mathrm{OH}}^{-}+2{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{CrO}}_4^{-2}+10{\mathrm{H}}_2\mathrm{O}+6{\mathrm{e}}^{-}.....(3)$
${\mathrm{IO}}_3^{-}+6{\mathrm{H}}_2\mathrm{O}+6{\mathrm{e}}^{-}\to {\mathrm{I}}^{-}+6{\mathrm{OH}}^{-}+3{\mathrm{H}}_2\mathrm{O}.....\left(4\right)$
eq(3) + eq(4)
$2\mathrm{Cr}(\mathrm{OH}{)}_3+{\mathrm{IO}}_3^{-}+8{\mathrm{H}}_2+4{\mathrm{OH}}^{-}\to 2{\mathrm{CrO}}_4^{-2}+{\mathrm{I}}^{-}+13{\mathrm{H}}_2\mathrm{O}$
Simplify the above equation
$2\mathrm{Cr}(\mathrm{OH}{)}_3+{\mathrm{IO}}_3^{-}+4{\mathrm{OH}}^{-}\to 2{\mathrm{CrO}}_4^{2-}+{\mathrm{I}}^{-}+5{\mathrm{H}}_2\mathrm{O}$