Balance the following redox reaction: Br- + MnO4- --> Br2 + Mn2+?

Answer 1

10Br- + 2(MnO4)- + 16H+ ----> 5Br2 + 2Mn2+ + 8H2O

Answer 2

You don't want an H+ ion in the end. I don't even know how the first guy attempted his answer.

But here's what I got:

6Br- + 4 H20 + MnO4 ---> 3 Br2 + Mn2+ +8 OH-

Here's the step by step:

From oxidation numbers, you find out that Br is being oxidized and Mn is being reduced, so you separate them, balance them and balance their charges. You then end up w/:

6Br- ----> 3 Br2 + 6 e-
6 e- + 8 H+ + MnO4 ----> Mn2+ + 4 H20

6 e-'s cancel out, you then combine them and proceed to make it basic by adding 8 OH- for the 8 H+ to both sides. You can cancel out the resulting 4H20 on the product side and subtract the same amount on the reactant side.. and you will end up w/:

6Br- + 4 H20 + MnO4 ---> 3 Br2 + Mn2+ +8 OH-

Oh, just noticed... this would only hold true if you meant Mn w/ a +2 charge, and not Mn2 w/ a +1 charge, which I doubt cause Mn is usually 2 or 3.

Answer 3

2Br- + 2MnO4 --> Br2 + Mn2+ + 2O2