Question

I would like to check my answer to the following question:

The permanganate ion oxidizes bromide in acidic conditions to form bromine, manganese(II) ion and water. Give the balanced equation of the reaction.

My answer:

3e- + 8H+ (aq) + MnO4- (aq) + 2Br- (aq)  Br2 (aq) + Mn2+ (aq) + 4H2O (l)

Do I have this correct? There's supposed to be an arrow between the Br- and the Br2

Answer 1

It may appear to be correct but it isn't. The atoms balance and the charge balances BUT there are two things wrong.
1. A balanced equation never has electrons on either side amd you have 3e on the left side.
2. The electrons lost do not balance the electrons gained. How can I help you with this?

You make an arrow this way ---> or ===>

Answer 2

Thanks on the arrow.

I don't know. In acid solutions I need the H+ ions. Charges have to balance. I balanced using the oxidation states of each species. I did half reactions. Perhaps the water is messing me up.

Answer 3

Should I use H3O+ rather than H+?

Answer 4

No, that won't work.

Answer 5

I'm lost. I really don't see what I am missing in this.

Answer 6

I think I have it...

10e- + 16H+ + 2MnO4- + 10 Br- --->

5 Br2 + 8H2O + 10 e-

Now remove the 10e- from both sides:

16H+ + 2MnO4- + 10 Br- --->

5 Br2 + 8H2O

Answer 7

I think your MnO4^- is right for acid solution. Br2 is also. Perhaps it the electrons. Let's check.
MnO4^- + 8H^+ + 5e ==> Mn^2+ + 4H2O
2Br- ==> Br2 + 2e

So multiply eqn 1 by 2 and eqn 2 by 5. Apparently you used some other numbers. This is what you should get.
2MnO4^- + 16H^+ + 10e + 10Br^- ==> 2Mn^2+ + 8H2O + 5Br2 + 10e

Cancel the 10e on each side and that's it.
I think what you did is add the to half equations (which were balanced correctly), then added those three electrons to the left to make the charge balance. But you see the electron change you have on the MnO4^- is from 7 to 2 for 5e and the electron change for 2Br^- to Br2 is 2e. So the electrons lost don't equal electrons gained and adding 3e to the left (which might seem to be right) isn't the way you balance the electrons. You make the e lost = e gained by multiplying each equation to equal some number. Hope this helps.

Answer 8

Ooops I left some out:

16H+ + 2MnO4- + 10 Br- --->

5 Br2 + 2Mn 2+ + 8H2O

:) Is this it???

Answer 9

We overlapped. Yes I got it. Thank you.

Answer 10

Ah, I see. You figures it out while I was typing. Good work.
Note that for a redox equation to balance it must balance three things.
1. atoms must balance (yours did)
2. charge must balance (yours did)
3. e lost must = e gained. (yours didn't. Yours had 5e gained and 2 electrons lost.)
4. Of course it goes without saying that a fully balanced equation can't have extra electrons floating around.

Answer 11

Yes. Thank you. I totally forgot to multiply the half reactions after balancing them. Lame. I appreciate the help.