KMnO4 + H2O2 + H2SO4 = O2 + MnSO4 + K2SO4 + H2O

Permanent hydrogen peroxide potassium and sulfuric acid (kmno4 h2o2 h2so4) are accessible in nearly all chemical laboratories around the world. KMnO4 and H2O2 react mainly in the medium of H2SO4. To balance the redox reaction, we use the ion-electron technique.

Hydrogen peroxide react with potassium permanganate and sulfuric acid

H2O2 + H2SO4 + KMnO4  = O2 + H2 + MnSO4 + K2SO4

or, After balancing the Redox reaction

5H₂O₂ + 3H₂SO₄ + 2KMnO₄  = 5O₂ + 8H₂O + 2MnSO₄ + K₂SO₄

Now by using the ino-electron method we can balance the Oxidation-Reduction reaction.

H2O2 + KMnO4 + H2SO4 = O2+ H2O + MnSO4 + K2SO4

is the skeleton reaction for the redox reaction of Hydrogen peroxide potassium permanganate and  sulfuric acid (kmno4 h2o2 h2so4)

Here,

The Oxidizing agent: KMnO₄ or MnO₄^-1

The Reducing agent: H₂O₂ or, O^-1

Reduction Half Reaction:

⇒ MnO₄ ^-1 + 8H⁺ +5e^– = 4H₂O + Mn^2- … …. …. …. (1)

Oxidation Half Reaction:

⇒ 2O^-1– 2e^– = O₂ … … … … (2)

Now,

equation (1)x2 + (2)x5,

2MnO₄ ^-1 + 16H⁺ +10e^– = 8H₂O + 2Mn^2-

10 O^-1– 10e^– = 5O₂

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⇒ 2MnO₄ ^-1 + 10 O^-1 + 16H⁺= 2Mn^2- + 8H₂O +5O₂

or, adding necessary ions and radicals we get,

⇒ 5H₂O₂ + 3H₂SO₄ + 2KMnO₄  = 5O₂ + 8H₂O + 2MnSO₄ + K₂SO₄

“Answer”

⇒ 5H₂O₂ + 3H₂SO₄ + 2KMnO₄ = 5O₂ + 8H₂O + 2MnSO₄ + K₂SO₄

Read more:

K2Cr2O7 + FeSO4 + H2SO4 = Cr2(SO4)3 + Fe2(SO4)3 + K2SO4 + H2O

FeSO4 + KMnO4 + H2SO4 = Fe2(SO4)3 + MnSO4 + H2O + K2SO4

NaCl + KMnO4 + H2SO4 = Cl2 + MnSO4 + Na2SO4 + K2SO4 + H2O