Sodium chloride react with potassium permanganate and sulfuric acid

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

NaCl + KMnO4 + H2SO4 react to produce the free chloride ions. This is a Redox reaction.

2KMnO4 + 10NaCl+ 8H2SO4  = 5Cl2 + 2MnSO4+ 5Na2SO4 + K2SO4+ 8H2O

Sodium chloride react with potassium permanganate and sulfuric acid

The skeleton formula of the reaction is-

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

or,

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

This is an oxidation-reduction (redox) reaction.

The KMnOis the Oxidizing agent and NaCl is the Reducing agent.

Sodium chloride react with potassium permanganate

Oxidizing agent: KMnOor MnO-1

Reducing agent: NaCl or Cl-1 

Read the Adsorption of acetic acid on charcoal and the isotherm

Oxidation Half Reaction:

MnO-1 +5e + 8H = Mn2- + 4H2O … …. …. …. (1)

Reduction Half Reaction

Cl-1 – e = Cl… … … … … (2)

Now equation (1) + (2)x5,

MnO-1 +5e + 8H = Mn2- + 4H2O

5Cl-1 – 5e = 5Cl0


MnO-1 + 5Cl-1+ 8H = 5Cl0 + Mn2- + 4H2O

or, MnO-1 + 5Cl-1+ 8H = 5/2 Cl2 + Mn2- + 4H2O

or, 2MnO-1 + 10Cl-1+ 16H = 5Cl2 + 2Mn2- + 8H2O

Now adding necessary ions and radicals we get,

2KMnO4 + 10NaCl+ 8H2SO4  = 5Cl2 + 2MnSO4+ 5Na2SO4 +K2SO4 + 8H2O

“Answer”

2KMnO4 + 10NaCl+ 8H2SO4  = 5Cl2 + 2MnSO4+ 5Na2SO4 +

K2SO4 + 8H2O

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