Balance the following redox reactions:

$I_{2} (s) + O {C l}^{-} (a q) ⟶ {I O}_{3}^{-} (a q) + {C l}^{-} (a q)$

I_{2} (s) + 5 O {C l}^{-} (a q) + H_{2} O (l) ⟶ {2 I O}_{3}^{-} (a q) + {5 C l}^{-} (a q) + {2 H}^{+} (a q)

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I_{2} (s) + 5 O {C l}^{-} (a q) + {2 H}_{2} O (l) ⟶ {I O}_{3}^{-} (a q) + {4 C l}^{-} (a q) + {2 H}^{+} (a q)

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I_{2} (s) + 3 O {C l}^{-} (a q) + {4 H}_{2} O (l) ⟶ {I O}_{3}^{-} (a q) + {4 C l}^{-} (a q) + {2 H}^{+} (a q)

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I_{2} (s) + 5 O {C l}^{-} (a q) + {7 H}_{2} O (l) ⟶ {I O}_{3}^{-} (a q) + {5 C l}^{-} (a q) + {2 H}^{+} (a q)

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Solution

The correct option is A $I_{2} (s) + 5 O {C l}^{-} (a q) + H_{2} O (l) ⟶ {2 I O}_{3}^{-} (a q) + {5 C l}^{-} (a q) + {2 H}^{+} (a q)$
The unbalanced chemical equation is as follows:
$I_{2} (s) + O {C l}^{-} (a q) ⟶ {I O}_{3}^{-} (a q) + {C l}^{-} (a q)$
Balance all atoms except H and O.
$I_{2} (s) + O {C l}^{-} (a q) ⟶ 2 {I O}_{3}^{-} (a q) + {C l}^{-} (a q)$
The oxidation number of iodine changes from 0 to 5. The change in the oxidation number per iodine atom is 5.
Total change in the oxidation number is 10.
The oxidation number of Cl changes from +1 to -1. The change in oxidation number is 2.
To balance the increase in oxidation number with decrease in oxidation number, multiply $O C l^{-}$ and $C l^{-}$ with 5.
$I_{2} (s) + 5 O {C l}^{-} (a q) ⟶ 2 {I O}_{3}^{-} (a q) + 5 {C l}^{-} (a q)$
To balance O atoms, add one water molecule on LHS.
$I_{2} (s) + 5 O {C l}^{-} (a q) + H_{2} O ⟶ 2 {I O}_{3}^{-} (a q) + 5 {C l}^{-} (a q)$
To balance H atoms, add two $H^{+}$ on the RHS.
$I_{2} (s) + 5 O {C l}^{-} (a q) + H_{2} O (l) ⟶ 2 {I O}_{3}^{-} (a q) + {5 C l}^{-} (a q) + {2 H}^{+} (a q)$
This is the balanced chemical equation.

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