2 MnO 4 - + bI - + cH 2 O → xI 2 + yMnO 2 + zOH - If the above equation is balanced with integer…

Reduction Half reaction

$2{\mathrm{MnO}}_4^{-}\to 2{\mathrm{MnO}}_2$

$2{\mathrm{MnO}}_4^{-}+4{\mathrm{H}}_2\mathrm{O}+6{\mathrm{e}}^{-}\to 2{\mathrm{MnO}}_2+8{\mathrm{OH}}^{-}$

Oxidation Half reaction

$2{\mathrm{I}}^{-}\to {\mathrm{I}}_2+2{\mathrm{e}}^{-}$

$6{\mathrm{I}}^{-}\to 3{\mathrm{I}}_2+6{\mathrm{e}}^{-}$

Adding oxidation half and reduction half, net reaction is

$2{\mathrm{MnO}}_4^{-}+6{\mathrm{I}}^{-}+4{\mathrm{H}}_2\mathrm{O}\to 3{\mathrm{I}}_2+2{\mathrm{MnO}}_2+8{\mathrm{OH}}^{-}$

$\Rightarrow \mathrm{z}=8$

$\Rightarrow \mathrm{Ans} 8$