Question

Cl2O7 + H2O2 ------->ClO2- +O2 + H+......balance the equation in basic medium by ion-electron method

Answer 1

Sorry for bad handwriting and the picture quality.

Please forgive me if you can !!

and in first page on the third point (on the left side) it's 3H₂O not 3H₂ .

Thanxx

Answer 2

1. The oxidation half equation is

  $H_{2}O_{2}$ ( aq ) → $O_{2}$ (g)

2 electrons are added to balance oxidation number

$H_{2}O_{2}$ ( aq ) → $O_{2}$ (g) + 2$e^{-}$

To balance the charge, 2 hydroxide ions are added.

$H_{2}O_{2}$ ( aq ) + 2$OH^{-}$→ $O_{2}$ (g) + 2$e^{-}$

2 water molecules are added to balance the O atoms.

$H_{2}O_{2}$ ( aq ) + 2$OH^{-}$→ $O_{2}$ (g) + 2$H_{2} O$ + 2$e^{-}$

2. The reduction half reaction is

$Cl_{2}O_{7}$ → $ClO_{2} ^{-}$ (aq)

The Cl atoms are balanced here

$Cl_{2}O_{7}$ →  2 $ClO_{2} ^{-}$ (aq)

The oxidation number is balanced by adding 8 electrons as:-

$Cl_{2}O_{7}$ + 8    $e^{-}$ →  2 $ClO_{2} ^{-}$ (aq)

The charge is balanced by adding 6OH- as:-

$Cl_{2}O_{7}$ + 8    $e^{-}$ →  2 $ClO_{2} ^{-}$ (aq) + 6$OH^{-}$

The oxygen atoms are balanced by adding 3H2O as:-

$Cl_{2}O_{7}$ + 8    $e^{-}$+ 3 $H_{2} O$ →  2 $ClO_{2} ^{-}$ (aq) + 6$OH^{-}$

Multiply the oxidation half reaction by 4

Then adding it to the reduction half reaction, we get the net balanced reaction equation as:-

 $Cl_{2}O_{7}$(g) + 4$H_{2}O_{2}$(aq) +  2$OH^{-}$(aq) →  2 $ClO_{2} ^{-}$(aq)  + 4 $O_{2}$ (g) + 5 $H_{2} O$(l)

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