Balance the following redox reaction in basic medium.
${C l}_{2} + I O_{3}^{-} + {O H}^{-} ⟶ I O_{4}^{-} + {C l}^{-}$

{C l}_{2} + I O_{3}^{-} + 3 {O H}^{-} ⟶ I O_{4}^{-} + 5 {C l}^{-} + H_{2} O

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

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{C l}_{2} + 2 I O_{3}^{-} + 2 {O H}^{-} ⟶ I O_{4}^{-} + 5 {C l}^{-} + H_{2} O

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None of the above

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Solution

The correct option is B ${C l}_{2} + I O_{3}^{-} + 2 {O H}^{-} ⟶ I O_{4}^{-} + 2 {C l}^{-} + H_{2} O$
The unbalanced redox equation is as follows:
${C l}_{2} + I O_{3}^{-} + {O H}^{-} ⟶ I O_{4}^{-} + {C l}^{-}$
Balance all atoms other than H and O.
${C l}_{2} + I O_{3}^{-} + {O H}^{-} ⟶ I O_{4}^{-} + 2 {C l}^{-}$
The oxidation number of $C l$ changes from 0 to -1. The change in the oxidation number of $C l$ per atom is 1. Total change in the oxidation number for 2 $C l$ atoms is 2.
The oxidation number of $I$ changes from 5 to 7. The change in the oxidation number of iodine is 2.
The increase in the oxidation number is balanced with a decrease in the oxidation number.
$O$ atoms are balanced.
To balance $H$ atoms, add one water molecule on RHS and one hydroxide ion on LHS.
${C l}_{2} + I O_{3}^{-} + 2 {O H}^{-} ⟶ I O_{4}^{-} + 2 {C l}^{-} + H_{2} O$
This is the balanced chemical equation.

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