[heysuzzz]

hi, a new member here:)
we did a bunch of redox labs at school, and now we need to write the equations and balance them...
however, i have no idea how we're supposed to know the products in a redox reaction, because the teacher never covered that... she just showed us how to balance already written equations.
so here it goes:


stock solution: KMnO4 + H2SO4 -> ? (doesn't look like double displacement..?)
to NaBr, add water and stock solution, then mineral oil. shake well. record the color of layers.

well, what i got is that the purple stock solution turned yellow when added to NaBr. there seemed to be only 2 layers, though; oil and aqueous solution. but the oil was not involved in the reaction.
and the lab says that Mno4 produces Mn2+ and Br is a brown liquid... but it doesn't help, since i still cannot figure out what the products are!


so please help... if i learn how to do this one, i think i'll be okay with the 9 problems that are left.
thanks in advance:D

[Dan]

Ok, so you know KMnO₄ in acid reacts with bromide ions to produce Mn²⁺ and bromine (Br₂)

So, the ionic equation is:

MnO₄^- + H⁺ + Br^-  -----> Mn²⁺ + Br₂ + __________

Work out the missing product (hint: what happens to the O), and balance.

[Will]

It is also a good idea to learn this half equation:

MnO₄^- + 8H⁺ + 5e^- --> Mn²⁺ + 4H₂O

The half equation for the bromide to bromine is 2Br^- --> Br₂ + 2e^-.

Now all you have to do is multiply each equation by a coefficient to make the electrons on each side equal (so you can cancel them out and put the equations together) (ie multiply manganate one by 2 and bromide one by 5); sorry if that gives too much away, Dan!

Borek's site ChemBuddy explains it well.

As a separate thing: Doesn't KMnO₄ react with conc. H₂SO₄ to give green-looking Mn₂O₇? (See bottom of this page from woelen's website.)

[heysuzzz]

thank you guys so much!! you've been a lot of help:D

to will17: that's what i read from other posts, but they all agree that the product is highly unstable and dangerous... hmm i wonder if that applies only for solid KMnO₄?