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

Potassium permanent hydrogen peroxide 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 reacts with potassium permanganate and sulfuric acid (KMnO4 H2O2 H2SO4)

The reaction is an electron exchanging reaction. The reactants in this reaction give and take a certain amount of electrons to make some product. Here, in this chemical reaction, the KMnO4 and H2O2 play the main role, as they exchange the electrons. On the other side, the sulfuric acid (H2SO4) just makes a suitable reaction environment. Since there is electron exchange takes place, we can understand that the oxidation number of the reactants must change. Therefore, the reaction equation in concern is an oxidation-reduction reaction equation.

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

or, After balancing the Redox reaction

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

As we talked in the above, the reaction is an Oxidation-Reduction reaction, therefore, by using the ino-electron method we can easily balance the this reaction.

Balancing the redox reaction by ion-electron method

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

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

If we carefully notice the above reaction, one of the reactants H₂O₂ releases an electron and another reactant KMnO₄ accepts the electron. Therefore, in this reaction the H₂O₂ acts as the reducing agent, on the other side, KMnO₄  acts as the oxidizing agent.

Here,

The Oxidizing agent: KMnO₄ or after excluding the spectrator ion, MnO₄^-1

The Reducing agent: H₂O₂ or after excluding the spectrator ion,  O^-1

Reduction Half Reaction

The oxidizing agent, KMnO₄, or after excluding the spectator ion, MnO₄^-1 takes up five electrons which are released by the reducing agent, H₂O₂, or after excluding the spectator ion,  O^-1, agent present in the reaction. After accepting the electrons oxidizing agent, KMnO₄ gets reduced by the reducing agent, H₂O₂.

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

Oxidation Half Reaction

As we all know the oxidation-reduction reaction is a simultaneous process, which means when the reduction half-reaction takes place,  at the same time, the oxidation half-reaction also occurred. Here, in this reaction, the reducing agent, H₂O₂ releases one electron for each oxygen atom and gets oxidized by the oxidizing agent, KMnO₄.

⇒ 2O^-1– 2e^– = O₂ … … … … (2)

Now, in the reduction half-reaction, five electrons are consumed, on the other hand, in the oxidation half-reaction, there are two electrons released to produce an oxygen molecule. To get the balanced full oxidation-reduction reaction we must multiply the oxidation half and reduction half-reaction with the corresponding number and add them together.

So, 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

At this stage we add necessary ions and radicals to the above ionic form of the full redox reaction,

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

Therefor, the balanced redox reaction is-

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

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