Balance the following equation in basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent.

$N_{2} H_{4} (l) + C l O_{3}^{-} (a q) \to N O (g) + C l^{-} (g)$

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Solution

The oxidation number of $N$ increases from $- 2$ to $+ 2$ . The oxidation number of $C l$ decreases from $+ 5$ to $- 1$ . Hence, hydrazine is the reducing agent and chlorate ion is the oxidizing agent.
Ion-electron method:
The oxidation half reaction is
$N_{2} H_{4} (l) \to N O (g)$
Balance the N atoms.
$N_{2} H_{4} (l) \to 2 N O (g)$
To balance oxidation number, add 8 electrons.
$N_{2} H_{4} (l) \to 2 N O (g) + 8 e^{-}$
Add 8 hydroxide ions are to balance the charge.
$N_{2} H_{4} (l) + 8 O H^{-} (a q) \to 2 N O (g) + 8 e^{-}$
The reduction half reaction is
$C l O_{3}^{-} (a q) \to C l^{-} (a q)$
Add 6 electrons to balance the oxidation number.
$C l O_{3}^{-} (a q) + 6 e^{-} \to C l^{-} (a q)$
Add 6 hydroxide ions to balance the charge.
$C l O_{3}^{-} (a q) + 6 e^{-} \to C l^{-} (a q) + 6 O H^{-} (a q)$
Multiply the oxidation half reaction by 3 and multiply the reduction half reaction by 2. Add two half reactions obtain the
balanced chemical equation.
$3 N_{2} H_{4} (a q) + 4 C l O_{3}^{-} (a q) \to 6 N O (s) + 4 C l^{-} (a q) + 6 H_{2} O (a q)$
Oxidation number method:
Total decrease in oxidation number of N is 8. Total increase in the oxidation number of Cl is 6.
Multiply $N_{2} H_{4}$ with 3 and multiply $C l O_{3}^{-}$ with 4.
$3 N_{2} H_{4} (l) + 4 C l O_{3}^{-} (a q) \to N O (g) + C l^{-} (a q)$
Balance N and Cl atoms.
$3 N_{2} H_{4} (l) + 4 C l O_{3}^{-} (a q) \to 6 N O (g) + 4 C l^{-} (a q)$
Balance the O atoms are by adding 6 water molecules.
$3 N_{2} H_{4} (l) + 4 C l O_{3}^{-} (a q) \to 6 N O (g) + 4 C l^{-} (a q) + 6 H_{2} O (l)$
This is the balanced chemical equation.

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Balance the following equation in basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent.

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