Question

balance the following equation by ion electron method. kmno4+feso4+h2so4=mnso4+fe2 (so4)3+k2so4+h2o any one answer for this question ​

Answer 1

Answer:

2KMNO4+10FeSO4+8H2SO4= K2SO4+2MNSO4+5Fe2(SO4)3+8H2O

FeSO4+ KMnO4+H2SO4-> Fe2(SO4)3+ MnSO4+ H2O

First of all, let's concentrate on what is really important - on the net ionic reaction:

Fe2++ MnO4-+ H+-> Fe3++ Mn2++ H2O

All other ions (K+and SO42-) are only spectators and we don't need them to balance the equation. Besides, charges are what is really important in the half reactions method.

At first glance you can see that in the reaction iron gets oxidized and permanganate gets reduced:

Fe2+-> Fe3+

MnO4--> Mn2+

We will start equation balancing with balancing these half reactions - using electrons to balance charge. In the case of iron oxidation half reaction atoms are already balanced, but charge is not. To balance charge we will add one electron on the right side:

Fe2+-> Fe3++ e-

We can use electrons safely, as the final step of balancing will be electron cancellation.

Now we have to balance permanganate reduction half reaction:

MnO4--> Mn2+

Before balancing charges we have to balance atoms. What to do with the oxygen? We know that the reaction takes place in acidic conditions (see sulfuric acid in the skeletal reaction at the beginning) - so we can add H+on the left and H2O on the right:

MnO4-+ H+-> Mn2++ H2O

Using simple balancing by inspection we will add two coefficients to balance atoms:

MnO4-+8H+-> Mn2++4H2O

Once the atoms are balanced it is time to balance charge with electrons - there is +7 charge on the left side of the equation and +2 on the right side. To balance charges we have to add 5 electrons on the left:

MnO4-+8H++5e--> Mn2++4H2O

At the moment we have two balanced half reactions. Here comes the final trick - we add these half reactions, multiplying them by such coefficients that electrons cancel out (easiest approach is to use just numbers of electrons, although if often means you will have to find lowest coefficients later). To have five electrons in both equations we have to multiply first equation by 5:

5Fe2+->5Fe3++5e-

and when we add both half reactions we will get

MnO4-+8H++5Fe2++5e--> Mn2++4H2O +5Fe3++5e-

Canceling out the electrons:

MnO4-+8H++5Fe2+-> Mn2++4H2O +5Fe3+

And the reaction is balanced.