Iron(II) sulfate where Iron has an oxidation number of +2 and Potassium permanganate where Mn has an oxidation number of +7, react with each other in an acidic condition. That means FeSO4, KMnO4, and H2SO4 react lớn produce the Iron(III) sulfate, Manganese sulfate, potassium sulfate as well as water. This is a Redox (oxidation-reduction) reaction.

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

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Now, if we concentrate on the chemical reaction equation which is our main fact of discussion in this article, we will notice that the Iron (Fe) in the reactant releases one electron in the meantime the Manganese takes up 5 electrons from the reaction. Therefore, it is an oxidation-reduction reaction as the electron transfer occurred in this reaction.

The balanced equation for the reaction discussed above is–

## 10FeSO_{4} + 2KMnO_{4} + 8H_{2}SO_{4} = 5Fe_{2}(SO_{4})_{3} + 2MnSO_{4} + K_{2}SO_{4} + 8H_{2}O

## Balancing the reaction among FeSO4, KMnO4, and H2SO4

From the above discussion, there is no doubt that the reaction is a redox reaction. As it is an oxidation-reduction reaction, it can easily be balanced by the ion-electron method where we take into concern the number of electron(s) transfers from the reducing agent lớn the oxidizing agent. So, further not tự any delay, let’s get started.

The skeleton formula of the reaction is-

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

## or,

## FeSO_{4} + KMnO_{4} + H_{2}SO_{4} = Fe_{2}(SO_{4})_{3} + MnSO_{4} + K_{2}SO_{4} + H_{2}O

Here FeSO_{4} is that the reductant because iron released electron and get oxidized in addition as KMnO4 takes electrons from the reducing agent and get reduced, is the oxidant.

**Reducing agent:** FeSO_{4 }or by excluding the spectrator ion(s) Fe^{2+}

**Oxidizing agent:** KMnO_{4} or by excluding the spectrator ion(s) MnO_{4}^{-1}

## Reduction Half Reaction

We should know that the oxidizing agents are responsible for the reduction reaction in a redox reaction. Here, in the above reaction manganese takes 5 electrons and the oxidation number becomes +2. So reduction takes place in terms of KMnO4 or MnO_{4}^{–}.

⇒ MnO_{4 }^{-1} + 5e^{–} + 8H^{+ } = 4H_{2}O + Mn^{2+ } … …. …. …. (1)

## Oxidation Half Reaction

On the other hand, the Iron in the reactant part of the reaction released only one electron and produced an iron (III) ion, as it is a reducing agent. The oxidation number of reducing agent iron(II) ion becomes +3 from +2.

⇒ Fe^{2+} – e^{–} = Fe^{3+ }… … … … … (2)

## Adding the oxidation as well as reduction reaction halves lớn get a full redox reaction

As the oxidizing agent, KMnO_{4} has lớn take 5 electrons lớn become Mn^{+2}, on the country, the reducing agent iron (II) sulfate releases only one electron lớn get oxidized. So there are five times reducing agents needed lớn reduce the KMnO_{4}. Therefore, we should multiply the number (2) equation by 5 and then add them together lớn get the full oxidation-reduction reaction.

**Now equation (1) + (2)x5,**

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MnO_{4 }^{-1} +5e^{–} + 8H^{+ } = 4H_{2}O + Mn^{2- }

5Fe^{2+} – 5e^{–} = 5Fe^{3+}

MnO_{4 }^{-1} + 5Fe^{2+}+ 8H^{+ } = Mn^{2- }+ 4H_{2}O + 5Fe^{3+}

**Now adding necessary ions and radicals we get,**

5Fe**SO _{4}**+

**K**MnO

_{4}+ 4H

_{2}

**SO**= Mn

_{4}^{ }**SO**

_{4}^{ }+ 5/2Fe

_{2}

**(SO4)**+ 4H

_{3}_{2}O +

**K**

_{2}SO_{4}

**Or, **

**10FeSO _{4} + 2KMnO_{4} + 8H_{2}SO_{4} = 5Fe_{2}(SO_{4})_{3} + 2MnSO_{4} + K_{2}SO_{4} + 8H_{2}O**

### “Answer”

## 10FeSO_{4} + 2KMnO_{4} + 8H_{2}SO_{4} = 5Fe_{2}(SO_{4})_{3} + 2MnSO_{4} + K_{2}SO_{4} + 8H_{2}O

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