[vort3x]

http://chemistry.about.com/od/generalchemistry/ss/redoxbal_4.htm

I'm currently balancing redox reactions using the half-reaction method. Everything seems work fine until I come to electrons.

In the above example, why are 2e added to the right for the first, and 5e to the left on the bottom?

For the first one... 2 I - , oxidation number and charge of -1, so why 2e on the right ride? Just not adding up for me.

[macman104]

So on the left hand side, you have an oxidation number of -1, and there are 2 moles of I^-, for a total of -2.  And on the right hand side, what is the oxidation state of I₂?  So, to balance the -2 on the left, how many electrons do you need to add on the right?

Same thing for the second equation, compare the total oxidation state on the right, with the left, where do you need to add electrons to balance them?

[vort3x]

OH, okay... so I do still follow oxidation numbers for that. I₂ in its elemental form is 0, so therefore 2e must be added on the right side.

For the second one:

8H⁺ oxidation +1
MnO4^- oxidation +7 (That's really the oxidation of +7 for Mn to make the compound be -1... so the oxidation IS +7 right?).
Left side total: 8+

Mn²⁺ oxidation 2+
4H₂O oxidation 0
Right side total: 2+

So I see where I'm getting confused. I must add 5e to the left to make it equal 2+ ... but sometimes I think I should add 6e to the right side to make it equal 8? BUT, I know oxidation numbers aren't electrons, and electrons are negative so it's removing charges on the left side... what's the best way to think of it?

[macman104]

OH, okay... so I do still follow oxidation numbers for that. I₂ in its elemental form is 0, so therefore 2e must be added on the right side.

Correct, and then for the second part, you need to add the oxidation states of EVERY atom.  Not just the positive ones.So...

For the second one:

8H⁺ oxidation +1

This is a total of +8, because there are 8 of them.

MnO4^- oxidation +7 (That's really the oxidation of +7 for Mn to make the compound be -1... so the oxidation IS +7 right?).

Right, Mn is +7, but O is -2, and there are 4 of those, so (+7) + 4(-2) = -1, so your left-side total is (-1) + (+8) = +7.

Mn²⁺ oxidation 2+
4H₂O oxidation 0
Right side total: 2+

You could also break the H₂O into H, which is a +1, and O, which is a -2, but yes, it's an overall of 0. 

Now is it clear why you add 5e? Because you have +7 on the left, and +2 on the right!  The best way to think about it is to add electrons to whichever side is more positive, and enough electrons to bring that positive value equal to the lower positive.

[vort3x]

For the 8H⁺ are you sure coefficients count? I thought it would just be a +1 not a +8 because you weren't supposed to count coefficients when doing oxidation numbers?

[macman104]

But you are balancing the oxidation the total transfer of electrons on each side.  So you need to count all of the oxidation states, and it is the total that is what is important.