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

$C l_{2} O_{7} (g) + H_{2} O_{2} (a q) \to C l O_{2}^{-} (a q) + O_{2} (g) + H^{+}$

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Solution

The oxidation number of chlorine decreases from +7 to +3 and the oxidation number of $O$ increases from -1 to zero. Thus, $C l_{2} O_{7}$ is oxidizing agent and $H_{2} O_{2}$ is the reducing agent.
Ion electron method:
The oxidation half equation is
$H_{2} O_{2} (a q) \to O_{2} (g)$
To balance oxidation number, 2 electrons are added.
$H_{2} O_{2} (a q) \to O_{2} (g) + 2 e^{-}$
2 hydroxide ions are added to balance the charge.
$H_{2} O_{2} (a q) + 2 O H^{-} \to O_{2} (g) + 2 e^{-}$
2 water molecules are added to balance the O atoms.
$H_{2} O_{2} (a q) + 2 O H^{-} \to O_{2} (g) + 2 H_{2} O (a q) + 2 e^{-}$
The reduction half reaction is $C l_{2} O_{7} \to C l O_{2}^{-} (a q)$
The Cl atoms are balanced
$C l_{2} O_{7} \to C l O_{2}^{-} (a q)$
8 electrons are added to balance the oxidation number.
$C l_{2} O_{7} + 8 e^{-} \to C l O_{2}^{-} (a q)$
6 hydroxide ions are added to balance the charge.
$C l_{2} O_{7} + 8 e^{-} \to C l O_{2}^{-} (a q) + 6 O H^{-} (a q)$
The oxidation half equation is multiplied with 4 and added to reduction
half equation.
$C I_{2} O_{7} (g) + 4 H_{2} O_{2} (a q) + 2 O H^{-} \to C I O_{2}^{-} (a q) + 4 O_{2} (g) + 5 H_{2} O (l)$
Oxidation number method:
Total decrease in oxidation number of $C l_{2} O_{7}$ is 8.
Total increase in oxidation number of $H_{2} O_{2}$ is 2.
$H_{2} O_{2}$ and $O_{2}$ are multiplied with 4
$C l_{2} O_{7} (g) + 4 H_{2} O_{2} (a q) \to C l O_{2}^{-} (a q) + 4 O_{2} (g)$
Chlorine atoms are balanced
$C l_{2} O_{7} (g) + 4 H_{2} O_{2} (a q) \to 2 C l O_{2}^{-} (a q) + 4 O_{2} (g)$
$O$ atoms are balanced by adding 3 water molecules.
$C l_{2} O_{7} (g) + 4 H_{2} O_{2} (a q) \to 2 C l O_{2}^{-} (a q) + 4 O_{2} (g) + 3 H_{2} O (l)$
$H$ atoms are balanced by adding 2 hydroxide ions and 2 water molecules.
$C I_{2} O_{7} (g) + 4 H_{2} O_{2} (a q) + 2 O H^{-} \to C I O_{2}^{-} (a q) + 4 O_{2} (g) + 5 H_{2} O (l)$

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Balance the following equation in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent. Cl2O7(g)+H2O2(aq)→ClO−2(aq)+O2(g)+H+

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

$C l_{2} O_{7} (g) + H_{2} O_{2} (a q) \to C l O_{2}^{-} (a q) + O_{2} (g) + H^{+}$