2 MnO _ 4 ^ - + bI ^ - + cH _ 2 O → xI _ 2 + yMnO _ 2 + zOH ^ - If...

Reduction Half reaction

$2{MnO}_{4}^{-} \to 2{MnO}_{2}$

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

Oxidation Half reaction

$2{I}^{-} \to {I}_{2} + 2{e}^{-}$

$6{I}^{-} \to 3{I}_{2} + 6{e}^{-}$

Adding oxidation half and reduction half, net reaction is

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

$⇒ z = 8$

$⇒ Ans8$