Question

Balance the following equations by ion electron method:
$I_2+O{H}^{-}\to {I{O}_3}^{-}+I^{-}+H_{2}O$.

Answer 1

$I_2+O{H}^{-}\to I{O}_3^{-}+I^{-}+H_{2}O$
$1$st step: Splitting into two half reactions,
$\underset{\left(Oxidationhalfreaction\right)}{I_2+O{H}^{-}\to I{O}_3^{-}+H_{2}O}$; $\underset{\left(Reductionhalfreaction\right)}{I_2\to I^{-}}$
$2$nd step: Adding $O{H}^{-}$ ions,
$I_2+12O{H}^{-}\to 2I{O}_3^{-}+6H_{2}O$
$3$rd step: Adding electrons to the sides deficient in electrons,
$I_2+12O{H}^{-}\to 2I{O}_3^{-}+6H_{2}O+10e;I_2+2e\to 2I^{-}$
$4$th step: Balancing electrons in both the half reactions,
$I_2+12O{H}^{-}\to 2I{O}_3^{-}+6H_{2}O+10e$
$5[I_2+2e^{-}\to 2I^{-}]$
$5$th step: Adding both the half reactions,
$6I_2+12O{H}^{-}\to 2I{O}_3^{-}+10I^{-}+6H_{2}O$
Dividing by $2$,
$3I_2+6O{H}^{-}\to I{O}_3^{-}+5I^{-}+3H_{2}O$.