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When balancing redox reactions we have always - apart from all the rules pertaining to balancing chemical equations - additional information about electrons moving. In the case of oxidation numbers method we assume electrons are transferred between atoms (which is only an approximation), in the case of half reactions method we assume there are two systems exchanging electrons (which is much closer to reality, although what we observe may be a multistep process with numerous intermediate reagents).
Let's try to use half reactions method to balance reaction equation for Fe(II) oxidation with permanganate in the acidic solution:
FeSO4 + KMnO4 + H2SO4 -> Fe2(SO4)3 + MnSO4 + H2O + K2SO4
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.
Balancing hydrogen and oxygen in the half reactions method requires knowledge about the conditions in which reaction takes place. To balance oxygen we can add H+ on the side where there is oxygen excess and water on the second, just as we did in the above example. But we can also use OH- and water to do the trick, for example half reaction:
ClO- -> Cl-
is not balanced, but once we add water and OH-:
ClO- + H2O -> Cl- + 2OH-
we have not only balanced atoms but we are also ready to balance charge by adding two electrons on the left:
ClO- + H2O + 2e- -> Cl- + 2OH-
and the half reaction is ready to be used. General rule says that if the reaction takes place in acidic conditions we use water and H+ to balance oxygens, and if the reaction takes place in basic conditions - we use OH- and water. Don't worry if it looks like the reaction produces H+ in the solution that was already acidic - while it will influence reaction equilibrium, we are concentrating on the equation balancing, not on the equilibrium right now.
Also note that in really hard cases, when you have no idea what is really going on in the solution, you may look for the half reactions (and amount of electrons consumed/produced) in the standard potentials tables.
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