Fe2+ + MnO-4 --> Fe3+ + Mn2+
Fe2+ = +2
Mn = +7
O-4 = 4(-2) = -8
Fe3+ = 3+
Mn2+ = 2+
ok so next I gave 8H to MnO-4 and 4 H20 to Mn2+
Next I starting adding my electrons. 1e- to Fe3+. I understand that.
This is where Im lost according to my book I give 5e- to the reactant side.
Can you explain why it gets 5e-
The simple answer is that Mn changed frm +7 to +2 so it must have gained 5e.
MnO4^- + 8H^+ + 5e ==> Mn^2+ + 4H2O
In the given chemical equation, the redox reaction involves the transfer of electrons between Fe2+ and MnO-4. To balance the charges, you need to make sure that the total charge on both sides of the equation is equal.
Fe2+ has a charge of +2, and it is being oxidized to Fe3+ with a charge of +3. Therefore, to balance the charge, 1 electron (e-) is added to the product side:
Fe2+ -> Fe3+ + 1e-
Now, let's consider the MnO-4 ion. It has a charge of -1, while the Mn2+ ion has a charge of +2. To balance the charge, you need to add 3 electrons (3e-) to the reactant side:
MnO-4 + 3e- -> Mn2+
By adding 3 electrons to the reactant side, you achieve charge balance. However, to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number gained in the reduction half-reaction, you need to multiply the whole equation by a suitable factor.
In this case, the factor is chosen so that the total number of electrons on both sides is the same. Since 3 electrons were added to the reactant side, you need to multiply the whole equation by 5 to have the same number of electrons on both sides:
5Fe2+ + MnO-4 + 8H+ -> 5Fe3+ + Mn2+ + 4H2O
By multiplying the equation by 5, the total number of electrons transferred on both sides is balanced, which is crucial for maintaining the overall charge neutrality of the reaction.