Write the balanced Ionic equation which represents the oxidation of Iodide (I^–) Ion by permanganate ion in basic medium to give iodine (I₂) and manganese dioxide (MnO₂)

Oxidation half reaction : ${\mathbf{I}}^{\mathbf{-}}\mathbf{\to }{\mathbf{I}}_{\mathbf{2}}$
1) Balancing atoms other than O and H
$2{\mathrm{I}}^{-}\to {\mathrm{I}}_2$
2) Balance of oxygen atoms
$2{\mathrm{I}}^{-}\to {\mathrm{I}}_2$
3) Balance of Hydrogen atom
$2{\mathrm{I}}^{-}\to {\mathrm{I}}_2$
4) Balance of charges $2{\mathrm{I}}^{-}\to {\mathrm{I}}_2+2{\mathrm{e}}^{-}......\left(1\right)$
Reduction half reaction ${\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\to }{\mathbf{MnO}}_{\mathbf{2}}$
1) Balancing atoms other than O & H
${\mathrm{MnO}}_4^{-}\to {\mathrm{MnO}}_2$
2) Balance of oxygen atom
${\mathrm{MnO}}_4^{-}\to {\mathrm{MnO}}_2+2{\mathrm{H}}_2\mathrm{O}$
3) Balance of Hydrogen atoms
${\mathrm{MnO}}_4^{-}+4{\mathrm{H}}_2\mathrm{O}\to {\mathrm{MnO}}_2+2{\mathrm{H}}_2\mathrm{O}+4{\mathrm{OH}}^{-}$
4) Balance of charges
${\mathrm{MnO}}_4^{-}+4{\mathrm{H}}_2\mathrm{O}+3{\mathrm{e}}^{-}\to$${\mathrm{MnO}}_2+4{\mathrm{OH}}^{-}+2{\mathrm{H}}_2\mathrm{O}.....\left(2\right)$
$1\times 3;6{\mathrm{I}}^{-}\to 3{\mathrm{I}}_2+6{\mathrm{e}}^{-} .....\left(3\right)$
$2\times 2;2{\mathrm{MnO}}_4^{-}+8{\mathrm{H}}_2\mathrm{O}+6{\mathrm{e}}^{-}\to 2{\mathrm{MnO}}_2+8{\mathrm{OH}}^{-}+4{\mathrm{H}}_2\mathrm{O} ....\left(4\right)$
(3)+(4)
$2{\mathrm{MnO}}_4^{-}+6{\mathrm{I}}^{-}+4{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{MnO}}_2+3{\mathrm{I}}_2+8{\mathrm{OH}}^{-}$