Balance the given reaction by oxidation number method in acidic medium.
$C r_{2} O_{7 (a q)}^{2 -} + I_{(a q)}^{-} \to C r_{(a q)}^{3 +} + I O_{3 (a q)}^{-}$ .

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Solution

$3 \times 2 = 6$
Balance all the atoms other than
$O$ & $H$
$C r_{2} O_{1}^{2 -} + I^{-} ⟶ 2 C r^{3 +} + T O_{3}^{-}$
Balance the $O$ by adding $H_{2} O$ molecule
$C r_{2} O_{1}^{2 -} + I^{-} + 8 H^{+} \to 2 C r^{3 +} + I O_{3}^{-} + 4 H_{2} O$

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