Balance the following equation by oxdiation number method.
$C l_{2} + {I O_{3}}^{-} + O H^{-} \to {I O_{4}}^{-} + C l^{-} + H_{2} O$ .

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Solution

Writing oxidation numbers of all atoms,
$0 C l_{2} + + 5 I - 2 O_{3}^{-} + - 2 O^{-} \to + 7 I - 2 O_{4}^{-} + - 1 C l^{-} +_{2} - 2 O$
Oxidation numbers of Cl and I have changed.
$0 C l_{2} \to - 1 2 C l^{-}$ .......(i)
$+ 5 I O_{3}^{-} \to + 7 I O_{4}^{-}$ ........(ii)
Decrease in Ox. no. of $C l = 2$ units per $C l_{2}$ molecule
Increase in Ox. no. of $I = 2$ units per $I O_{3}^{-}$ molecule
$C l_{2} + I O_{3}^{-} \to I O_{4}^{-} + 2 C l^{-}$
To balance oxygen $2 O H^{-}$ ions be added on LHS and one $H_{2} O$ molecule on RHS. Hence, the balanced equation is
$C l_{2} + I O_{3}^{-} + 2 O H^{-} \to I O_{4}^{-} + 2 C l^{-} + H_{2} O$ .

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C l_{2} + {I O_{3}}^{-} + O H^{-} \to {I O_{4}}^{-} + C l^{-} + H_{2} O

{C l}_{2} + I O_{3}^{-} + {O H}^{-} ⟶ I O_{4}^{-} + {C l}^{-}

C l_{2} + O H^{-} \to C l^{-} + C l O^{-}_{3} + H_{2} O

H_{2} O_{2} + C l O_{2} + O H^{-} \to C l^{-} + O_{2} + H_{2} O

{C l}_{2} + I O_{3}^{⊖} ⟶ {C l}^{⊖} + I O_{4}^{⊖}

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