Balance the following redox equation by ion-eletron method taking place in acidic medium.
$C r_{2} O_{7}^{2 -} + N O_{2}^{-} \to C r^{3 +} + N O_{3}^{-}$

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Solution

In acidic medium:
Reduction: $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} ⟶ 2 C r^{3 +} + 7 H_{2} O \to (1)$
Oxidation: $N O_{2}^{-} + H_{2} O - 2 e^{-} ⟶ N O_{3}^{-} + 2 H^{+} \to (2)$
Now applying equation $(1)$ + equation $(2) \times 3$
We will get,
$C r_{2} O_{7}^{2 -} + 14 H^{+} + {3 N O_{2}}^{-} + 3 H_{2} O ⟶ 2 C r^{3 +} + 7 H_{2} O + {3 N O_{3}}^{-} + 6 H^{+}$
or,
$C r_{2} O_{7}^{2 -} + 8 H^{+} + {3 N O_{2}}^{-} ⟶ 2 C r^{3 +} + 4 H_{2} O + {3 N O_{3}}^{-}$
This is the balanced redox equation.

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