Separating a redox reaction into two half reactions

[Sara Chalhoub]

Hi, I am having trouble understanding how to separate a redox reaction.

For example, in this reaction from class:
8H+ + MnO4- + 5Fe2+ --> Mn2+ + 5Fe3+ + 4H2O

I understand that the oxidation reaction is 5Fe2+ --> 5Fe3+ + 5e- because Fe2+ lost an electron;
but I don't understand the reduction equation 8H+ + MnO4- + 5e- --> Mn2+ + 4H2O

Why is the 8H+ to 4H2O in the reduction equation and not the oxidation one? And if Mn is turning into Mn2+ doesn't that mean that it is an oxidation reaction since it is losing 2 electrons?

[Dylan Do]

It is actually going from to meaning it gains electrons. How did I know that? Well...
let's say charge of Mn reactant is 'X'
Now we know the total charge of has to be -1.
So now use algebra to find Mn charge (X):
-1 = X + (4*-2)

[Andy Sun 2I]

The oxidation number of Mn in MnO4- is +7, so the reduction reaction changes Mn+7 into Mn+2.
Also, the 8H and 4H2O are included in the reduction reaction because the the O from the H2O is from the MnO4-, therefore it must be included in the balanced equation.