FeSO4 + KMnO4 + H2SO4

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

FeSO4 + KMnO4 + H2SO4 react to produce the Iron(II) sulfate ions. This is a Redox reaction.

Iron(II) sulfate react with potassium permanganate and sulfuric acid

the balanced formula for the reaction is –

10FeSO4 + 2KMnO4 + 8H2SO4 = 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2O

The skeleton formula of the reaction is-

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

or,

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

Here FeSO4 is the Reducing agent as well as KMnO4 is the Oxidizing agent.

FeSO4 + KMnO4 + H2SO4

Reducing agent: FeSO4 ­or Fe2+

Oxidizing agent: KMnO4 or MnO4-1

Rear More: NaCl + KMnO4 + H2SO4 = Cl2 + MnSO4 + Na2SO4 + K2SO4 + H2O

Reduction Half Reaction:

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

Oxidation Half Reaction:

Fe2+ – e = Fe3+ … … … … … (2)

Now equation (1) + (2)x5,

MnO-1 +5e + 8H = Mn2- + 4H2O

5Fe2+ – 5e = 5Fe3+


MnO-1 + 5Fe2++ 8H = Mn2- + 5Fe3+ + 4H2O

Now adding necessary ions and radicals we get,

 5FeSO4+ KMnO4 + 4H2SO4  = MnSO4 + 5/2Fe2(SO4)3 + 4H2O +K2SO4

Or,  

10FeSO4 + 2KMnO4 + 8H2SO4 = 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2O

“Answer”

10FeSO4 + 2KMnO4 + 8H2SO4 = 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2O

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