Question

10. Write balance chemical equation for reaction between H2O2 and MnO4 in acidic and basic medium.​

Answer 1

Answer:

The unbalanced redox equation is as follows:

H2O2+MnO4−⟶O2+MnO2(basic)

All atoms other than H and O are balanced.

The oxidation number of Mn changes from +7 to +4. The change in the oxidation number is 3.

The oxidation number of O changes from -1 to 0. The change in the oxidation number per O atom  is 1. The total change in oxidation number for 2 O atoms is 2.

To balance the increase in the oxidation number with decrease in the oxidation number multiply MnO4− and MnO2 with 2 and multiply H2O2 and O2 with 3..

3H2O2+2MnO4−⟶3O2+2MnO2(basic)

To balance O atoms, add four water molecules on RHS.

3H2O2+2MnO4−⟶3O2+2MnO2+4H2O(basic)

To balance H atoms, add 2H+ on LHS.

3H2O2+2MnO4−+2H+⟶3O2

Answer 2

Answer:

The balanced chemical equation in acidic medium:

$2MnO_{4}^{-} +5H_{2}O_{2}+6H^{+}$  →  $2Mn^{2+}+5O_{2}+8H_{2}O$

In basic medium:

$3H_{2}O_{2}+2MnO_{4}^{-}$   →   $2MnO_{2}+3O_{2}+2OH^{-}+2H_{2}O$

Explanation:

• Reaction of MnO₄⁻ with H₂O₂ in acidic medium:

     $H_{2}O_{2}$  →  $O_{2}+ 2H^{+}+2e^{-}$                                      ]×2

     $MnO_{4}^{-} +8H^{+}+5e^{-}$   →  $Mn^{2+}+4H_{2}O$                ]×5

 -------------------------------------------------------------------------------------

 $2MnO_{4}^{-} +5H_{2}O_{2}+16H^{+}$  →  $2Mn^{2+}+5O_{2}+8H_{2}O+10H^{+}$

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More simplified version:

   $2MnO_{4}^{-} +5H_{2}O_{2}+6H^{+}$  →  $2Mn^{2+}+5O_{2}+8H_{2}O$

• Reaction of MnO₄⁻ with H₂O₂ in basic medium:

         $3H_{2}O_{2}+2MnO_{4}^{-}$   →   $2MnO_{2}+3O_{2}+4H_{2}O$

Add 2 H⁺ ,to balance the H-atoms on LHS and doe to basic medium add 2OH⁻  on both sides:

     $3H_{2}O_{2}+2MnO_{4}^{-}+2H^{+}+2OH^{-}$  →   $2MnO_{2}+3O_{2}+4H_{2}O +2OH^{-}$

Now,  2H⁺ and 2OH⁻ combine and form water molecule which cancel with H₂O on RHS.

              $3H_{2}O_{2}+2MnO_{4}^{-}$   →   $2MnO_{2}+3O_{2}+2OH^{-}+2H_{2}O$