Balancing redox equations
balance each equation using both the half reaction and oxidation number method. on a separate piece of paper, re-write each question, and show all your steps for each question.
a)fe2 mno4-fe3 mn2 (acidic)

Ok,

(A) Fe 2+ + MnO4 - -> Fe 3+ + Mn2+

First step : Asign the oxidation numbers

Fe ( Iron ) has an oxidation number of +2 as it's equal to it's charge.

MnO4 ( Permanganate )'s oxidation number is the oxidation number of Mn as it's unknown :

The over all oxidation number has to equal to -1 and we know from the rules that at most times Oxygen has an oxidation number of -2

Mn + 4 ( O ) = -1
Mn +4(-2) =-1
Mn + (-8) = -1
Mn=+8 + ( -1)
Mn=+7

The oxidation number of Fe +3 is equal to it's charge i.e +3
Same for the oxidation number of Mn is equal to +2.

Write the oxidation numbers under the equation.

Fe2+ + MnO4 - -> Fe 3+ + Mn 2+

+2           +7              +3          +2

Now identify what is  oxidised and reduced.
Oxidation is the loss of electrons where Reduction is the gain of electrons.

Ok so oxidation number of Fe +2 goes to +3 that means it looses one electron ( e- ).
That means it's oxdised.
The oxidation number of MnO4 - goes from +7 to +2 that means it gains five electrons i.e it's reduced.

To balance the equation we know that in order for MnO4 - to be reduced we need 5 electrons. When something is reduced something has to be oxidised.
Therefore we need 5 Fe +2 to give up 5 electrons.

The ratio is 5:1

5Fe 2+ + MnO4 -  -> 5Fe 3+ + Mn +2

Hope this helps the reaction is not complete though as you gave me the proper reaction is as fallows.

5Fe 2+ + MnO4 - + 8H + -> 5Fe +3 + Mn 2+ + 4H2O